INTRODUCTION
学生のときにこういう方面に凄くはまってた時期がありました。 その時期に導出した公式とかをノートにまとめたものです。 あまり厳密なことは気にせずにやってたのであしからずです。
積分表示された関数の級数表示の1例
様々な積分表示された関数を級数で表すこと。
\begin{eqnarray} \int ^1_0\frac {(1-x)^{s-1}}{(1-ux)^{\alpha +1}}x^{\beta -1}dx&=& \int ^1_0(1-x)^{s-1}x^{\beta -1}\sum _{n=0}^\infty \frac {(\alpha +1)(\alpha +2)\cdots (\alpha +n)}{ n!}u^nx^ndx \nonumber \\ &=&\sum _{n=0}^\infty \frac {(\alpha +1)\cdots (\alpha +n)}{n!}u^n\int ^1_0(1-x)^{s-1} x^{\beta +n-1}dx \nonumber \\ &=&\sum _{n=0}^\infty \frac {(\alpha +1)\cdots (\alpha +n)}{n!}\frac {\beta (\beta +1)\cdots (\beta +n-1)}{(s+\beta )(s+\beta +1)\cdots (s+\beta +n-1)}u^n\frac {\Gamma (s)\Gamma (\beta )}{ \Gamma (s+\beta )} \end{eqnarray}
これに\(u\rightarrow 1\)の極限を取ればまとめると
\begin{eqnarray} \frac {\Gamma (s+\beta )\Gamma (s-\alpha )}{\Gamma (s)\Gamma (s-\alpha +\beta )} =1+\frac {\alpha \beta }{1!(s+\beta )}+\frac {\alpha (\alpha +1)\beta (\beta +1)}{2! (s+\beta )(s+\beta +1)}+\frac {\alpha (\alpha +1)(\alpha +2)\beta (\beta +1)(\beta +2)}{3! (s+\beta )(s+\beta +1)(s+\beta +2)}+\cdots \end{eqnarray}
を得る。同様な変形をして
\begin{eqnarray} \int ^1_0x^{s-1}(1-ux)^{\beta -1}dx=\frac {1}{s}-\frac {\beta -1}{1!(s+1)}u +\frac {(\beta -1)(\beta -2)}{2!(s+2)}u^2-\frac {(\beta -1)(\beta -2)(\beta -3)}{3!(s+3)}u^3+\cdots \end{eqnarray}
よって\(u\rightarrow 1\)の極限を取れば
\begin{eqnarray} \frac {\Gamma (s)\Gamma (\beta )}{\Gamma (s+\beta )}= \frac {1}{s}-\frac {(\beta -1)}{1!(s+1)}+\frac {(\beta -1)(\beta -2)}{2!(s+2)} -\frac {(\beta -1)(\beta -2)(\beta -3)}{3!(s+3)}+\cdots \end{eqnarray}
などを得る。例えば\(s=\beta =\frac {\pi }{2}\)を代入して
\begin{eqnarray} \pi =2+\frac {1}{1!\cdot 3}+\frac {1\cdot 3}{2!\cdot 5\cdot 2}+\frac {1\cdot 3\cdot 5}{3!\cdot 7\cdot 2^2}+\frac {1\cdot 3\cdot 5\cdot 7}{4!\cdot 9\cdot 2^3}+\cdots \end{eqnarray}
などが得られる。また\(\beta =1-s\)を代入すれば
\begin{eqnarray} \frac {\pi }{\sin \pi s}=\frac {1}{s}+\frac {s}{1!(s+1)}+\frac {s(s+1)}{2!(s+2)}+\frac {s(s+1)(s+2)}{ 3!(s+3)}+\cdots \end{eqnarray}
を得る。 etc...
オイラー・マクローリンの公式
関数\(F(x+h)\)のテイラー展開は形式的には
\begin{eqnarray} F(x+h)&=&\sum _{n=0}^\infty \frac {1}{n!}\left (\frac {d}{dx}\right )^nF(x) \nonumber \\ &=&\exp \left (h\frac {d}{dx}\right )F(x) \end{eqnarray}
と書ける。このことに留意して
\begin{eqnarray} F(x)+F(x+h)+F(x+2h)+\cdots +F(x+Nh)&=&\sum _{n=0}^Ne^{nh\frac {d}{dx}}F(x) \nonumber \\ &=&\frac {1-e^{(N+1)h\frac {d}{dx}}}{1-e^{h\frac {d}{dx}}}F(x) \end{eqnarray}
と形式的に書ける。ここで”複素解析のお話”noteによれば
\begin{eqnarray} \frac {x}{1-e^x}=-1+\frac {1}{2}x-\frac {B_1}{2!}x^2+\frac {B_2}{4!}x^4 -\frac {B_3}{6!}x^6+\cdots \end{eqnarray}
とかけた。従ってこの式に形式的に\(x\)に\(\frac {d}{dx}\)を”代入”すると、 \(\frac {dF}{dx}(x)=f(x)\)と置けば
\begin{eqnarray} \sum _{n=0}^Nf(x+nh)&=&\int ^{x+(N+1)h}_x\!\!\!\!\!\!\! f(y)dy-\frac {1}{2}\biggl (f(x+(N+1)h)-f(x)\biggr )+\frac {B_1}{2!}\frac {d}{dx} \biggl (f(x+(N+1)h)-f(x)\biggr ) \nonumber \\ &&-\frac {B_2}{4!}\frac {d^3}{dx^3}\biggl (f(x+(N+1)h)-f(x)\biggr ) +\frac {B_3}{6!}\frac {d^5}{dx^5}\biggl (f(x+(N+1)h)-f(x)\biggr )-\cdots \end{eqnarray}
となる。これをオイラー・マクローリンの公式という。 これを用いて簡単な例を計算してみる。
まずオイラー・マクローリンの公式を用いれば
\begin{eqnarray} S^k(N):=\sum _{n=1}^Nn^k \end{eqnarray}
が計算できる。\(f(x)=x^k\)と置き、最後に\(x=0\)、\(h=1\)と置けばよい。計算すると
\begin{eqnarray} S^k(N)&=&\frac {(N+1)^{k+1}}{k+1}-\frac {1}{2}(N+1)^k+\frac {B_1}{2!}k(N+1)^{k-1} -\frac {B_2}{4!}k(k-1)(k-2)(N+1)^{k-3}+\cdots \end{eqnarray}
となる。ここで和は\(N+1\)の1次以上の項までとる。\(k=1\)の時は
\begin{eqnarray} S^1(N)=\frac {N(N+1)}{2} \end{eqnarray}
\(k=2\)の時は
\begin{eqnarray} S^2(N)=\frac {N(N+1)(2N+1)}{6} \end{eqnarray}
などとよく知られた公式が得られる。
他にも応用例はあるが、ここでは省略する。また工夫すれば\(\sum _n(-1)^nf(n)\)のような ものも同様な計算をして公式を得ることが出来るが、それも省略する。
冪級数表示の1例
”解析の話発展編”noteによれば
\begin{eqnarray} \int ^\infty _0f(x)x^{s-1}dx=\Gamma (s)\phi (s) \end{eqnarray}
の時
\begin{eqnarray} f(x)=\sum _{n=0}^\infty (-1)^n\frac {\phi (-n)}{\Gamma (1+n)}x^n \end{eqnarray}
となる。この関係式を使って得られるいくつかの公式
\begin{eqnarray} \sin x+\sin 2x+\sin 3x+\sin 4x+\sin 5x\cdots &=&\frac {1}{2}\cot \frac {x}{2} \nonumber \\ \sin x-\sin 2x+\sin 3x-\sin 4x+\cdots &=&\frac {B_1}{2!}(2^2-1)x+\frac {B_2}{4!}(2^4-1)x^3 +\frac {B_3}{6!}(2^6-1)x^5+\cdots \nonumber \\ &=&\frac {1}{2}\tan \frac {x}{2} \end{eqnarray}
他には
\begin{eqnarray} \frac {e^x}{(e^x-1)^2}=\frac {1}{x^2}-\frac {1}{2}\frac {B_1}{0!}+\frac {1}{4}\frac {B_2}{2!}x^2 -\frac {1}{6}\frac {B_3}{4!}x^4+\cdots \end{eqnarray}
など得られる。これは実際に\(\frac {d}{dx}(\frac {1}{e^x-1})=\frac {-e^x}{(e^x-1)^2}\)を用いて計算 した結果と一致する。 この調子で計算した結果を列挙していく。
\begin{eqnarray} &&\sin x+\frac {1}{2^a}\sin 2x+\frac {1}{3^a}\sin 3x+\frac {1}{4^a}\sin 4x+\cdots \nonumber \\ &&=\frac {\pi }{2}-\frac {1}{2}x\ \ \ \ \ (a=1) \nonumber \\ &&=\pi ^2B_1x-\frac {\pi }{4}x^2+\frac {1}{3!\cdot 2}x^3\ \ \ \ \ (a=3) \nonumber \\ &&=\frac {2^3\pi ^4}{4!}B_2x-\frac {\pi ^2}{3!\cdot 6}x^3+\frac {\pi }{4!\cdot 2}x^4-\frac {1}{2\cdot 5!}x^5\ \ \ \ \ (a=5) \nonumber \\ &&=\frac {2^5\pi ^6}{6!}B_3x-\frac {2^3\pi ^4}{3!\cdot 4!}B_2x^3+\frac {2\pi ^2}{5!\cdot 2!}B_1x^5 -\frac {\pi }{6!\cdot 2}x^6+\frac {1}{2\cdot 7!}x^7\ \ \ \ \ (a=7) \nonumber \\ &&=\frac {2^7\pi ^8}{8!}B_4x-\frac {2^5\pi ^6}{3!\cdot 6!}B_3x^3+\frac {2^3\pi ^4}{5!\cdot 4!}B_2 x^5-\frac {2\pi ^2}{7!\cdot 2!}B_1x^7+\frac {\pi }{8!\cdot 2}x^8-\frac {1}{2\cdot 9!}x^9 \ \ \ \ \ (a=9) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=(-1)^c\left (\frac {(2c)!}{x^{2c+1}}-\frac {B_{c+1}}{1!\cdot (2c+2)}x-\frac {B_{c+2}}{3!\cdot (2c+4)}x^3 -\frac {B_{c+3}}{5!\cdot (2c+6)}x^5-\cdots \right )\ \ \ (a=-2c,\ c=1,2,3,\cdots ) \nonumber \\ &&\left (=\frac {1}{2}\cot \frac {x}{2}\ \ \ (a=0)\right ) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \label {11} \end{eqnarray}
これらは\(a\)が正の偶数の場合には係数が発散する。\(a=-1,-3,-5,\cdots \)の場合には\(=0\)となる。
\begin{eqnarray} &&\sin x-\frac {1}{2^a}\sin 2x+\frac {1}{3^a}\sin 3x-\frac {1}{4^a}\sin 4x+\cdots \nonumber \\ &&=\frac {1}{2}\tan \frac {x}{2}\ \ \ (a=0) \nonumber \\ &&=\ln 2\cdot x+(1-2^2)\frac {B_1}{2\cdot 3!}x^3+(1-2^4)\frac {B_2}{4\cdot 5!}x^5+(1-2^6) \frac {B_3}{6\cdot 7!}x^7+\cdots \ \ \ (a=2) \nonumber \\ &&=\frac {3}{4}\zeta (3)x-\frac {\ln 2}{6}x^3-(1-2^2)\frac {B_1}{5!\cdot 2}x^5-(1-2^4) \frac {B_2}{7!\cdot 4}x^7-(1-2^6)\frac {B_3}{9!\cdot 6}x^9-\cdots \ \ \ (a=4) \nonumber \\ &&=\frac {15}{16}\zeta (5)x-\frac {1}{8}\zeta (3)x^3+\frac {\ln 2}{5!}x^5+(1-2^2) \frac {B_1}{7!\cdot 2}x^7+(1-2^4)\frac {B_2}{9!\cdot 4}x^9+\cdots \ \ \ (a=6) \nonumber \\ &&=\frac {2^6-1}{2^6}\zeta (7)x-\frac {2^4-1}{3!\cdot 2^4}\zeta (5)x^3+\frac {2^2-1}{5!\cdot 2^2} \zeta (3)x^5-\frac {\ln 2}{7!}x^7-(1-2^2)\frac {B_1}{9!\cdot 2}x^9-(1-2^4)\frac {B_2}{11!\cdot 4}x^{11}-\cdots \ \ \ (a=8) \nonumber \\ &&=\frac {2^8-1}{2^8}\zeta (9)x-\frac {2^6-1}{3!\cdot 2^6}\zeta (7)x^3+\frac {2^4-1}{5!\cdot 2^4} \zeta (5)x^5-\frac {2^2-1}{7!\cdot 2^2}\zeta (3)x^7+\frac {\ln 2}{9!}x^9+(1-2^2)\frac {B_1}{11!\cdot 2}x^{11} \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+(1-2^4)\frac {B_1}{13!\cdot 4}x^{13}-\cdots \ \ \ (a=10) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=\frac {1}{2}x\ \ \ \ \ (a=1) \nonumber \\ &&=\frac {(2-1)\pi ^2}{1!\cdot 2!}B_1x-\frac {1}{2\cdot 3!}x^3 \ \ \ (a=3) \nonumber \\ &&=\frac {(2^3-1)\pi ^4}{1!\cdot 4!}B_2x-\frac {(2-1)\pi ^2}{3!\cdot 2!}B_1x^3+\frac {1}{2\cdot 5!} x^5\ \ \ (a=5) \nonumber \\ &&=\frac {(2^5-1)\pi ^6}{1!\cdot 6!}B_3x-\frac {(2^3-1)\pi ^4}{3!\cdot 4!}B_2x^3 +\frac {(2-1)\pi ^2}{5!\cdot 2!}B_1x^5-\frac {1}{2\cdot 7!}x^7\ \ \ (a=7) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=(-1)^c\left (\frac {(4^{c+1}-1)B_{c+1}}{1!\cdot (2c+2)}x+\frac {(4^{c+2}-1)B_{c+2}}{3!\cdot (2c+4)}x^3 +\frac {(4^{c+3}-1)B_{c+3}}{5!\cdot (2c+6)}x^5+\cdots \right )\ \ \ (a=-2c,\ c=0,1,2,3,\cdots )\nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \end{eqnarray}
これらは\(a=-1,-3,-5,\cdots \)で\(=0\)となる。
\begin{eqnarray} &&\sin x+\frac {1}{3^a}\sin 3x+\frac {1}{5^a}\sin 5x+\frac {1}{7^a}\sin 7x+\frac {1}{9^a}\sin 9x \cdots \nonumber \\ &&=\frac {\pi }{4}\ \ \ (a=1) \nonumber \\ &&=-\frac {\pi }{4\cdot 2!}x^2+\frac {\pi ^2}{8}x \ \ \ (a=3) \nonumber \\ &&=\frac {\pi }{4\cdot 4!}X^4-\frac {\pi ^2}{3!\cdot 8}x^3+\frac {(2^4-1)\pi ^4}{4!\cdot 2}B_2x \ \ \ (a=5) \nonumber \\ &&=-\frac {\pi }{6!\cdot 4}x^6+\frac {\pi ^2}{5!\cdot 8}x^5-\frac {(2^4-1)\pi ^4}{3!\cdot 4\cdot 2} B_2x^3+\frac {(2^6-1)\pi ^6}{6!\cdot 2}B_3x\ \ \ (a=7) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=(-1)^c\left (\frac {1}{2}\frac {(2c)!}{x^{2c+1}}+\frac {(2^{2c+1}-1)B_{c+1}}{1!\cdot (2c+2)}x +\frac {(2^{2c+3}-1)B_{c+2}}{3!\cdot (2c+4)}x^3+\frac {(2^{2c+5}-1)B_{c+3}}{5!\cdot (2c+6)}x^5+\cdots \right ) \ \ \ \ (a=-2c,\ c=0,1,2,3,\cdots ) \nonumber \\ &&\left (=\frac {1}{2\sin x}\ \ \ (a=0) \right )\nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \vdots \end{eqnarray}
これらは\(a=-1,-3,-5,-7,\cdots \)で\(=0\)となる。\(a=2,4,6,\cdots \)で係数が発散する。
\begin{eqnarray} &&\sin x-\frac {1}{3^a}\sin 3x+\frac {1}{5^a}\sin 5x-\frac {1}{7^a}\sin 7x+\cdots \nonumber \\ &&=0\ \ \ (a=0) \nonumber \\ &&=\frac {\pi }{4}x\ \ \ (a=2) \nonumber \\ &&=\frac {\left (\frac {\pi }{2}\right )^3}{2\cdot 2!}E_1x-\frac {\left (\frac {\pi }{2}\right )}{2 \cdot 3!\cdot 0!}E_0x^3 \ \ \ (a=4) \nonumber \\ &&=\frac {\left (\frac {\pi }{2}\right )^5}{2!\cdot 1!\cdot 4!}E_2x-\frac {\left (\frac {\pi }{2} \right )^3}{2\cdot 3!\cdot 2!}E_1x^3+\frac {\left (\frac {\pi }{2}\right )}{2\cdot 5!\cdot 0!} E_0x^5 \ \ \ (a=6) \nonumber \\ &&=\frac {\left (\frac {\pi }{2}\right )^7}{2\cdot 1!\cdot 6!}E_3x -\frac {\left (\frac {\pi }{2}\right )^5}{2\cdot 3!\cdot 4!}E_2x^3 +\frac {\left (\frac {\pi }{2}\right )^3}{2\cdot 5!\cdot 2!}E_1x^5 -\frac {\left (\frac {\pi }{2}\right )}{2\cdot 7!\cdot 0!}E_0x^7\ \ \ (a=8) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=\mu (0)x+\frac {E_1}{2\cdot 3!}x^3+\frac {E_2}{2\cdot 5!}x^5+\frac {E_3}{2\cdot 7!}x^7+\cdots \ \ \ (a=1) \nonumber \\ &&=\mu (2)x-\frac {\mu (0)}{3!}x^3-\frac {E_1}{2\cdot 5!}x^5-\frac {E_2}{2\cdot 7!}x^7-\cdots \ \ \ (a=3) \nonumber \\ &&=\mu (4)x-\frac {\mu (2)}{3!}x^3+\frac {\mu (0)}{5!}x^5+\frac {E_1}{2\cdot 7!}x^7+\cdots \ \ \
(a=5) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=(-1)^{c+1}\left (\frac {E_{c+1}}{1!\cdot 2}x+\frac {E_{c+2}}{3!\cdot 2}x^3+\frac {E_{c+3}}{5!\cdot 2}x^5 +\cdots \right )\ \ \ (a=-(2c+1),\ c=-1,0,1,2,3,\cdots )\nonumber \\ \end{eqnarray}
これは\(a=0,-2,-4,-6,\cdots \)で\(=0\)となる。
\begin{eqnarray} &&\cos x+\frac {1}{2^a}\cos 2x+\frac {1}{3^a}\cos 3x+\frac {1}{4^a}\cos 4x+\frac {1}{5^a}\cos 5x +\cdots \nonumber \\ &&=\frac {2\pi ^2}{0!\cdot 2!}B_1-\frac {\pi }{1!\cdot 2}x+\frac {1}{2\cdot 2!}x^2 \ \ \ (a=2) \nonumber \\ &&=\frac {2^3\pi ^4}{0!\cdot 4!}B_2-\frac {2\pi ^2}{2!\cdot 2!}B_1x^2+\frac {\pi }{3!\cdot 2}x^3 -\frac {1}{2\cdot 4!}x^4 \ \ \ (a=4) \nonumber \\ &&=\frac {2^5\pi ^6}{01\cdot 6!}B_3-\frac {2^3\pi ^4}{2!\cdot 4!}B_2x^2+\frac {2\pi ^2}{4!\cdot 2!} B_1x^4-\frac {\pi }{2\cdot 5!}x^5+\frac {1}{2\cdot 6!}x^6 \ \ \ (a=6) \nonumber \\ &&=\frac {2^7\pi ^8}{0!\cdot 8!}B_4-\frac {2^5\pi ^6}{2!\cdot 6!}B_3x^2+\frac {2^3\pi ^4}{4!\cdot 4!} B_2x^4-\frac {2\pi ^2}{6!\cdot 2!}B_1x^6+\frac {\pi }{2\cdot 7!}x^7-\frac {1}{2\cdot 8!}x^8 \ \ \ (a=8) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=-\frac {1}{x^2}-\frac {B_1}{2\cdot 0!}-\frac {B_2}{4\cdot 2!}x^2-\frac {B_3}{6\cdot 4!}x^4 -\frac {B_4}{8\cdot 6!}x^6-\cdots \ \ \ (a=-1) \nonumber \\ &&=\frac {3!}{x^4}+\frac {B_2}{4\cdot 0!}+\frac {B_3}{6\cdot 2!}x^2+\frac {B_4}{8\cdot 4!}x^4 +\frac {B_5}{10\cdot 6!}x^6+\cdots \ \ \ (a=-3) \nonumber \\ &&=-\frac {5!}{x^6}-\frac {B_3}{6\cdot 0!}-\frac {B_4}{8\cdot 2!}x^2-\frac {B_5}{10\cdot 4!}x^4 -\frac {B_6}{12\cdot 6!}x^6-\cdots \ \ \ (a=-5) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \ \vdots \end{eqnarray}
これは\(a=-2,-4,-6,\cdots \)で\(=0\)、\(a=0\)で\(=-\frac {1}{2}\)となる。\(a=1,3,5,7,\cdots \)で係数が発散する。
\begin{eqnarray} &&\cos x+\frac {1}{3^a}\cos 3x+\frac {1}{5^a}\cos 5x+\frac {1}{7^a}\cos 7x+\cdots \nonumber \\ &&=\frac {(2^2-1)\pi ^2}{2\cdot 0!\cdot 2!}B_1-\frac {\pi }{4\cdot 1!}x \ \ \ (a=2) \nonumber \\ &&=\frac {(2^4-1)\pi ^4}{2\cdot 0!\cdot 4!}B_2-\frac {(2^2-1)\pi ^2}{2\cdot 2!\cdot 2!}B_1x^2 +\frac {\pi }{4\cdot 3!}x^3 \ \ \ (a=4) \nonumber \\ &&=\frac {(2^6-1)\pi ^6}{2\cdot 0!\cdot 6!}B_3-\frac {(2^4-1)\pi ^4}{2\cdot 2!\cdot 4!}B_2x^2 +\frac {(2^2-1)\pi ^2}{2\cdot 4!\cdot 2!}B_1x^4-\frac {\pi }{4\cdot 5!} \ \ \ (a=6) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=-\frac {1}{2x^2}+\frac {B_1}{2\cdot 0!}+\frac {(2^3-1)}{4\cdot 2!}B_2x^2 +\frac {(2^5-1)}{6\cdot 4!}B_3x^4+\frac {(2^7-1)}{8\cdot 6!}B_4x^6+\cdots \ \ \ (a=-1) \nonumber \\ &&=\frac {3!}{2x^4}-\frac {(2^3-1)}{4\cdot 0!}B_2-\frac {(2^5-1)}{6\cdot 2!}B_3x^2 -\frac {(2^7-1)}{8\cdot 4!}B_4x^4-\frac {(2^9-1)}{10\cdot 6!}B_5x^6-\cdots \ \ \ (a=-3) \nonumber \\ &&=-\frac {5!}{2x^6}+\frac {(2^5-1)}{6\cdot 0!}B_3+\frac {(2^7-1)}{8\cdot 2!}B_4x^2 +\frac {(2^9-1)}{10\cdot 4!}B_5x^4+\frac {(2^{11}-1)}{12\cdot 6!}B_6x^6+\cdots \ \ \ (a=-5) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \vdots \end{eqnarray}
これは\(a=0,-2,-4,-6,\cdots \)で\(=0\)となる。\(a=1,3,5,7,\cdots \)で係数が発散する。
\begin{eqnarray} &&\cos x-\frac {1}{2^a}\cos 2x+\frac {1}{3^a}\cos 3x-\frac {1}{4^a}\cos 4x+\frac {1}{5^a}\cos 5x -\cdots \nonumber \\ &&=\frac {\pi ^2}{0!\cdot 2!}B_1-\frac {1}{2\cdot 2!}x^2 \ \ \ (a=2) \nonumber \\ &&=\frac {(2^3-1)\pi ^4}{0!\cdot 4!}B_2-\frac {\pi ^2}{2!\cdot 2!}B_1x^2+\frac {1}{2\cdot 4!}x^4 \ \ \ (a=4) \nonumber \\ &&=\frac {(2^5-1)\pi ^6}{0!\cdot 6!}B_3-\frac {(2^3-1)\pi ^4}{2!\cdot 4!}B_2x^2 +\frac {\pi ^2}{4!\cdot 2!}B_1x^4-\frac {1}{2\cdot 6!}x^6 \ \ \ (a=6) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=\ln 2-\frac {2^2-1}{2\cdot 2!}B_1x^2-\frac {2^4-1}{4\cdot 4!}B_2x^4-\frac {2^6-1}{6\cdot 6!} B_3x^6-\cdots \ \ \ (a=1) \nonumber \\ &&=\frac {2^2-1}{2^2}\zeta (3)-\frac {\ln 2}{2!}x^2+\frac {2^2-1}{2\cdot 4!}B_1x^4 +\frac {2^4-1}{4\cdot 6!}B_2x^6+\cdots \ \ \ (a=3) \nonumber \\ &&=\frac {2^4-1}{2^4}\zeta (5)-\frac {2^2-1}{2^2\cdot 2!}\zeta (3)x^2+\frac {\ln 2}{4!}x^4 -\frac {2^2-1}{2\cdot 6!}B_1x^6-\cdots \ \ \ (a=5) \nonumber \\ &&=\frac {2^6-1}{2^6}\zeta (7)-\frac {2^4-1}{2^4\cdot 2!}\zeta (5)x^2+\frac {2^2-1}{2^2\cdot 4!} \zeta (3)x^4-\frac {\ln 2}{6!}x^6+\cdots \ \ \ (a=7) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=\frac {2^2-1}{2\cdot 0!}B_1+\frac {2^4-1}{4\cdot 2!}B_2x^2+\frac {2^6-1}{6\cdot 4!}B_3x^4 +\frac {2^8-1}{8\cdot 6!}B_4x^6+\cdots \ \ \ (a=-1) \nonumber \\ &&=-\frac {2^4-1}{4\cdot 0!}B_2-\frac {2^6-1}{6\cdot 2!}B_3x^2-\frac {2^8-1}{8\cdot 4!}B_4x^4 -\frac {2^{10}-1}{10\cdot 6!}B_5x^6-\cdots \ \ \ (a=-3) \nonumber \\ &&=\frac {2^6-1}{6\cdot 0!}B_3+\frac {2^8-1}{8\cdot 2!}B_4x^2+\frac {2^{10}-1}{10\cdot 4!}x^4 +\frac {2^{12}-1}{12\cdot 6!}B_6x^6+\cdots \ \ \ (a=-5) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \end{eqnarray}
これは\(a=-2,-4,-6,\cdots \)で\(=0\)となり、\(a=0\)で\(=\frac {1}{2}\)となる。
\begin{eqnarray} &&\cos x-\frac {1}{3^a}\cos 3x+\frac {1}{5^a}\cos 5x-\frac {1}{7^a}\cos 7x+\frac {1}{9^a}\cos 9x -\cdots \nonumber \\ &&=\frac {\pi }{4}\ \ \ (a=1) \nonumber \\ &&=\frac {\pi ^3}{2^4\cdot 2!}E_1-\frac {\pi }{4\cdot 2!}x^2 \ \ \ (a=3) \nonumber \\ &&=\frac {\pi ^5}{2^6\cdot 4!}E_2-\frac {\pi ^3}{2^4\cdot 2!\cdot 2!}E_1x^2+\frac {\pi }{4\cdot 4!} x^4 \ \ \ (a=5) \nonumber \\ &&=\frac {\pi ^7}{2^8\cdot 6!}E_3-\frac {\pi ^5}{2^6\cdot 4!\cdot 2!}E_2x^2+\frac {\pi ^3}{2^4\cdot 2!\cdot 4!}E_1x^4-\frac {\pi }{4\cdot 6!}x^6 \ \ \ (a=7) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=\mu (2)-\frac {1}{2\cdot 2!}x^2-\frac {1}{2\cdot 2!}x^2-\frac {E_1}{2\cdot 4!}x^4 -\frac {E_2}{2\cdot 6!}x^6-\frac {E_3}{2\cdot 8!}x^8-\cdots \ \ \ (a=2) \nonumber \\ &&=\mu (4)-\frac {1}{2!}\mu (2)x^2+\frac {1}{2\cdot 4!}x^4+\frac {E_1}{2\cdot 6!}x^6 +\frac {E_2}{2\cdot 8!}x^8+\cdots \ \ \ (a=4) \nonumber \\ &&=\mu (6)-\frac {1}{2!}\mu (4)x^2+\frac {1}{4!}\mu (2)x^4-\frac {1}{2\cdot 6!}x^6-\frac {E_1}{2\cdot 8!}x^8-\cdots \ \ \ (a=6) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \nonumber \\ &&=\frac {1}{2}+\frac {E_1}{2\cdot 2!}x^2+\frac {E_2}{2\cdot 4!}x^4+\frac {E_3}{2\cdot 6!}x^6 +\frac {E_4}{2\cdot 8!}x^8+\cdots \ \ \ (a=0) \nonumber \\ &&=-\frac {E_1}{2}-\frac {E_2}{2\cdot 2!}x^2-\frac {E_3}{2\cdot 4!}x^4-\frac {E_4}{2\cdot 6!}x^6 -\frac {E_5}{2\cdot 8!}x^8-\cdots \ \ \ (a=-2) \nonumber \\ &&=\frac {E_1}{2}+\frac {E_3}{2\cdot 2!}x^2+\frac {E_4}{2\cdot 4!}x^4+\frac {E_5}{2\cdot 6!}x^6 +\frac {E_6}{2\cdot 8!}x^8+\cdots \ \ \ (a=-4) \nonumber \\ &&\ \ \ \ \ \ \ \ \ \ \vdots \end{eqnarray}
これは\(a=-1,-3,-5,\cdots \)で\(=0\)となる。
ディリクレ級数の1例
上で求めた公式をさらに利用して
\begin{eqnarray} \frac {\sin x}{1-u^2}+\frac {2\sin 2x}{2^2-u^2}+\frac {3\sin 3x}{3^2-u^2} +\frac {4\sin 4x}{4^2-u^2}+\frac {5\sin 5x}{5^2-u^2}+\cdots &=&\frac {\pi }{2}\cos xu-\frac {\pi }{2}\cot \pi u\cdot \sin xu \nonumber \\ &=&\frac {\pi }{2}\frac {\sin \{(\pi -x)u\}}{\sin \pi u} \end{eqnarray}
これは
\begin{eqnarray} \sum _{n=1}\frac {n\sin nx}{n^2-u^2}=\sum _{m=0}^\infty u^{2m}\sum _{n=1}^\infty \frac {\sin nx}{n^{2m+1}} \end{eqnarray}
を(19)を用いて\(u\)と\(x\)の冪級数で表し、それらから係数を比較し右辺を求める。 同様の計算で
\begin{eqnarray} \frac {\sin x}{1-u^2}-\frac {2\sin 2x}{2^2-u^2}+\frac {3\sin 3x}{3^2-u^2} -\frac {4\sin 4x}{4^2-u^2}+\frac {5\sin 5x}{5^2-u^2}-\cdots =\frac {\pi }{2}\frac {\sin xu}{\sin \pi u} \end{eqnarray}
\begin{eqnarray} \frac {\sin x}{1-u^2}+\frac {3\sin 3x}{3^2-u^2}+\frac {5\sin 5x}{5^2-u^2} +\frac {7\sin 7x}{7^2-u^2}+\frac {9\sin 9x}{9^2-u^2}+\cdots &=&\frac {\pi }{4}\cos xu+\frac {\pi }{4}\tan \frac {\pi }{2}u\cdot \sin ux \nonumber \\ &=&\frac {\pi }{4}\frac {\cos \{(x-\frac {\pi }{2})u\}}{\cos \frac {\pi }{2}u} \end{eqnarray}
\begin{eqnarray} \frac {\sin x}{1-u^2}-\frac {\sin 3x}{3^2-u^2}+\frac {\sin 5x}{5^2-u^2} -\frac {\sin 7x}{7^2-u^2}+\frac {\sin 9x}{9^2-u^2}-\cdots =\frac {\pi \sin xu}{4u\cos \frac {\pi }{2}u} \end{eqnarray}
\begin{eqnarray} \frac {\cos x}{1-u^2}+\frac {2^2\cos 2x}{2^2-u^2}+\frac {3^2\cos 3x}{3^2-u^2} +\frac {4^2\cos 4x}{4^2-u^2}+\frac {5^2\cos 5x}{5^2-u^2}+\cdots &=&-\frac {\pi u}{2}\biggl (\sin xu+\cot \pi u\cdot \cos xu\biggr ) \nonumber \\ &=&-\frac {\pi u}{2}\frac {\cos \{(x-\pi )u\}}{\sin \pi u} \end{eqnarray}
\begin{eqnarray} \frac {\cos x}{1-u^2}-\frac {2^2\cos 2x}{2^2-u^2}+\frac {3^2\cos 3x}{3^2-u^2} -\frac {4^2\cos 4x}{4^2-u^2}+\frac {5^2\cos 5x}{5^2-u^2}-\cdots =\frac {\pi u}{2}\frac {\cos xu}{\sin \pi u} \end{eqnarray}
\begin{eqnarray} \frac {\cos x}{1-u^2}+\frac {\cos 3x}{3^2-u^2}+\frac {\cos 5x}{5^2-u^2} +\frac {\cos 7x}{7^2-u^2}+\frac {\cos 9x}{9^2-u^2}+\cdots &=&\frac {\pi }{4u}\biggl (\tan \frac {\pi u}{2}\cdot \cos xu-\sin xu\biggr ) \nonumber \\ &=&\frac {\pi }{4u}\frac {\sin \{(\frac {\pi }{2}-x)u\}}{\cos \frac {\pi }{2}u} \end{eqnarray}
\begin{eqnarray} \frac {\cos x}{1-u^2}-\frac {3\cos 3x}{3^2-u^2}+\frac {5\cos 5x}{5^2-u^2} -\frac {7\cos 7x}{7^2-u^2}+\frac {9\cos 9x}{9^2-u^2}-\cdots =\frac {\pi }{4}\frac {\cos xu}{\cos \frac {\pi }{2}u} \end{eqnarray}
次にゼータ関数の積分表示を応用して級数の積分表示を求める。
\begin{eqnarray} \int ^\infty _0\frac {x^{s-1}}{e^{bx}(e^{ax}-1)}dx=\Gamma (s) \biggl (\frac {1}{(b+a)^s}+\frac {1}{(b+2a)^s}+\frac {1}{(b+3a)^s}+\cdots \biggr ) \label {b-1} \end{eqnarray}
\begin{eqnarray} \int ^\infty _0\frac {e^{bx}x^{s-1}}{e^{ax}-e^{-ax}}dx=\Gamma (s)\biggl ( \frac {1}{(a-b)^s}+\frac {1}{(3a-b)^s}+\frac {1}{(5a-b)^s}+\cdots \biggr ) \end{eqnarray}
\begin{eqnarray} \int ^\infty _0\frac {\cosh bx}{\sinh ax}x^{s-1}dx &=& \int ^\infty _0\frac {e^{bx}+e^{-bx}}{e^{ax}-e^{-ax}}x^{s-1}dx \nonumber \\ &=&\Gamma (s)\biggl (\frac {1}{(a-b)^s}+\frac {1}{(3a-b)^s}+\frac {1}{(5a-b)^s}+\cdots \nonumber \\ &&\ \ \ \ \ \ \ +\frac {1}{(a+b)^s}+\frac {1}{(3a+b)^s}+\frac {1}{(5a+b)^s}+\cdots \biggr ) \end{eqnarray}
\begin{eqnarray} \int ^\infty _0\frac {\sinh bx}{\sinh ax}x^{s-1}dx &=&\Gamma (s)\biggl (\frac {1}{(a-b)^s}+\frac {1}{(3a-b)^s}+\frac {1}{(5a-b)^s}+\cdots \nonumber \\ &&\ \ \ \ \ \ \ -\frac {1}{(a+b)^2}-\frac {1}{(3a+b)^s}-\frac {1}{(5a+b)^s}-\cdots \biggr ) \label {b-2} \end{eqnarray}
などを得る。
次に上で求めた公式からゼータ関数の形の級数の積分表示を求める。例えば
\begin{eqnarray} \int ^\infty _0u^{s-1}\frac {\sinh \{(\pi -x)u\}}{\sinh \pi u}du &=&\frac {2}{\pi }\int ^\infty _0u^{s-1}\sum _{n=1}\frac {n\sin nx}{n^2+u^2}du \nonumber \\ &=&\frac {2}{\pi }\sum _{n=1}n\sin nx\int ^\infty _0\frac {u^{s-1}}{n^2+u^2}du \nonumber \\ &=&\frac {1}{\sin \frac {\pi }{2}s}\sum _{n=1}^\infty \frac {\sin nx}{n^{1-s}} \end{eqnarray}
として求められる。同様の計算をすることで似た形の他のゼータ関数の形の級数の積分表示を 求められる。まとめると
\begin{eqnarray} &&\phi _1(s)=\frac {\sin x}{1^s}+\frac {\sin 2x}{2^s}+\frac {\sin 3x}{3^s}+\frac {\sin 4x}{4^s} +\cdots \\ &&\phi _2(s)=\frac {\sin x}{1^s}-\frac {\sin 2x}{2^s}+\frac {\sin 3x}{3^s}-\frac {\sin 4x}{4^s} +\cdots \\ &&\phi _3(s)=\frac {\sin x}{1^s}+\frac {\sin 3x}{3^s}+\frac {\sin 5x}{5^s}+\frac {\sin 7x}{7^s} +\cdots \\ &&\phi _4(s)=\frac {\sin x}{1^s}-\frac {\sin 3x}{3^s}+\frac {\sin 5x}{5^s}-\frac {\sin 7x}{7^s} +\cdots \\ &&\ \ \ and \nonumber \\ &&\psi _1(s)=\frac {\cos x}{1^s}+\frac {\cos 2x}{2^s}+\frac {\cos 3x}{3^s}+\frac {\cos 4x}{4^s} +\cdots \\ &&\psi _2(s)=\frac {\cos x}{1^s}-\frac {\cos 2x}{2^s}+\frac {\cos 3x}{3^s}-\frac {\cos 4x}{4^s} +\cdots \\ &&\psi _3(s)=\frac {\cos x}{1^s}+\frac {\cos 3x}{3^s}+\frac {\cos 5x}{5^s}+\frac {\cos 7x}{7^s} +\cdots \\ &&\psi _4(s)=\frac {\cos x}{1^s}-\frac {\cos 3x}{3^s}+\frac {\cos 5x}{5^s}-\frac {\cos 7x}{7^s} +\cdots \\ \end{eqnarray}
と置いた時
\begin{eqnarray} &&\phi _1(1-s)=\sin \frac {\pi }{2}s\int ^\infty _0u^{s-1}\frac {\sinh \{(\pi -x)u\}}{\sinh \pi u}du \label {a-1} \\ &&\phi _2(1-s)=\sin \frac {\pi }{2}s\int ^\infty _0u^{s-1}\frac {\sinh xu}{\sinh \pi u}du \\ &&\phi _3(1-s)=\frac {\sin \frac {\pi }{2}s}{2}\int ^\infty _0u^{s-1} \frac {\cosh \{(x-\frac {\pi }{2})u\}}{\cosh \frac {\pi }{2}u}du \\ &&\phi _4(1-s)=\frac {\cos \frac {\pi }{2}s}{2}\int ^\infty _0u^{s-1} \frac {\sinh xu}{\cosh \frac {\pi }{2}u}du \\ &&\ \ \ \ \ and \nonumber \\ &&\psi _1(1-s)=\cos \frac {\pi }{2}s\int ^\infty _0u^{s-1} \frac {\cosh \{(x-\pi )u\}}{\sinh \pi u}du \\ &&\psi _2(1-s)=-\cos \frac {\pi }{2}s\int ^\infty _0u^{s-1}\frac {\cosh xu}{\sinh \pi u}du \\ &&\psi _3(1-s)=\frac {\cos \frac {\pi }{2}s}{2}\int ^\infty _0u^{s-1} \frac {\sinh \{(\frac {\pi }{2}-x)u\}}{\cosh \frac {\pi }{2}u}du \\ &&\psi _4(1-s)=\frac {\sin \frac {\pi }{2}s}{2}\int ^\infty _0u^{s-1}\frac {\cosh xu}{\cosh \frac {\pi }{2}u}du \label {a-2} \end{eqnarray}
次に
\begin{eqnarray} \psi _1(s)+i\phi _1(s)&=&\sum _{n=1}^\infty \frac {e^{inx}}{n^s} \nonumber \\ &=&\sum _{n=1}^\infty e^{inx}\frac {1}{\Gamma (s)}\int ^\infty _0e^{-nu}u^{s-1}du \nonumber \\ &=&\frac {1}{\Gamma (s)}\int ^\infty _0\frac {e^{ix-u}}{1-e^{ix-u}}u^{s-1}du \nonumber \\ &=&\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {e^{ix}-e^{-u}}{e^u-2\cos x+e^{-u}}du \end{eqnarray}
となることより、実部と虚部を比較すれば\(\phi _1(s)\)と\(\psi _1(s)\)の積分表示も得られる。 他も同様である。まとめて書くと
\begin{eqnarray} &&\phi _1(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\sin x}{e^u-2\cos x+e^{-u}}du \\ &&\phi _2(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\sin x}{e^u+2\cos x+e^{-u}}du \\ &&\phi _3(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\sin x(e^u+e^{-u})}{e^{2u}-2\cos 2x+e^{-2u}}du \\ &&\phi _4(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\sin x(e^u-e^{-u})}{e^{2u}+2\cos 2x+e^{-2u}}du \\ &&\ \ \ \ and \nonumber \\ &&\psi _1(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\cos x-e^{-u}}{e^u-2\cos x+e^{-u}}du\\ &&\psi _2(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\cos x+e^{-u}}{e^u+2\cos x+e^{-u}} du \\ &&\psi _3(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\cos x(e^u-e^{-u})}{e^{2u}- 2\cos 2x+e^{-2u}}du \\ &&\psi _4(s)=\frac {1}{\Gamma (s)}\int ^\infty _0u^{s-1}\frac {\cos x(e^u+e^{-u})}{e^{2u}+2 \cos 2x+e^{-2u}}du \end{eqnarray}
となる。 ここで(50)から(57)式までに対して、(36)から(39)などを 用いると、
\begin{eqnarray} &&\phi _1(1-s)=\Gamma (s)\sin \frac {\pi }{2}s\biggl ( \frac {1}{x^s}+\frac {1}{(2\pi +x)^s}-\frac {1}{(2\pi -x)^s}+\frac {1}{(4\pi +x)^s} -\frac {1}{(4\pi -x)^s}+\cdots \biggr ) \\ &&\phi _2(1-s)=\Gamma (s)\sin \frac {\pi }{2}s\biggl ( \frac {1}{(\pi -x)^s}-\frac {1}{(\pi +x)^s}+\frac {1}{(3\pi -x)^s}-\frac {1}{(3\pi +x)^s} +\frac {1}{(5\pi -x)^s}-\frac {1}{(5\pi +x)^s}+\cdots \biggr ) \\ &&\phi _3(1-s)=\frac {1}{2}\Gamma (s)\sin \frac {\pi }{2}s\biggl ( \frac {1}{x^s}-\frac {1}{(\pi +x)^s}+\frac {1}{(\pi -x)^s}+\frac {1}{(2\pi +x)^s} -\frac {1}{(2\pi -x)^s}+\cdots \biggr ) \\ &&\phi _4(1-s)=\frac {1}{2}\Gamma (s)\cos \frac {\pi }{2}s\biggl ( \frac {1}{\left (\frac {\pi }{2}-x\right )^s}-\frac {1}{\left (\frac {\pi }{2}+x\right )} -\frac {1}{\left (\frac {3}{2}\pi -x\right )^s}+\frac {1}{\left (\frac {3}{2}\pi +x\right )^s} +\frac {1}{\left (\frac {5}{2}\pi -x\right )^s}-\frac {1}{\left (\frac {5}{2}\pi +x\right )^s} +\cdots \biggr ) \nonumber \\ \\ &&\ \ \ and \nonumber \\ &&\psi _1(1-s)=\Gamma (s)\cos \frac {\pi }{2}s\biggl ( \frac {1}{x^s}+\frac {1}{(2\pi -x)^s}+\frac {1}{(2\pi +x)^s}+\frac {1}{(4\pi -x)^s} +\frac {1}{(4\pi +x)^s}+\cdots \biggr ) \\ &&\psi _2(1-s)=-\Gamma (s)\cos \frac {\pi }{2}s\biggl ( \frac {1}{(\pi -x)^s}+\frac {1}{(\pi +x)^s}+\frac {1}{(3\pi -x)^s}+\frac {1}{(3\pi +x)^s} +\frac {1}{(5\pi -x)^s}+\frac {1}{(5\pi +x)^s}+\cdots \biggr ) \nonumber \\ \\ &&\psi _3(1-s)=\frac {1}{2}\Gamma (s)\cos \frac {\pi }{2}s\biggl ( \frac {1}{x^s}-\frac {1}{(\pi +x)^s}-\frac {1}{(\pi -x)^s}+\frac {1}{(2\pi +x)^s} +\frac {1}{(2\pi -x)^s}+\cdots \biggr ) \\ &&\psi _4(1-s)=\frac {1}{2}\Gamma (s)\sin \frac {\pi }{2}s\biggl ( \frac {1}{\left (\frac {\pi }{2}+x\right )^s}+\frac {1}{\left (\frac {\pi }{2}-x\right )^s} -\frac {1}{\left (\frac {3}{2}\pi +x\right )^s}-\frac {1}{\left (\frac {3}{2}\pi -x\right )^s} +\frac {1}{\left (\frac {5}{2}\pi +x\right )^s}+\frac {1}{\left (\frac {5}{2}\pi -x\right )^s} +\cdots \biggr ) \nonumber \\ \end{eqnarray}
を得る。
いろいろな級数を多項式やベキ級数で表現する
ここまでのことを踏まえていろいろな級数を多項式で表現していく。
\begin{eqnarray} &&\frac {1}{e^x-1}+\frac {2^a}{e^{2x}-1}+\frac {3^a}{e^{3x}-1}+\frac {4^a}{e^{4x}-1}+\frac {5^a}{e^{5x}-1}+\dots \nonumber \\ &&= \frac {(2\pi )^2}{2\cdot 2}B_1\frac {1}{x^2}-\frac {1}{2x}+\frac {1}{2\cdot 2}B_1\ \ \ (a=1)\nonumber \\ &&=\frac {(2\pi )^4}{2\cdot 4}B_2\frac {1}{x^4}-\frac {1}{2\cdot 4}B_2\ \ \ (a=3)\nonumber \\ &&=\frac {(2\pi )^6}{2\cdot 6}B_3\frac {1}{x^6}+\frac {1}{2\cdot 6}B_3\ \ \ (a=5)\nonumber \\ &&=\frac {(2\pi )^8}{2\cdot 8}B_4\frac {1}{x^8}-\frac {1}{2\cdot 8}B_4\ \ \ (a=7)\nonumber \\ &&\vdots \nonumber \\ &&=2!\cdot \zeta (3)\frac {1}{x^3}-\frac {B_1}{2x}+\frac {B_1B_2}{1!\cdot 2\cdot 4}x+\frac {B_2B_3}{3!\cdot 4\cdot 6}x^3+\frac {B_3B_4}{5!\cdot 6\cdot 8}x^5+\dots \ \ \ (a=2)\nonumber \\ &&=4!\cdot \zeta (5)\frac {1}{x^5}+\frac {B_2}{4x}-\frac {B_1B_3}{1!\cdot 2\cdot 6}x-\frac {B_2B_4}{3!\cdot 4\cdot 8}x^3-\frac {B_3B_5}{5!\cdot 6\cdot 10}x^5-\dots \ \ \ (a=4)\nonumber \\ &&=6!\cdot \zeta (7)\frac {1}{x^7}-\frac {B_3}{6x}+\frac {B_1B_4}{1!\cdot 2\cdot 8}x+\frac {B_2B_5}{3!\cdot 4\cdot 10}x^3+\frac {B_3B_6}{5!\cdot 6\cdot 12}x^5+\dots \ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&=\frac {(2\pi )^4}{2\cdot 4!}B_2\frac {1}{x}-\frac {1}{2}\zeta (3)+\frac {(2\pi )^2B_1^2}{2\cdot 2\cdot 2!\cdot 1!}x+\frac {B_2}{2\cdot 4 \cdot 3!}x^3-\frac {\zeta (3)}{2\cdot (2\pi )^2}x^2\ \ \ (a=-3)\nonumber \\ &&=\frac {(2\pi )^6}{2\cdot 6!}B_3\frac {1}{x}-\frac {1}{2}\zeta (5)+\frac {(2\pi )^4B_1B_2}{2\cdot 2\cdot 4!\cdot 1!}x-\frac {(2\pi )^2B_1B_2}{2\cdot 4\cdot 2!\cdot 3!}x^3-\frac {B_3}{2\cdot 6\cdot 5!}x^5+\frac {\zeta (5)}{2\cdot (2\pi )^4}x^4\ \ \ (a=-5)\nonumber \\ &&=\frac {(2\pi )^8}{2\cdot 8!}B_4\frac {1}{x}-\frac {1}{2}\zeta (7)+\frac {(2\pi )^6B_1B_3}{2\cdot 2\cdot 6!\cdot 1!}x-\frac {(2\pi )^4B_2^2}{2\cdot 4\cdot 3!\cdot 4!}x^3+\frac {(2\pi )^2B_1B_3}{2\cdot 6\cdot 2!\cdot 5!}x^5+\frac {B_4}{2\cdot 8\cdot 7!}x^7-\frac {\zeta (7)}{2\cdot (2\pi )^6}x^6\ \ \ (a=-7)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=-1\)と\(a=0,-2,-4,-6,\cdots \)で係数が発散する。
\begin{eqnarray} &&\frac {1}{e^x-1}-\frac {2^a}{e^{2x}-1}+\frac {3^a}{e^{3x}-1}-\frac {4^a}{e^{4x}-1}+\frac {5^a}{e^{5x}-1}-\dots \nonumber \\ &&= \frac {1}{2x}-\frac {2^2-1}{2\cdot 2}B_1\ \ \ (a=1)\nonumber \\ &&= \frac {2^4-1}{2\cdot 4}B_2\ \ \ (a=3)\nonumber \\ &&=-\frac {2^6-1}{2\cdot 6}B_3\ \ \ (a=5)\nonumber \\ &&= \frac {2^8-1}{2\cdot 8}B_4\ \ \ (a=7)\nonumber \\ &&\vdots \nonumber \\ &&= \ln 2\cdot \frac {1}{x}-\frac {1}{4}+\frac {2^2-1}{2^2}B_1^2x+\frac {2^4-1}{3!\cdot 4^2}B_2^2x^3+\frac {2^6-1}{5!\cdot 6^2}B_3^2x^5+\dots \ \ \ (a=0)\nonumber \\ &&=\frac {2^2-1}{2}B_1\frac {1}{x}-\frac {(2^4-1)B_1B_2}{2\cdot 4\cdot 1!}x-\frac {(2^6-1)B_2B_3}{4\cdot 6\cdot 3!}x^3-\frac {(2^8-1)B_3B_4}{6\cdot 8\cdot 5!}x^5-\dots \ \ \ (a=2)\nonumber \\ &&= -\frac {2^4-1}{4}B_2\frac {1}{x}+\frac {(2^6-1)B_1B_3}{2\cdot 6\cdot 1!}x+\frac {(2^8-1)B_2B_4}{4\cdot 8\cdot 3!}x^3+\frac {(2^{10}-1)B_3B_5}{6\cdot 10\cdot 5!}x^5+\dots \ \ \ (a=4)\nonumber \\ &&= \frac {2^6-1}{6}B_3\frac {1}{x}-\frac {(2^8-1)B_1B_4}{2\cdot 8\cdot 1!}x-\frac {(2^{10}-1)B_2B_5}{4\cdot 10\cdot 3!}x^3-\frac {(2^{12}-1)B_3B_6}{6\cdot 12\cdot 5!}x^7-\dots \ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&=\frac {(2-1)\pi ^2}{2!}B_1\frac {1}{x}-\frac {\ln 2}{2}+\frac {(2-1)B_1}{2\cdot 2\cdot 1!}x\ \ \ (a=-1)\nonumber \\ &&=\frac {(2^3-1)\pi ^4}{4!}B_2\frac {1}{x}-\frac {2^2-1}{2^3}\zeta (3)+\frac {(2-1)\pi ^2B_1B_1}{2\cdot 1!\cdot 2!}x-\frac {B_2}{2\cdot 4\cdot 3!}x^3\ \ \ (a=-3)\nonumber \\ &&=\frac {(2^5-1)\pi ^6}{6!}B_3\frac {1}{x}-\frac {2^4-1}{2^5}\zeta (5)+\frac {(2^3-1)\pi ^4B_1B_2}{2\cdot 1!\cdot 4!}x-\frac {\pi ^2B_2B_1}{4\cdot 3!\cdot 2!}x^3+\frac {B_3}{2\cdot 6\cdot 5!}x^5\ \ \ (a=-5)\nonumber \\ &&= \frac {(2^7-1)\pi ^8}{8!}B_4\frac {1}{x}-\frac {2^6-1}{2^7}\zeta (7)+\frac {(2^5-1)\pi ^6B_1B_3}{2\cdot 1!\cdot 6!}x-\frac {(2^3-1)\pi ^4B_2B_2}{4\cdot 3!\cdot 4!}x^3+\frac {\pi ^2B_3B_1}{6\cdot 2!\cdot 5!}x^5-\frac {B_4}{2\cdot 8\cdot 7!}x^7\ \ \ (a=-7)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {2^7-1}{2^2}\zeta (3)\frac {1}{x}-\frac {\pi ^2B_1}{2\cdot 2!}+\frac {\ln 2\cdot B_1}{2\cdot 1!}x-\frac {(2^2-1)B_1B_2}{2\cdot 4\cdot 3!}x^3-\frac {(2^4-1)B_2B_3}{4\cdot 6\cdot 5!}x^5-\dots \ \ \ (a=-2)\nonumber \\ &&=\frac {2^4-1}{2^4}\zeta (5)\frac {1}{x}-\frac {(2^3-1)\pi ^4B_2}{2\cdot 4!}+\frac {(2^2-1)B_1}{2\cdot 2^2}\zeta (3)x-\frac {\ln 2\cdot B_2}{4\cdot 3!}x^3+\frac {(2^2-1)B_1B_3}{2\cdot 6\cdot 5!}x^5+\frac {(2^4-1)B_2B_4}{4\cdot 8\cdot 7!}x^7\dots \ \ \ (a=-4)\nonumber \\ &&= \frac {2^6-1}{2^6}\zeta (7)\frac {1}{x}-\frac {(2^5-1)\pi ^6B_3}{2\cdot 6!}+\frac {(2^4-1)B_1}{2\cdot 2^4}\zeta (5)x-\frac {(2^2-1)B_2}{4\cdot 2^2\cdot 3!}\zeta (3)x^3+\frac {\ln 2\cdot B_3}{6\cdot 5!}x^5-\frac {(2^2-1)B_1B_4}{2\cdot 8\cdot 7!}x^7\nonumber \\ &&\hspace {13cm} -\frac {(2^4-1)B_2B_5}{4\cdot 10\cdot 9!}x^9-\dots \ \ \ (a=-6)\nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x-1}+\frac {3^a}{e^{3x}-1}+\frac {5^a}{e^{5x}-1}+\frac {7^a}{e^{7x}-1} +\frac {9^a}{e^{9x}-1}+\cdots \nonumber \\ &&= \frac {1}{2}\frac {(2\pi )^2}{2\cdot 2}B_1\frac {1}{x^2}-\frac {B_1}{2\cdot 2}\ \ \ (a=1) \nonumber \\ &&= \frac {1}{2}\frac {(2\pi )^4}{2\cdot 4}B_2\frac {1}{x^4}+\frac {2^3-1}{2}\frac {B_2}{4}\ \ \ (a=3) \nonumber \\ &&= \frac {1}{2}\frac {(2\pi )^6}{2\cdot 6}B_3\frac {1}{x^6}-\frac {2^5-1}{2}\frac {B_3}{6}\ \ \ (a=5) \nonumber \\ &&= \frac {1}{2}\frac {(2\pi )^8}{2\cdot 8}B_4\frac {1}{x^8}+\frac {2^7-1}{2}\frac {B_4}{8}\ \ \ (a=7)\ \ \ \ \nonumber \\ &&\vdots \nonumber \\ &&= \frac {2!}{2}\zeta (3)\frac {1}{x^3}+\frac {B_1}{2}\frac {1}{x} -\frac {(2^3-1)B_1B_2}{2\cdot 4\cdot 1!}x-\frac {(2^5-1)B_2B_3}{4\cdot 6\cdot 3!}x^3-\cdots \ \ \ (a=2) \nonumber \\ &&= \frac {4!}{2}\zeta (5)\frac {1}{x^5}-\frac {2^3-1}{4}B_2\frac {1}{x} +\frac {(2^5-1)B_1B_3}{2\cdot 6\cdot 1!}x+\frac {(2^7-1)B_2B_4}{4\cdot 8\cdot 3!}x^3+\cdots \ \ \ (a=4) \nonumber \\ &&= \frac {6!}{2}\zeta (7)\frac {1}{x^7}+\frac {2^5-1}{6}B_3\frac {1}{x} -\frac {(2^7-1)B_1B_4}{2\cdot 8\cdot 1!}x-\frac {(2^9-1)B_2B_5}{2\cdot 10\cdot 3!}x^3-\cdots \ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {(2^4-1)\pi ^4}{2\cdot 4!}B_2\frac {1}{x}-\frac {2^3-1}{2^4}\zeta (3) +\frac {(2\pi )^2B_1B_1}{2\cdot 2\cdot 1!\cdot 2!}\frac {2^2-1}{2^2}x -\frac {1}{2^4\pi ^2}x^2\ \ \ (a=-3) \nonumber \\ &&= \frac {(2^6-1)\pi ^6}{2\cdot 6!}B_3\frac {1}{x}-\frac {2^5-1}{2^6}\zeta (5) +\frac {(2\pi )^4B_1B_2}{2\cdot 2\cdot 1!\cdot 4!}\frac {2^4-1}{2^4}x -\frac {(2\pi )^2B_2B_1}{2\cdot 4\cdot 3!\cdot 2!}\frac {2^2-1}{2^2}x^3 +\frac {1}{2^6\pi ^4}x^4\ \ \ (a=-5)\nonumber \\ &&= \frac {(2^8-1)\pi ^8}{2\cdot 8!}B_4\frac {1}{x}-\frac {2^7-1}{2^8}\zeta (7) +\frac {(2\pi )^6B_1B_3}{2\cdot 2\cdot 1!\cdot 6!}\frac {2^6-1}{2^6}x -\frac {(2\pi )^4B_2B_2}{2\cdot 4\cdot 3!\cdot 4!}\frac {2^4-1}{2^4}x^3 +\frac {(2\pi )^2B_3B_1}{2\cdot 6\cdot 5!\cdot 2!}\frac {2^2-1}{2^2}x^5 -\frac {1}{2^8\pi ^6}x^6\ \ \ (a=-7)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=-1\)と\(a=0,-2,-4,-6,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x-1}-\frac {3^a}{e^{3x}-1}+\frac {5^a}{e^{5x}-1}-\frac {7^a}{e^{7x}-1} +\frac {9^a}{e^{9x}-1}-\cdots \nonumber \\ &&= \frac {E_0}{2}\frac {1}{x}-\frac {B_1E_1}{2\cdot 2\cdot 1!}x-\frac {B_2E_2}{2\cdot 4\cdot 3!}x^3 -\frac {B_3E_3}{2\cdot 6\cdot 5!}x^5-\cdots \ \ \ (a=1)\nonumber \\ &&= -\frac {E_1}{2}\frac {1}{x}+\frac {B_1E_2}{2\cdot 2\cdot 1!}x+\frac {B_2E_3}{2\cdot 4\cdot 3!}x^3 +\frac {B_3E_4}{2\cdot 6\cdot 5!}x^5+\cdots \ \ \ (a=3)\nonumber \\ &&= \frac {E_2}{2}\frac {1}{x}-\frac {B_1E_3}{2\cdot 2\cdot 1!}x-\frac {B_2E_4}{2\cdot 4\cdot 3!}x^3 -\frac {B_3E_5}{2\cdot 6\cdot 5!}x^5-\cdots \ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\pi }{4}\frac {1}{x}-\frac {1}{4}\ \ \ (a=0)\nonumber \\ &&= \frac {E_1}{4}\ \ \ (a=2)\nonumber \\ &&= -\frac {E_2}{4}\ \ \ (a=4)\nonumber \\ &&= \frac {E_3}{4}\ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \mu (2)\frac {1}{x}-\frac {\pi }{8}+\frac {B_1E_0}{2\cdot 2\cdot 1!}x +\frac {B_2E_1}{2\cdot 4\cdot 3!}x^3+\frac {B_3E_2}{2\cdot 6\cdot 5!}x^5+\cdots \ \ \ (a=-1)\nonumber \\ &&= \mu (4)\frac {1}{x}-\frac {\left (\frac {\pi }{2}\right )^3}{4\cdot 2!}E_1 +\frac {B_1}{2\cdot 1!}\mu (2)x-\frac {B_2E_0}{2\cdot 4\cdot 3!}x^3-\frac {B_3E_1}{2\cdot 6\cdot 5!}x^5-\cdots \ \ \ (a=-3)\nonumber \\ &&= \mu (6)\frac {1}{x}-\frac {\left (\frac {\pi }{2}\right )^5}{4\cdot 4!}E_2 +\frac {B_1}{2\cdot 1!}\mu (4)x-\frac {B_2}{4\cdot 3!}\mu (2)x^3 +\frac {B_3E_0}{2\cdot 6\cdot 5!}x^5+\frac {B_4E_1}{2\cdot 8\cdot 7!}x^7+\cdots \ \ \ (a=-5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^3}{2\cdot 2!}E_1\frac {1}{x}-\frac {1}{2}\mu (2) +\frac {\left (\frac {\pi }{2}\right )B_1E_0}{2\cdot 2\cdot 0!\cdot 1!}x\ \ \ (a=-2)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^5}{2\cdot 4!}E_2\frac {1}{x}-\frac {1}{2}\mu (4) +\frac {\left (\frac {\pi }{2}\right )^3B_1E_1}{2\cdot 2\cdot 2!\cdot 1!}x -\frac {\left (\frac {\pi }{2}\right )B_2E_0}{2\cdot 4\cdot 0!\cdot 3!}x^3\ \ \ (a=-4)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^7}{2\cdot 6!}E_3\frac {1}{x}-\frac {1}{2}\mu (6) +\frac {\left (\frac {\pi }{2}\right )^5B_1E_2}{2\cdot 2\cdot 4!\cdot 1!}x -\frac {\left (\frac {\pi }{2}\right )^3B_2E_1}{2\cdot 4\cdot 2!\cdot 3!}x^3 +\frac {\left (\frac {\pi }{2}\right )B_3E_0}{2\cdot 6\cdot 0!\cdot 5!}x^5 \ \ \ (a=-6)\nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x+1}+\frac {2^a}{e^{2x}+1}+\frac {3^a}{e^{3x}+1}+\frac {4^a}{e^{4x}+1} +\frac {5^a}{e^{5x}+1}+\cdots \nonumber \\ &&= \frac {\pi ^2}{2}B_1\frac {1}{x^2}-\frac {B_1}{2\cdot 2}\ \ \ (a=1) \nonumber \\ &&= \frac {(2^3-1)\pi ^4}{4}B_2\frac {1}{x^4}+\frac {B_2}{2\cdot 4}\ \ \ (a=3)\nonumber \\ &&= \frac {(2^5-1)\pi ^6}{6}B_3\frac {1}{x^6}-\frac {B_3}{2\cdot 6}\ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \ln 2\frac {1}{x}-\frac {1}{4}+\frac {(2^2-1)B_1B_1}{2\cdot 2\cdot 1!}x +\frac {(2^4-1)B_2B_2}{4\cdot 4\cdot 3!}x^3+\frac {(2^6-1)B_3B_3}{6\cdot 6\cdot 5!}x^5+\cdots \ \ \ (a=0)\nonumber \\ &&= \frac {(2^2-1)2!}{2^2}\zeta (3)\frac {1}{x^3}-\frac {(2^2-1)B_1B_2}{2\cdot 4\cdot 1!}x -\frac {(2^4-1)B_2B_3}{4\cdot 6\cdot 3!}x^3-\frac {(2^6-1)B_3B_4}{6\cdot 8\cdot 5!}x^5-\cdots \ \ \ (=2)\nonumber \\ &&= \frac {(2^4-1)4!}{2^4}\zeta (5)\frac {1}{x^5}+\frac {(2^2-1)B_1B_3}{2\cdot 6\cdot 1!}x +\frac {(2^4-1)B_2B_4}{4\cdot 8\cdot 3!}x^3+\frac {(2^6-1)B_3B_5}{6\cdot 10\cdot 5!}x^5+\cdots \ \ \ (a=4)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{2}\zeta (3)-\frac {(2^2-1)(2\pi )^2}{2\cdot 2\cdot 1!\cdot 2!}B_1B_1x +\frac {2^3-1}{2^3\pi ^2}\zeta (3)x^2-\frac {(2^4-1)B_2}{2\cdot 4\cdot 3!}x^3\ \ \ (a=-3)\nonumber \\ &&= \frac {1}{2}\zeta (5)-\frac {(2^2-1)(2\pi )^4}{2\cdot 2\cdot 1!\cdot 4!}B_1B_2x +\frac {(2^4-1)(2\pi )^2}{2\cdot 4\cdot 3!\cdot 2!}B_2B_1x^3 -\frac {2^5-1}{2^5\pi ^4}\zeta (5)x^4+\frac {(2^6-1)B_3}{2\cdot 6\cdot 5!}x^5\ \ \ (a=-5)\nonumber \\ &&= \frac {1}{2}\zeta (7)-\frac {(2^2-1)(2\pi )^6}{2\cdot 2\cdot 1!\cdot 6!}B_1B_3x +\frac {(2^4-1)(2\pi )^4}{2\cdot 4\cdot 3!\cdot 4!}B_2B_2x^3 -\frac {(2^6-1)(2\pi )^2B_3B_1}{2\cdot 6\cdot 5!\cdot 2!}x^5 +\frac {2^7-1}{2^7\pi ^6}\zeta (7)x^6-\frac {(2^8-1)B_4}{2\cdot 8\cdot 7!}x^7\ \ \ (a=-7)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=-1\)と\(a=-2,-4,-6,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x+1}-\frac {2^a}{e^{2x}+1}+\frac {3^a}{e^{3x}+1}-\frac {4^a}{e^{4x}+1} +\frac {5^a}{e^{5x}+1}-\cdots \nonumber \\ &&= \frac {2^2-1}{2\cdot 2}B_1\ \ \ (a=1)\nonumber \\ &&= -\frac {2^4-1}{2\cdot 4}B_2\ \ \ (a=3)\nonumber \\ &&= \frac {2^6-1}{2\cdot 6}B_3\ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{4}-\frac {(2^2-1)^2}{2\cdot 2\cdot 1!}B_1B_1x -\frac {(2^4-1)^2}{4\cdot 4\cdot 3!}B_2B_2x^3-\frac {(2^6-1)^2}{6\cdot 6\cdot 5!}B_3B_3x^5-\cdots \ \ \ (a=0)\nonumber \\ &&= \frac {(2^4-1)(2^2-1)}{2\cdot 4\cdot 1!}B_1B_2x+\frac {(2^6-1)(2^4-1)}{4\cdot 6\cdot 3!}B_2B_3x^3+\frac {(2^8-1)(2^6-1)}{6\cdot 8\cdot 5!}B_3B_4x^5+\cdots \ \ \ (a=2)\nonumber \\ &&= -\frac {(2^6-1)(2^2-1)}{2\cdot 6\cdot 1!}B_1B_3x-\frac {(2^8-1)(2^4-1)}{4\cdot 8\cdot 3!}B_2B_4x^3-\frac {(2^{10}-1)(2^6-1)}{6\cdot 10\cdot 5!}B_3B_5x^5-\cdots \ \ \ (a=4)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\ln 2}{2}-\frac {2^2-1}{2\cdot 2\cdot 1!}B_1x\ \ \ (a=-1)\nonumber \\ &&= \frac {2^2-1}{2^3}\zeta (3)-\frac {(2^2-1)\pi ^2}{2\cdot 1!\cdot 2!}B_1B_1x +\frac {(2^4-1)B_2}{2\cdot 4\cdot 3!}x^3\ \ \ (a=-3)\nonumber \\ &&= \frac {2^4-1}{2^5}\zeta (5)-\frac {(2^3-1)(2^2-1)\pi ^4}{2\cdot 1!\cdot 4!}B_2B_1x +\frac {(2-1)(2^4-1)\pi ^2}{4\cdot 3!\cdot 2!}B_1B_2x^3 -\frac {(2^6-1)B_3}{2\cdot 6\cdot 5!}x^5\ \ \ (a=-5)\nonumber \\ &&= \frac {2^6-1}{2^7}\zeta (7)-\frac {(2^5-1)(2^2-1)\pi ^6}{2\cdot 1!\cdot 6!}B_1B_3x +\frac {(2^3-1)(2^4-1)\pi ^4}{4\cdot 3!\cdot 4!}B_2B_2x^3 -\frac {(2-1)(2^6-1)\pi ^2}{6\cdot 5!\cdot 2!}B_3B_1x^5 \nonumber \\ && \hspace {38em}+\frac {(2^8-1)B_4}{2\cdot 8\cdot 7!}x^7 \ \ \ (a=-7)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\pi ^2}{2\cdot 2!}B_1-\frac {(2^2-1)\ln 2}{2\cdot 1!}B_1x +\frac {(2^2-1)(2^4-1)}{2\cdot 4\cdot 3!}B_2B_1x^3 +\frac {(2^4-1)(2^6-1)}{4\cdot 6\cdot 5!}B_3B_2x^5+\cdots \ \ \ (a=-2)\nonumber \\ &&= \frac {(2^3-1)\pi ^4}{2\cdot 4!}B_2-\frac {(2^2-1)(2^2-1)B_1}{2\cdot 2^2\cdot 1!}\zeta (3)x +\frac {(2^4-1)\ln 2}{4\cdot 3!}B_2x^3 -\frac {(2^2-1)(2^6-1)}{2\cdot 6\cdot 5!}B_1B_3x^5 \nonumber \\ && \hspace {30em} -\frac {(2^4-1)(2^8-1)}{4\cdot 8\cdot 7!}B_2B_4x^7-\cdots \ \ \ (a=-4)\nonumber \\ &&= \frac {(2^5-1)\pi ^6}{2\cdot 6!}B_3-\frac {(2^4-1)(2^2-1)B_1}{2\cdot 2^4\cdot 1!}\zeta (5)x +\frac {(2^2-1)(2^4-1)B_2}{4\cdot 2^2\cdot 3!}\zeta (3)x^3 -\frac {(2^6-1)\ln 2}{6\cdot 5!}B_3x^5+\frac {(2^2-1)(2^8-1)}{2\cdot 8\cdot 7!}B_1B_4x^7 \nonumber \\ && \hspace {30em}+\frac {(2^4-1)(2^{10}-1)}{4\cdot 10\cdot 9!}B_2B_5x^9+\cdots \ \ \ (a=-6) \nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x+1}+\frac {3^a}{e^{3x}+1}+\frac {5^a}{e^{5x}+1}+\frac {7^a}{e^{7x}+1}+\cdots \nonumber \\ &&= \frac {\pi ^2}{2\cdot 2}B_1\frac {1}{x^2}+\frac {B_1}{2\cdot 2}\ \ \ (a=1)\nonumber \\ &&= \frac {(2^3-1)\pi ^4}{2\cdot 4}B_2\frac {1}{x^4}-\frac {2^3-1}{2\cdot 4}B_2\ \ \ (a=3)\nonumber \\ &&= \frac {(2^5-1)\pi ^6}{2\cdot 6}B_3\frac {1}{x^6}+\frac {2^5-1}{2\cdot 6}B_3\ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\ln 2}{2}\frac {1}{x}-\frac {(2-1)(2^2-1)}{2\cdot 2\cdot 1!}B_1B_1x -\frac {(2^3-1)(2^4-1)}{4\cdot 4\cdot 3!}B_2B_2x^3-\cdots \ \ \ (a=0)\nonumber \\ &&= \frac {(2^2-1)2!}{2^3}\zeta (3)\frac {1}{x^3}+\frac {(2^3-1)(2^2-1)}{2\cdot 4\cdot 1!}B_2B_1x +\frac {(2^5-1)(2^4-1)}{4\cdot 6\cdot 3!}B_3B_2x^3+\cdots \ \ \ (a=2)\nonumber \\ &&= \frac {(2^4-1)4!}{2^5}\zeta (5)\frac {1}{x^5}-\frac {(2^5-1)(2^2-1)}{2\cdot 6\cdot 1!}B_3B_1x -\frac {(2^7-1)(2^4-1)}{4\cdot 8\cdot 3!}B_4B_2x^3-\cdots \ \ \ (a=4)\nonumber \\ &&= \frac {(2^6-1)6!}{2^7}\zeta (7)\frac {1}{x^7}+\frac {(2^7-1)(2^2-1)}{2\cdot 8\cdot 1!}B_4B_1x +\frac {(2^9-1)(2^4-1)}{4\cdot 10\cdot 3!}B_5B_2x^3+\cdots \ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {2^3-1}{2^4}\zeta (3)-\frac {(2^2-1)^2\pi ^2}{2\cdot 2\cdot 1!\cdot 2!}B_1B_1x +\frac {2^3-1}{2^4\pi ^2}\zeta (3)x^2\ \ \ (a=-3)\nonumber \\ &&= \frac {2^5-1}{2^6}\zeta (5)-\frac {(2^4-1)(2^2-1)\pi ^4}{2\cdot 2\cdot 1!\cdot 4!}B_2B_1x +\frac {(2^2-1)(2^4-1)\pi ^2}{2\cdot 4\cdot 3!\cdot 2!}B_1B_2x^3 -\frac {2^5-1}{2^6\pi ^4}\zeta (5)x^4\ \ \ (a=-5)\nonumber \\ &&= \frac {2^7-1}{2^8}\zeta (7)-\frac {(2^6-1)(2^2-1)\pi ^6}{2\cdot 2\cdot 1!\cdot 6!}B_3B_1x +\frac {(2^4-1)(2^4-1)\pi ^4}{2\cdot 4\cdot 3!\cdot 4!}B_2B_2x^3 -\frac {(2^2-1)(2^6-1)\pi ^2}{2\cdot 6\cdot 5!\cdot 2!}B_1B_3x^5\nonumber \\ && \hspace {35em} +\frac {2^7-1}{2^8\pi ^6}\zeta (7)x^6\ \ \ (a=-7)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=-1\)と\(a=-2,-4,-6,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x+1}-\frac {3^a}{e^{3x}+1}+\frac {5^a}{e^{5x}+1}-\frac {7^a}{e^{7x}+1}+\cdots \nonumber \\ &&= \frac {2^2-1}{2\cdot 2\cdot 1!}E_1B_1x+\frac {2^4-1}{2\cdot 4\cdot 3!}E_2B_2x^3 +\frac {2^6-1}{2\cdot 6\cdot 5!}E_3B_3x^5+\cdots \ \ \ (a=1)\nonumber \\ &&= -\frac {2^2-1}{2\cdot 2\cdot 1!}B_1E_2x-\frac {2^4-1}{2\cdot 4\cdot 3!}B_2E_3x^3 -\frac {2^6-1}{2\cdot 6\cdot 5!}B_3E_4x^5-\cdots \ \ \ (a=3)\nonumber \\ &&= \frac {2^2-1}{2\cdot 2\cdot 1!}E_3B_1x+\frac {2^4-1}{2\cdot 4\cdot 3!}E_4B_2x^3 +\frac {2^6-1}{2\cdot 6\cdot 5!}E_5B_3x^5+\cdots \ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {E_0}{2\cdot 2}\ \ \ (a=0)\nonumber \\ &&= -\frac {E_1}{2\cdot 2}\ \ \ (a=2)\nonumber \\ &&= \frac {E_2}{2\cdot 2}\ \ \ (a=4)\nonumber \\ &&= -\frac {E_3}{2\cdot 2}\ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\pi }{2^3\cdot 0!}E_0-\frac {2^2-1}{2\cdot 2\cdot 1!}B_1E_0x -\frac {2^4-1}{2\cdot 4\cdot 3!}B_2E_1x^3-\frac {2^6-1}{2\cdot 6\cdot 5!}B_3E_2x^5-\cdots \ \ \ (a=-1)\nonumber \\ &&= \frac {\pi ^3}{2^5\cdot 2!}E_1-\frac {2^2-1}{2\cdot 1!}B_1\mu (2)x+\frac {2^4-1}{2\cdot 4\cdot 3!}B_2E_0x^3+\frac {2^6-1}{2\cdot 6\cdot 5!}B_3E_1x^5+\cdots \ \ \ (a=-3)\nonumber \\ &&= \frac {\pi ^5}{2^7\cdot 4!}E_2-\frac {2^2-1}{2\cdot 1!}B_1\mu (4)x +\frac {2^4-1}{4\cdot 3!}B_2\mu (2)x^3-\frac {2^6-1}{2\cdot 6\cdot 5!}B_3E_0x^5 -\frac {2^8-1}{2\cdot 8\cdot 7!}B_4E_1x^7-\cdots \ \ \ (a=-5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{2}\mu (2)-\frac {(2^2-1)\left (\frac {\pi }{2}\right )}{2\cdot 2\cdot 1!\cdot 0!}B_1E_0x \ \ \ (a=-2)\nonumber \\ &&= \frac {1}{2}\mu (4)-\frac {(2^2-1)\left (\frac {\pi }{2}\right )^3}{2\cdot 2\cdot 1!\cdot 2!}B_1E_1x +\frac {(2^4-1)\left (\frac {\pi }{2}\right )}{2\cdot 4\cdot 3!\cdot 0!}B_2E_0x^3\ \ \ (a=-4)\nonumber \\ &&= \frac {1}{2}\mu (6)-\frac {(2^2-1)\left (\frac {\pi }{2}\right )^5}{2\cdot 2\cdot 1!\cdot 4!}B_1E_2x +\frac {(2^4-1)\left (\frac {\pi }{2}\right )^3}{2\cdot 4\cdot 3!\cdot 2!}B_2E_1x^3 -\frac {(2^6-1)\left (\frac {\pi }{2}\right )}{2\cdot 6\cdot 5!\cdot 0!}B_3E_0x^5\ \ \ (a=-6)\nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x-e^{-x}}+\frac {2^a}{e^{2x}-e^{-2x}}+\frac {3^a}{e^{3x}-e^{-3x}} +\frac {4^a}{e^{4x}-e^{-4x}}+\frac {5^a}{e^{5x}-e^{-5x}}+\cdots \nonumber \\ &&= \frac {(2^2-1)\pi ^2}{2\cdot 2}B_1\frac {1}{x^2}-\frac {1}{4x}\ \ \ (a=1)\nonumber \\ &&= \frac {(2^4-1)\pi ^4}{2\cdot 4}B_2\frac {1}{x^4}\ \ \ (a=3)\nonumber \\ &&= \frac {(2^6-1)\pi ^6}{2\cdot 6}B_3\frac {1}{x^6}\ \ \ (a=5)\nonumber \\ &&= \frac {(2^8-1)\pi ^8}{2\cdot 8}B_4\frac {1}{x^8}\ \ \ (a=7)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {(2^3-1)2!}{2^3}\zeta (3)\frac {1}{x^3}-\frac {B_1}{2\cdot 2}\frac {1}{x} -\frac {B_1B_2}{2\cdot 4\cdot 1!}x-\frac {(2^3-1)B_2B_3}{4\cdot 6\cdot 3!}x^3 -\frac {(2^5-1)B_3B_4}{6\cdot 8\cdot 5!}x^5-\cdots \ \ \ (a=2)\nonumber \\ &&= \frac {(2^5-1)4!}{2^5}\zeta (5)\frac {1}{x^5}+\frac {B_2}{2\cdot 4}\frac {1}{x} +\frac {B_1B_3}{2\cdot 6\cdot 1!}x+\frac {(2^3-1)B_2B_4}{4\cdot 8\cdot 3!}x^3 +\frac {(2^5-1)B_3B_5}{6\cdot 10\cdot 5!}x^5+\cdots \ \ \ (a=4)\nonumber \\ &&= \frac {(2^7-1)6!}{2^7}\zeta (7)\frac {1}{x^7}-\frac {B_3}{2\cdot 6}\frac {1}{x} -\frac {B_1B_4}{2\cdot 8\cdot 1!}x-\frac {(2^3-1)B_2B_5}{4\cdot 10\cdot 3!}x^3 -\frac {(2^5-1)B_3B_6}{6\cdot 12\cdot 5!}x^5-\cdots \ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\pi ^2}{2!}B_1\frac {1}{x}-\frac {\ln 2}{2}+\frac {B_1}{2\cdot 2\cdot 1!}x\ \ \ (a=-1)\nonumber \\ &&= \frac {2^2\pi ^4}{4!}B_2\frac {1}{x}-\frac {(2-1)(2\pi )^2}{2\cdot 2\cdot 1!\cdot 2!}B_1B_1x +\frac {2^2-1}{2(2\pi )^2}\zeta (3)x^2-\frac {2^3-1}{2\cdot 4\cdot 3!}B_2x^3\ \ \ (a=-3)\nonumber \\ &&= \frac {2^4\pi ^6}{6!}B_3\frac {1}{x}-\frac {(2-1)(2\pi )^4}{2\cdot 2\cdot 1!\cdot 4!}B_1B_2x +\frac {(2^3-1)(2\pi )^2}{2\cdot 4\cdot 3!\cdot 2!}B_2B_1x^3 -\frac {2^4-1}{2(2\pi )^4}\zeta (5)x^4+\frac {2^5-1}{2\cdot 6\cdot 5!}B_3x^5\ \ \ (a=-5)\nonumber \\ &&= \frac {2^6\pi ^8}{8!}B_4\frac {1}{x}-\frac {(2-1)(2\pi )^6}{2\cdot 2\cdot 1!\cdot 6!}B_1B_3x +\frac {(2^3-1)(2\pi )^4}{2\cdot 4\cdot 3!\cdot 4!}B_2B_2x^3 -\frac {(2^5-1)(2\pi )^2}{2\cdot 6\cdot 5\cdot 2!}B_3B_1x^5 +\frac {2^6-1}{2(2\pi )^6}\zeta (7)x^6\nonumber \\ &&\hspace {35em} -\frac {2^7-1}{2\cdot 8\cdot 7!}B_4x^7\ \ \ (a=-7)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=0,-2,-4,-6,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x-e^{-x}}-\frac {2^a}{e^{2x}-e^{-2x}}+\frac {3^a}{e^{3x}-e^{-3x}} -\frac {4^a}{e^{4x}-e^{-4x}}+\cdots \nonumber \\ &&= \frac {\ln 2}{2}\frac {1}{x}-\frac {(2-1)(2^2-1)}{2\cdot 2\cdot 1!}B_1B_1x -\frac {(2^3-1)(2^4-1)}{4\cdot 4\cdot 3!}B_2B_2x^3 -\frac {(2^5-1)(2^6-1)}{6\cdot 6\cdot 5!}B_3B_3x^5-\cdots \ \ \ (a=0)\nonumber \\ &&= \frac {2^2-1}{2\cdot 2}B_1\frac {1}{x}+\frac {(2-1)(2^4-1)}{2\cdot 4\cdot 1!}B_1B_2x +\frac {(2^3-1)(2^6-1)}{4\cdot 6\cdot 3!}B_2B_3x^3 +\frac {(2^5-1)(2^8-1)}{6\cdot 8\cdot 5!}B_3B_4x^5+\cdots \ \ \ (a=2)\nonumber \\ &&= -\frac {2^4-1}{2\cdot 4}B_2\frac {1}{x}-\frac {(2-1)(2^6-1)}{2\cdot 6\cdot 1!}B_1B_3x -\frac {(2^3-1)(2^8-1)}{4\cdot 8\cdot 3!}B_2B_4x^3 -\frac {(2^5-1)(2^{10}-1)}{6\cdot 10\cdot 5!}B_3B_5x^5-\cdots \ \ \ (a=4)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{4x}\ \ \ (a=1)\nonumber \\ &&= 0\ \ \ (a=3,5,7,9,\cdots )\nonumber \\ &&= \frac {\pi ^2}{2\cdot 2!}B_1\frac {1}{x}-\frac {B_1}{2\cdot 2\cdot 1!}x\ \ \ (a=-1)\nonumber \\ &&= \frac {(2^3-1)\pi ^4}{2\cdot 4!}B_2\frac {1}{x}-\frac {(2-1)(2-1)\pi ^2}{2\cdot 1!\cdot 2!}B_1B_1x +\frac {2^3-1}{2\cdot 4\cdot 3!}B_2x^3\ \ \ (a=-3)\nonumber \\ &&= \frac {(2^5-1)\pi ^6}{2\cdot 6!}B_3\frac {1}{x}-\frac {(2-1)(2^3-1)\pi ^4}{2\cdot 1!\cdot 4!}B_1B_2x +\frac {(2^3-1)(2-1)\pi ^2}{4\cdot 3!\cdot 2!}B_2B_1x^3 -\frac {2^5-1}{2\cdot 6\cdot 5!}B_3x^5\ \ \ (a=-5)\nonumber \\ &&= \frac {(2^7-1)\pi ^8}{2\cdot 8!}B_4\frac {1}{x}-\frac {(2-1)(2^5-1)\pi ^6}{2\cdot 1!\cdot 6!}B_1B_3x +\frac {(2^3-1)(2^3-1)\pi ^4}{4\cdot 3!\cdot 4!}B_2B_2x^3 -\frac {(2^5-1)(2-1)\pi ^2}{6\cdot 5!\cdot 2!}B_3B_1x^5\nonumber \\ &&\hspace {35em}+\frac {2^7-1}{2\cdot 8\cdot 7!}B_4x^7 \ \ \ (a=-7)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {2^2-1}{2^3}\zeta (3)\frac {1}{x}-\frac {(2-1)\ln 2}{2\cdot 1!}B_1x +\frac {(2^3-1)(2^2-1)}{2\cdot 4\cdot 3!}B_2B_1x^3 +\frac {(2^5-1)(2^4-1)}{4\cdot 6\cdot 5!}B_3B_2x^5\nonumber \\ &&\hspace {33em} +\frac {(2^7-1)(2^6-1)}{6\cdot 8\cdot 7!}B_4B_3x^7+\cdots \ \ \ (a=-2)\nonumber \\ &&= \frac {2^4-1}{2^5}\zeta (5)\frac {1}{x}-\frac {(2-1)(2^2-1)}{2^3\cdot 1!}B_1\zeta (3)x +\frac {(2^3-1)\ln 2}{4\cdot 3!}B_2x^3-\frac {(2^5-1)(2^2-1)}{2\cdot 6\cdot 5!}B_3B_1x^5 \nonumber \\ &&\hspace {33em} -\frac {(2^7-1)(2^4-1)}{4\cdot 8\cdot 7!}B_4B_2x^7-\cdots \ \ \ \ (a=-4)\nonumber \\ &&= \frac {2^6-1}{2^7}\zeta (7)\frac {1}{x}-\frac {(2-1)(2^4-1)}{2\cdot 2^4\cdot 1!}B_1\zeta (5)x +\frac {(2^3-1)(2^2-1)}{4\cdot 2^2\cdot 3!}B_2\zeta (3)x^3 -\frac {(2^5-1)\ln 2}{6\cdot 5!}B_3x^5\nonumber \\ &&\hspace {22em}+\frac {(2^7-1)(2^2-1)}{2\cdot 8\cdot 7!}B_4B_1x^7 +\frac {(2^9-1)(2^4-1)}{4\cdot 10\cdot 9!}B_5B_2x^9+\cdots \ \ \ (a=-6)\nonumber \\ &&= \frac {2^8-1}{2^9}\zeta (9)\frac {1}{x}-\frac {(2-1)(2^6-1)}{2\cdot 2^6\cdot 1!}B_1\zeta (7)x +\frac {(2^3-1)(2^4-1)}{4\cdot 2^4\cdot 3!}B_2\zeta (5)x^3 -\frac {(2^5-1)(2^2-1)}{6\cdot 2^2\cdot 5!}B_3\zeta (3)x^5\nonumber \\ &&\hspace {22em} +\frac {(2^7-1)\ln 2}{8\cdot 7!}B_4x^7-\frac {(2^9-1)(2^2-1)}{2\cdot 10\cdot 9!}B_5B_1x^9 -\cdots \ \ \ (a=-8)\nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x-e^{-x}}+\frac {3^a}{e^{3x}-e^{-3x}}+\frac {5^a}{e^{5x}-e^{-5x}} +\frac {7^a}{e^{7x}-e^{-7x}}+\cdots \nonumber \\ &&= \frac {(2^2-1)\pi ^2}{4\cdot 2}B_1\frac {1}{x^2}\ \ \ (a=1)\nonumber \\ &&= \frac {(2^4-1)\pi ^4}{4\cdot 4}B_2\frac {1}{x^4}\ \ \ (a=3)\nonumber \\ &&= \frac {(2^6-1)\pi ^6}{4\cdot 6}B_3\frac {1}{x^6}\ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {(2^3-1)2!}{2^4}\zeta (3)\frac {1}{x^3}+\frac {B_1}{2\cdot 2}\frac {1}{x} +\frac {(2-1)(2^3-1)}{2\cdot 4\cdot 1!}B_1B_2x +\frac {(2^3-1)(2^5-1)}{4\cdot 6\cdot 3!}B_2B_3x^3\nonumber \\ &&\hspace {33em} +\frac {(2^5-1)(2^7-1)}{6\cdot 8\cdot 5!}B_3B_4x^5+\cdots \ \ \ (a=2)\nonumber \\ &&= \frac {(2^5-1)4!}{2^6}\zeta (5)\frac {1}{x^5}-\frac {2^3-1}{2\cdot 4}B_2\frac {1}{x} -\frac {(2-1)(2^5-1)}{2\cdot 6\cdot 1!}B_1B_3x-\frac {(2^3-1)(2^7-1)}{4\cdot 8\cdot 3!}B_2B_4x^3\nonumber \\ &&\hspace {33em} -\frac {(2^5-1)(2^9-1)}{6\cdot 10\cdot 5!}B_3B_5x^5-\cdots \ \ \ (a=4)\nonumber \\ &&= \frac {(2^7-1)6!}{2^8}\zeta (7)\frac {1}{x^7}+\frac {2^5-1}{2\cdot 6}B_3\frac {1}{x} +\frac {(2-1)(2^7-1)}{2\cdot 8\cdot 1!}B_1B_4x+\frac {(2^3-1)(2^9-1)}{4\cdot 10\cdot 3!}B_2B_5x^3\nonumber \\ &&\hspace {33em} +\frac {(2^5-1)(2^{11}-1)}{6\cdot 12\cdot 5!}B_3B_6x^5+\cdots \ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {(2^2-1)\pi ^2}{4\cdot 2!}B_1\frac {1}{x}-\frac {\ln 2}{4}\ \ \ (a=-1)\nonumber \\ &&= \frac {(2^4-1)\pi ^4}{4\cdot 4!}B_2\frac {1}{x}-\frac {(2-1)(2^2-1)\pi ^2}{2\cdot 2\cdot 1!\cdot 2!} B_1B_1x+\frac {2^2-1}{2^4\pi ^2}\zeta (3)x^2\ \ \ (a=-3)\nonumber \\ &&= \frac {(2^6-1)\pi ^6}{4\cdot 6!}B_3\frac {1}{x}-\frac {(2-1)(2^4-1)\pi ^4}{2\cdot 2\cdot 1!\cdot 4!} B_1B_2x+\frac {(2^3-1)(2^2-1)\pi ^2}{2\cdot 4\cdot 3!\cdot 2!}B_2B_1x^3 -\frac {2^4-1}{2^6\pi ^4}\zeta (5)x^4\ \ \ (a=-5)\nonumber \\ &&= \frac {(2^8-1)\pi ^8}{4\cdot 8!}B_4\frac {1}{x}-\frac {(2-1)(2^6-1)\pi ^6}{2\cdot 2\cdot 1!\cdot 6!} B_1B_3x+\frac {(2^3-1)(2^4-1)\pi ^4}{2\cdot 4\cdot 3!\cdot 4!}B_2B_2x^3 -\frac {(2^5-1)(2^2-1)\pi ^2}{2\cdot 6\cdot 5!\cdot 2!}B_3B_1x^5\nonumber \\ &&\hspace {33em} +\frac {2^6-1}{2^8\pi ^6}\zeta (7)x^6\ \ \ (a=-7)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=0,-2,-4,-6,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x-e^{-x}}-\frac {3^a}{e^{3x}-e^{-3x}}+\frac {5^a}{e^{5x}-e^{-5x}} -\frac {7^a}{e^{7x}-e^{-7x}}+\cdots \nonumber \\ &&= \frac {E_0}{4}\frac {1}{x}+\frac {2-1}{2\cdot 2\cdot 1!}B_1E_1x+\frac {2^3-1}{2\cdot 4\cdot 3!}B_2E_2x^3+\frac {2^5-1}{2\cdot 6\cdot 5!}B_3E_3x^5+\cdots \ \ \ (a=1)\nonumber \\ &&= -\frac {E_1}{4}\frac {1}{x}-\frac {2-1}{2\cdot 2\cdot 1!}B_1E_2x-\frac {2^3-1}{2\cdot 4\cdot 3!}B_2E_3x^3 -\frac {2^5-1}{2\cdot 6\cdot 5!}B_3E_4x^5-\cdots \ \ \ (a=3)\nonumber \\ &&= \frac {E_2}{4}\frac {1}{x}+\frac {2-1}{2\cdot 2\cdot 1!}B_1E_3x +\frac {2^3-1}{2\cdot 4\cdot 3!}B_2E_4x^3+\frac {2^5-1}{2\cdot 6\cdot 5!}B_3E_5x^5+\cdots \ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\pi }{8}\frac {1}{x}\ \ \ (a=0)\nonumber \\ &&= 0\ \ \ (a=2,4,6,8,\cdots )\nonumber \\ &&= \frac {1}{2}\mu (2)\frac {1}{x}-\frac {2-1}{2\cdot 2\cdot 1!}B_1E_0x -\frac {2^3-1}{2\cdot 4\cdot 3!}B_2E_1x^3-\frac {2^5-1}{2\cdot 6\cdot 5!}B_3E_2x^5-\cdots \ \ \ (a=-1)\nonumber \\ &&= \frac {1}{2}\mu (4)\frac {1}{x}-\frac {2-1}{2\cdot 1!}B_1\mu (2)x+\frac {2^3-1}{2\cdot 4\cdot 3!}B_2E_0x^3+\frac {2^5-1}{2\cdot 6\cdot 5!}B_3E_1x^5+\cdots \ \ \ (a=-3)\nonumber \\ &&= \frac {1}{2}\mu (6)\frac {1}{x}-\frac {2-1}{2\cdot 1!}B_1\mu (4)x +\frac {2^3-1}{4\cdot 3!}B_2\mu (2)x^3-\frac {2^5-1}{2\cdot 6\cdot 5!}B_3E_0x^5-\cdots \ \ \ (a=-5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^3}{4\cdot 2!}E_1\frac {1}{x} -\frac {(2-1)\left (\frac {\pi }{2}\right )}{2\cdot 2\cdot 0!\cdot 1!}B_1E_0x\ \ \ (a=-2)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^5}{4\cdot 4!}E_2\frac {1}{x} -\frac {(2-1)\left (\frac {\pi }{2}\right )^3}{2\cdot 2\cdot 1!\cdot 2!}B_1E_1x +\frac {(2^3-1)\left (\frac {\pi }{2}\right )}{2\cdot 4\cdot 3!\cdot 0!}B_2E_0x^3\ \ \ (a=-4)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^7}{4\cdot 6!}E_3\frac {1}{x} -\frac {(2-1)\left (\frac {\pi }{2}\right )^5}{2\cdot 2\cdot 1!\cdot 4!}B_1E_2x +\frac {(2^3-1)\left (\frac {\pi }{2}\right )^3}{2\cdot 4\cdot 3!\cdot 2!}B_2E_1x^3 -\frac {(2^5-1)\left (\frac {\pi }{2}\right )}{2\cdot 6\cdot 5!\cdot 0!}B_3E_0x^5\ \ \ (a=-6)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^9}{4\cdot 8!}E_4\frac {1}{x} -\frac {(2-1)\left (\frac {\pi }{2}\right )^7}{2\cdot 2\cdot 1!\cdot 6!}B_1E_3x +\frac {(2^3-1)\left (\frac {\pi }{2}\right )^5}{2\cdot 4\cdot 3!\cdot 4!}B_2E_2x^3 -\frac {(2^5-1)\left (\frac {\pi }{2}\right )^3}{2\cdot 6\cdot 5!\cdot 2!}B_3E_1x^5 +\frac {(2^7-1)\left (\frac {\pi }{2}\right )}{2\cdot 8\cdot 7!\cdot 0!}B_4E_0x^7\ \ \ (a=-8)\nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x+e^{-x}}+\frac {2^a}{e^{2x}+e^{-2x}}+\frac {3^a}{e^{3x}+e^{-3x}} +\frac {4^a}{e^{4x}+e^{-4x}}+\frac {5^a}{e^{5x}+e^{-5x}}+\cdots \nonumber \\ &&= 1!\cdot \mu (2)\frac {1}{x^2}-\frac {E_0B_1}{2\cdot 2\cdot 0!}-\frac {E_1B_2}{2\cdot 4\cdot 2!}x^2 -\frac {E_2B_3}{2\cdot 6\cdot 4!}x^4-\cdots \ \ \ (a=1)\nonumber \\ &&= 3!\cdot \mu (4)\frac {1}{x^4}+\frac {E_0B_2}{2\cdot 4\cdot 0!}+\frac {E_1B_3}{2\cdot 6\cdot 2!}x^2+\frac {E_2B_4}{2\cdot 8\cdot 4!}x^4+\cdots \ \ \ (a=3)\nonumber \\ &&= 5!\cdot \mu (6)\frac {1}{x^6}-\frac {E_0B_3}{2\cdot 6\cdot 0!} -\frac {E_1B_4}{2\cdot 8\cdot 2!}x^2-\frac {E_2B_5}{2\cdot 10\cdot 4!}x^4-\cdots \ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{2}\left (\frac {\pi }{2}\right )E_0\frac {1}{x}-\frac {1}{4}\ \ \ (a=0)\nonumber \\ &&= \frac {1}{2}\left (\frac {\pi }{2}\right )^3E_1\frac {1}{x^3}\ \ \ (a=2)\nonumber \\ &&= \frac {1}{2}\left (\frac {\pi }{2}\right )^5E_2\frac {1}{x^5}\ \ \ (a=4)\nonumber \\ &&= \frac {1}{2}\left (\frac {\pi }{2}\right )^7E_3\frac {1}{x^7}\ \ \ (a=6)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {(2\pi )^2}{2\cdot 2\cdot 0!\cdot 2!}E_0B_1-\frac {2}{\pi }\mu (2)x +\frac {E_1}{2\cdot 2\cdot 2!}x^2\ \ \ (a=-2)\nonumber \\ &&= \frac {(2\pi )^4}{2\cdot 2\cdot 0!\cdot 4!}E_0B_2-\frac {(2\pi )^2}{2\cdot 2\cdot 2!\cdot 2!}E_1B_1x^2+\left (\frac {2}{\pi }\right )^3\mu (4)x^3 -\frac {E_2}{2\cdot 2\cdot 4!}x^4\ \ \ (a=-2)\nonumber \\ &&= \frac {(2\pi )^6}{2\cdot 2\cdot 0!\cdot 6!}E_0B_3-\frac {(2\pi )^4}{2\cdot 2\cdot 2!\cdot 4!}E_1B_2x^2+\frac {(2\pi )^2}{2\cdot 2\cdot 4!\cdot 2!}E_2B_1x^4 -\left (\frac {2}{\pi }\right )^5\mu (6)x^5+\frac {E_3}{2\cdot 2\cdot 6!}x^6\ \ \ (a=-6)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=-1,-3,-5,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x+e^{-x}}-\frac {2^a}{e^{2x}+e^{-2x}}+\frac {3^a}{e^{3x}+e^{-3x}} -\frac {4^a}{e^{4x}+e^{-4x}}+\frac {5^a}{e^{5x}+e^{-5x}}-\cdots \nonumber \\ &&= \frac {2^2-1}{2\cdot 2\cdot 0!}E_0B_1+\frac {2^4-1}{2\cdot 4\cdot 2!}E_1B_2x^2 +\frac {2^6-1}{2\cdot 6\cdot 4!}E_2B_3x^4+\cdots \ \ \ (a=1)\nonumber \\ &&= -\frac {2^4-1}{2\cdot 4\cdot 0!}E_0B_2-\frac {2^6-1}{2\cdot 6\cdot 2!}E_1B_3x^2 -\frac {2^8-1}{2\cdot 8\cdot 4!}E_2B_4x^4-\cdots \ \ \ (a=3)\nonumber \\ &&= \frac {2^6-1}{2\cdot 6\cdot 0!}E_0B_3+\frac {2^8-1}{2\cdot 8\cdot 2!}E_1B_4x^2 +\frac {2^{10}-1}{2\cdot 10\cdot 4!}E_2B_5x^4+\cdots \ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{4}\ \ \ (a=0)\nonumber \\ &&= 0\ \ \ (a=2,4,6,8,10,\cdots )\nonumber \\ &&= \frac {\ln 2}{2}-\frac {2^2-1}{2\cdot 2\cdot 2!}E_1B_1x^2 -\frac {2^4-1}{2\cdot 4\cdot 4!}E_2B_2x^4-\frac {2^6-1}{2\cdot 6\cdot 6!}E_3B_3x^6-\cdots \ \ \ (a=-1)\nonumber \\ &&= \frac {2^2-1}{2^3}E_0\zeta (3)-\frac {\ln 2}{2\cdot 2!}E_1x^2 +\frac {2^2-1}{2\cdot 2\cdot 4!}E_2B_1x^4+\frac {2^4-1}{2\cdot 4\cdot 6!}E_3B_2x^6+\cdots \ \ \ (a=-3)\nonumber \\ &&= \frac {2^4-1}{2^5}E_0\zeta (5)-\frac {2^2-1}{2^3\cdot 2!}E_1\zeta (3)x^2 +\frac {\ln 2}{2\cdot 4!}E_2x^4-\frac {2^2-1}{2\cdot 2\cdot 6!}E_3B_1x^6-\cdots \ \ \ (a=-5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\pi ^2}{2\cdot 0!\cdot 2!}E_0B_1-\frac {E_1}{4\cdot 2!}x^2\ \ \ (a=-2)\nonumber \\ &&= \frac {(2^3-1)\pi ^4}{2\cdot 0!\cdot 4!}E_0B_2-\frac {\pi ^2}{2\cdot 2!\cdot 2!}E_1B_1x^2 +\frac {E_2}{4\cdot 4!}x^4\ \ \ (a=-4)\nonumber \\ &&= \frac {(2^5-1)\pi ^6}{2\cdot 0!\cdot 6!}E_0B_3-\frac {(2^3-1)\pi ^4}{2\cdot 2!\cdot 4!}E_1B_2x^2 +\frac {\pi ^2}{2\cdot 4!\cdot 2!}E_2B_1x^4-\frac {E_3}{4\cdot 6!}x^6\ \ \ (a=-6)\nonumber \\ &&= \frac {(2^7-1)\pi ^8}{2\cdot 0!\cdot 8!}E_0B_4-\frac {(2^5-1)\pi ^6}{2\cdot 2!\cdot 6!}E_1B_3x^2 +\frac {(2^3-1)\pi ^4}{2\cdot 4!\cdot 4!}E_2B_2x^4 -\frac {\pi ^2}{2\cdot 6!\cdot 2!}E_3B_1x^6+\frac {E_4}{4\cdot 8!}x^8\ \ \ (a=-8)\nonumber \\ &&\vdots \end{eqnarray}
\begin{eqnarray} && \frac {1}{e^x+e^{-x}}+\frac {3^a}{e^{3x}+e^{-3x}}+\frac {5^a}{e^{5x}+e^{-5x}} +\frac {7^a}{e^{7x}+e^{-7x}}+\frac {9^a}{e^{9x}+e^{-9x}}+\cdots \nonumber \\ &&= \frac {1}{2}\mu (2)\frac {1}{x^2}+\frac {2-1}{2\cdot 2\cdot 0!}E_0B_1 +\frac {2^3-1}{2\cdot 4\cdot 2!}E_1B_2x^2+\frac {2^5-1}{2\cdot 6\cdot 4!}E_2B_3x^4 +\cdots \ \ \ (a=1)\nonumber \\ &&= \frac {3!}{2}\mu (4)\frac {1}{x^4}-\frac {2^3-1}{2\cdot 4\cdot 0!}E_0B_2 -\frac {2^5-1}{2\cdot 6\cdot 2!}E_1B_3x^2-\frac {2^7-1}{2\cdot 8\cdot 4!}E_2B_4x^4 -\cdots \ \ \ (a=3)\nonumber \\ &&= \frac {5!}{2}\mu (6)\frac {1}{x^6}+\frac {2^5-1}{2\cdot 6\cdot 0!}E_0B_3 +\frac {2^7-1}{2\cdot 8\cdot 2!}E_1B_4x^2+\frac {2^9-1}{2\cdot 10\cdot 4!}E_2B_5x^4 +\cdots \ \ \ (a=5)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {1}{4}\left (\frac {\pi }{2}\right )E_0\frac {1}{x}\ \ \ (a=0)\nonumber \\ &&= \frac {1}{4}\left (\frac {\pi }{2}\right )^3E_1\frac {1}{x^3}\ \ \ (a=2)\nonumber \\ &&= \frac {1}{4}\left (\frac {\pi }{2}\right )^5E_2\frac {1}{x^5}\ \ \ (a=4)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {(2^2-1)\pi ^2}{2\cdot 2\cdot 0!\cdot 2!}E_0B_1-\frac {1}{2}\left (\frac {2}{\pi }\right )\mu (2)x \ \ \ (a=-2)\nonumber \\ &&= \frac {(2^4-1)\pi ^4}{2\cdot 2\cdot 0!\cdot 4!}E_0B_2 -\frac {(2^2-1)\pi ^2}{2\cdot 2\cdot 2!\cdot 2!}E_1B_1x^2+\frac {1}{2}\left (\frac {2}{\pi }\right )^3 \mu (4)x^3\ \ \ (a=-4)\nonumber \\ &&= \frac {(2^6-1)\pi ^6}{2\cdot 2\cdot 0!\cdot 6!}E_0B_3-\frac {(2^4-1)\pi ^4}{2\cdot 2\cdot 2!\cdot 4!}E_1B_2x^2+\frac {(2^2-1)\pi ^2}{2\cdot 2\cdot 4!\cdot 2!}E_2B_1x^4 -\frac {1}{2}\left (\frac {2}{\pi }\right )^5\mu (6)x^5\ \ \ (a=-6)\nonumber \\ &&\vdots \end{eqnarray}
これらは\(a=-1,-3,-5,\cdots \)で係数が発散する。
\begin{eqnarray} && \frac {1}{e^x+e^{-x}}-\frac {3^a}{e^{3x}+e^{-3x}}+\frac {5^a}{e^{5x}+e^{-5x}} -\frac {7^a}{e^{7x}+e^{-7x}}+\frac {9^a}{e^{9x}+e^{-9x}}-\cdots \nonumber \\ &&= \frac {E_0E_0}{4\cdot 0!}+\frac {E_1E_1}{4\cdot 2!}x^2+\frac {E_2E_2}{4\cdot 4!}x^4 +\frac {E_3E_3}{4\cdot 6!}x^6+\cdots \ \ \ (a=0)\nonumber \\ &&= -\frac {E_0E_1}{4\cdot 0!}-\frac {E_1E_2}{4\cdot 2!}x^2-\frac {E_2E_3}{4\cdot 4!}x^4 -\frac {E_3E_4}{4\cdot 6!}x^6-\cdots \ \ \ (a=2)\nonumber \\ &&= \frac {E_0E_2}{4\cdot 0!}+\frac {E_1E_3}{4\cdot 2!}x^2+\frac {E_2E_4}{4\cdot 4!}x^4 +\frac {E_3E_5}{4\cdot 6!}x^6+\cdots \ \ \ (a=4)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )E_0E_0}{2\cdot 2\cdot 0!\cdot 0!}\ \ \ (a=-1)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^3E_0E_1}{2\cdot 2\cdot 0!\cdot 2!} -\frac {\left (\frac {\pi }{2}\right )E_1E_0}{2\cdot 2\cdot 2!\cdot 0!}x^2\ \ \ (a=-1)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^5E_0E_2}{2\cdot 2\cdot 0!\cdot 4!} -\frac {\left (\frac {\pi }{2}\right )^3E_1E_1}{2\cdot 2\cdot 2!\cdot 2!}x^2 +\frac {\left (\frac {\pi }{2}\right )E_2E_0}{2\cdot 2\cdot 4!\cdot 0!}x^4\ \ \ (a=-5)\nonumber \\ &&= \frac {\left (\frac {\pi }{2}\right )^7E_0E_3}{2\cdot 2\cdot 0!\cdot 6!} -\frac {\left (\frac {\pi }{2}\right )^5E_1E_2}{2\cdot 2\cdot 2!\cdot 4!}x^2 +\frac {\left (\frac {\pi }{2}\right )^3E_2E_1}{2\cdot 2\cdot 4!\cdot 2!}x^4 -\frac {\left (\frac {\pi }{2}\right )E_3E_0}{2\cdot 2\cdot 6!\cdot 0!}x^6\ \ \ (a=-7)\nonumber \\ &&\vdots \nonumber \\ &&= \frac {E_0}{2\cdot 0!}\mu (2)-\frac {E_1E_0}{4\cdot 2!}x^2-\frac {E_2E_1}{4\cdot 4!}x^4 -\frac {E_3E_2}{4\cdot 6!}x^6-\cdots \ \ \ (a=-2)\nonumber \\ &&= \frac {E_0}{2\cdot 0!}\mu (4)-\frac {E_1}{2\cdot 2!}\mu (2)x^2+\frac {E_2E_0}{4\cdot 4!}x^4 +\frac {E_3E_1}{4\cdot 6!}x^6+\cdots \ \ \ (a=-4)\nonumber \\ &&= \frac {E_0}{2\cdot 0!}\mu (6)-\frac {E_1}{2\cdot 2!}\mu (4)x^2+\frac {E_2}{2\cdot 4!}\mu (2)x^4 -\frac {E_3E_0}{4\cdot 6!}x^6-\cdots \ \ \ (a=-6)\nonumber \\ &&\vdots \nonumber \\ &&=0\ \ \ (a=1,3,5,7,\cdots ) \end{eqnarray}