Gamma関数をとある無限積を被積分関数に持つ積分に応用してみた☆

\begin{eqnarray} \Gamma(z)&\coloneqq&\lim_{n\rightarrow\infty}\frac{n^{z}n!}{z(z+1)\cdots(z+n)}\\ &=&\lim_{n\rightarrow\infty}\frac{e^{(\log{n}-H_{n})z}}{z}\prod_{k=1}^{n}\frac{e^{\frac{z}{k}}}{1+\frac{z}{k}}\\ &=&\frac{1}{ze^{\gamma z}}\prod_{k=1}^{\infty}\frac{e^{\frac{z}{k}}}{1+\frac{z}{k}} \end{eqnarray}

\begin{eqnarray} z^{n}-1&=&(z-1)(z-\omega)\cdots(z-\omega^{n-1})\\ &=&z^{n}+\sum_{k=1}^{n-1}(-1)^{k}z^{n-k}\sum_{0\leq m_{1}\lt m_{2}\lt\cdots\lt m_{k}\leq n-1}\omega^{m_{1}}\omega^{m_{2}}\cdots \omega^{m_{k}}-(-1)^{n-1}\omega^{\frac{n(n-1)}{2}} \end{eqnarray}

\begin{eqnarray} \Gamma(z)\Gamma(\omega z)\cdots \Gamma(\omega^{n-1} z)&=&\frac{1}{\omega^{\frac{n(n-1)}{2}}z^{n}e^{\gamma z\sum_{k=0}^{n-1}\omega^{k}}}\prod_{k=1}^{\infty}\frac{e^{-\frac{z\sum_{l=0}^{n-1}\omega^{l}}{k}}}{(1+\frac{z}{k})(1+\omega\frac{z}{k})\cdots(1+\omega^{n-1}\frac{z}{k})}\\ &=&\frac{(-1)^{n-1}}{z^{n}}\prod_{k=1}^{\infty}\frac{1}{1+(-1)^{n-1}\frac{z^{n}}{k^{n}}} \end{eqnarray}
ゆえに
\begin{equation} z^{n}\Gamma(z)\Gamma(\omega z)\cdots \Gamma(\omega^{n-1} z)=(-1)^{n-1}\prod_{k=1}^{\infty}\frac{1}{1+(-1)^{n-1}\frac{z^{n}}{k^{n}}} \end{equation}

[1]
\begin{eqnarray} \lim_{z\rightarrow -\frac{m}{\omega^{k}e^{i\theta}}}(z+\frac{m}{\omega^{k}e^{i\theta}})(e^{i\theta}z)^{n}\Gamma(\omega e^{i\theta}z)\cdots\Gamma(\omega^{n-1} e^{i\theta}z)&=&\frac{1}{e^{i\theta}\omega^{k}}(-\frac{m}{\omega^{k}})^{n}\Gamma(-\frac{m}{\omega^{k-1}})\Gamma(-\frac{m}{\omega^{k-2}})\cdots\Gamma(-\frac{m}{\omega})\Gamma(-m\omega)\cdots\Gamma(-m\omega^{n-k-1})\lim_{z\rightarrow -\frac{m}{\omega^{k}e^{i\theta}}}(\omega^{k}e^{i\theta} z+m)\Gamma(\omega^{k}e^{i\theta}z) \end{eqnarray}
[2]
\begin{eqnarray} \lim_{z\rightarrow -\frac{m}{\omega^{k}e^{i\theta}}}(\omega^{k}e^{i\theta} z+m)\Gamma(\omega^{k}e^{i\theta}z)&=&\lim_{z\rightarrow -\frac{m}{\omega^{k}e^{i\theta}}}\lim_{N\rightarrow\infty}(\omega^{k}e^{i\theta} z+m)\frac{N^{\omega^{k}e^{i\theta}z}N!}{\omega^{k}e^{i\theta}z(\omega^{k}e^{i\theta}z+1)\cdots(\omega^{k}e^{i\theta}z+N)}\\ &=&\frac{(-1)^{m}}{m!}\lim_{N\rightarrow\infty}\frac{N!}{N^{m}(N-m)!}\\ &=&\frac{(-1)^{m}}{m!}\lim_{N\rightarrow\infty}\frac{N}{N}\frac{N-1}{N}\cdots\frac{N-m+1}{N}\\ &=&\frac{(-1)^{m}}{m!} \end{eqnarray}
[3]以上の計算より
\begin{eqnarray} \lim_{z\rightarrow -\frac{m}{\omega^{k}e^{i\theta}}}(z+\frac{m}{\omega^{k}e^{i\theta}})(e^{i\theta}z)^{n}\Gamma(\omega e^{i\theta}z)\cdots\Gamma(\omega^{n-1} e^{i\theta}z)&=&\frac{1}{e^{i\theta}\omega^{k}}(-\frac{m}{\omega^{k}})^{n}\frac{(-1)^{m}}{m!}\Gamma(-\frac{m}{\omega^{k-1}})\Gamma(-\frac{m}{\omega^{k-2}})\cdots\Gamma(-\frac{m}{\omega})\Gamma(-m\omega)\cdots\Gamma(-m\omega^{n-k-1})\\ &=&\frac{(-1)^{m+n}m^{n}}{\omega^{k}e^{i\theta}m!}\Gamma(-\frac{m}{\omega^{k-1}})\Gamma(-\frac{m}{\omega^{k-2}})\cdots\Gamma(-\frac{m}{\omega})\Gamma(-m\omega)\cdots\Gamma(-m\omega^{n-k-1})\\ &=&\frac{(-1)^{m+n}m^{n}}{\omega^{k}e^{i\theta}m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{n-1} m) \end{eqnarray}

[1]$n\not\in 2\mathbb{N}$の場合
\begin{eqnarray} -\int_{I_{\pi-(n-1)\frac{2\pi}{2n-1}}}f(z)dz+\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}&=&-P.V.\int_{I_{\pi-(n-1)\frac{2\pi}{2n-1}}}f(z)dz+\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}\\ &=&-e^{i\frac{\pi}{2n-1}}P.V.\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1-\frac{x^{2n-1}}{k^{2n-1}})}+\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}\\ &=&i\pi\sum_{m=1}^{\infty}\lim_{z\rightarrow+\frac{m}{\omega^{n-1}}}(z+\frac{m}{\omega^{n-1}})z^{2n-1}\Gamma(z)\Gamma(\omega z)\cdots \Gamma(\omega^{2n-2} z)\\ &=&-i\pi\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n-1}}{\omega^{n-1}m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-2} m) \end{eqnarray}
\begin{eqnarray} -\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}+ \int_{I_{\pi-n\frac{2\pi}{2n-1}}}f(z)dz&=& -\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}+P.V. \int_{I_{\pi-n\frac{2\pi}{2n-1}}}f(z)dz\\ &=&-\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}+e^{-i\frac{\pi}{2n-1}}P.V. \int_{0}^{\infty}\frac{1}{\prod_{k=1}^{\infty}(1-\frac{z^{2n-1}}{k^{2n-1}})}dz\\ &=&i\pi\sum_{m=1}^{\infty}\lim_{z\rightarrow-\frac{m}{\omega^{n}}}(z+\frac{m}{\omega^{n}})z^{2n-1}\Gamma(z)\Gamma(\omega z)\cdots \Gamma(\omega^{2n-2} z)\\ &=&-i\pi\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n-1}}{\omega^{n}m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-2} m) \end{eqnarray}
\begin{eqnarray} (-e^{i\frac{\pi}{2n-1}}+e^{-i\frac{\pi}{2n-1}})\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}&=&-2i\sin{\frac{\pi}{2n-1}}\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}\\ &=&-i\pi(\frac{e^{i\frac{\pi}{2n-1}}}{\omega^{n}}+\frac{e^{-i\frac{\pi}{2n-1}}}{\omega^{n-1}})\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n-1}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-2} m)\\ &=&i2\pi\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n-1}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-2} m) \end{eqnarray}
\begin{equation} \int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n-1}}{k^{2n-1}})}=\frac{\pi}{\sin{\frac{\pi}{2n-1}}}\sum_{m=1}^{\infty}(-1)^{m-1}\frac{m^{2n-1}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-2} m) \end{equation}
[2]$n\in2\mathbb{N}$
\begin{eqnarray} -\int_{I_{\pi-\frac{\pi}{2n}-(n-1)\frac{\pi}{n}}}f(z)dz+\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}&=&-e^{i\frac{\pi}{2n}}P.V.\int_{I_{\pi+\frac{\pi}{2n}+(n-1)\frac{\pi}{n}}}f(z)dz+\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}\\ &=&-i\pi\sum_{m=1}^{\infty}\lim_{z\rightarrow-\frac{m}{\omega^{n-1}e^{i\frac{\pi}{2n}}}}(z+\frac{m}{\omega^{n-1}e^{i\frac{\pi}{2n}}})z^{2n}\Gamma(\omega e^{i\frac{\pi}{2n}}z)\Gamma(\omega^{2} e^{i\frac{\pi}{2n}}z)\cdots \Gamma(\omega^{2n-1} e^{i\frac{\pi}{2n}}z)\\ &=&-i\pi\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n}}{\omega^{n-1}e^{i\frac{\pi}{2n}}m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-1} m) \end{eqnarray}
\begin{eqnarray} -\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}+\int_{I_{\pi-\frac{\pi}{2n}-n\frac{\pi}{n}}}f(z)dz&=&-\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}+e^{-i\frac{\pi}{2n}}P.V.\int_{I_{\pi+\frac{\pi}{2n}+n\frac{\pi}{n}}}f(z)dz\\ &=&i\pi\sum_{z\rightarrow-\frac{m}{\omega^{n}e^{i\frac{\pi}{2n}}}}(z+\frac{m}{\omega^{n}e^{i\frac{\pi}{2n}}})z^{2n}\Gamma(\omega e^{i\frac{\pi}{2n}}z)\Gamma(\omega^{2} e^{i\frac{\pi}{2n}}z)\cdots \Gamma(\omega^{2n-1} e^{i\frac{\pi}{2n}}z)\\ &=&-i\pi\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n}}{\omega^{n}e^{i\frac{\pi}{2n}}m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-1} m) \end{eqnarray}
\begin{eqnarray} (-e^{i\frac{\pi}{2n}}+e^{-i\frac{\pi}{2n}})\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}&=&-2i\sin{\frac{\pi}{2n}}\int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}\\ &=&-i\pi(\frac{1}{\omega^{n}}+\frac{e^{-i\frac{\pi}{n}}}{\omega^{n-1}})\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-1} m)\\ &=&2i\pi\sum_{m=1}^{\infty}(-1)^{m}\frac{m^{2n}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-1} m) \end{eqnarray}
\begin{equation} \int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{2n}}{k^{2n}})}=\frac{\pi}{\sin{\frac{\pi}{2n}}}\sum_{m=1}^{\infty}(-1)^{m-1}\frac{m^{2n}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{2n-1} m) \end{equation}
[3]上記計算をまとめると
\begin{equation} \int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{n}}{k^{n}})}=\frac{\pi}{\sin{\frac{\pi}{n}}}\sum_{m=1}^{\infty}(-1)^{m-1}\frac{m^{n}}{m!}\Gamma(-\omega m)\Gamma(-\omega^{2} m)\cdots\Gamma(-\omega^{n-1} m) \end{equation}

[$\Rightarrow$]以下の式が成り立つと仮定する。
\begin{eqnarray} \left\{ \begin{array}{l} f^{(k}(\lambda)=0\quad(k=0,1,2,...,m-1)\\ f^{(m)}(\lambda)\neq 0 \end{array} \right. \end{eqnarray}
すると仮定より$f(\lambda)=0$なので$\exists g_{0}(x)\in\mathbb{C}[x]\ s.t.\ f(x)=(x-\lambda)g_{0}(x)$が成り立つ。
両辺$x$で微分すると$f^{(1)}(x)=g_{0}(x)+(x-\lambda)g_{0}^{'}(x)$なので$f^{(1)}(\lambda)=g_{0}(\lambda)=0$なので先と同様$\exists g_{1}(x)\in\mathbb{C}[x]\ s.t.\ g_{0}(x)=(x-\lambda)g_{1}(x)$となる事が分かる。
これを繰り返す事で、$\exists g_{m}(x)\in\mathbb{C}[x]\ s.t.\ f(x)=(x-\lambda)^{m}g_{m}(x)$が成り立つ事が分かる。
さらに以下の様に計算できるので$g_{m}(\lambda)\neq 0$である事が分かる。
\begin{equation} f^{(m)}(x)=m!g_{m}(x)+(x-\lambda)\sum_{k=1}^{m}\begin{pmatrix}m\\k\end{pmatrix}\frac{m!}{k!}(x-\lambda)^{k-1}g_{m}^{(k)}(x) \end{equation}
ゆえに$x=\lambda$は$m$重根。
[$\Leftarrow$]$x=\lambda$が$m$重根とすると$\exists g(x)\in\mathbb{C}[x](g(\lambda)\neq 0)\ s.t.\ f(x)=(x-\lambda)^{m}g(x)$が成り立つ。
両辺微分すると
\begin{equation} f^{(k)}(x)=\sum_{n=0}^{k}\begin{pmatrix}k\\n\end{pmatrix}\frac{m!}{n!}(x-\lambda)^{m-k+n}g^{(n)}(x) \end{equation}
より
\begin{eqnarray} \left\{ \begin{array}{l} f^{(k)}(\lambda)=0\quad(k=0,1,2,...,m-1)\\ f^{(m)}(\lambda)=m!g(\lambda)\neq0 \end{array} \right. \end{eqnarray}

[1]$\sin{n\pi}=0\quad(n\in\mathbb{Z})\land\cos{n\pi}=(-1)^{n}\quad(n\in\mathbb{Z})$なので$x=n\pi$が$\sin{x}$の一重根となる事が分かる。
\begin{eqnarray} \sin{x}&=&\lim_{N\rightarrow\infty}C_{N}x\prod_{\substack{-N\leq n\leq N\\n\neq 0}}(x-n\pi)\\ &=&\lim_{N\rightarrow\infty}(-1)^{N}(n\pi)^{2N}C_{N}x\prod_{\substack{-N\leq n\leq N\\n\neq 0}}(1-\frac{x}{n\pi}) \end{eqnarray}
[2]$\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1$より$C_{N}=\frac{(-1)^{N}}{(n\pi)^{2N}}$なので
\begin{equation} \sin{x}=x\prod_{n=1}^{\infty}(1-\frac{x^{2}}{n^{2}\pi^{2}}) \end{equation}

\begin{eqnarray} \Gamma(z)\Gamma(1-z)&=&\frac{1}{z}\prod_{k=1}^{\infty}\frac{1}{1-\frac{z^{2}}{k^{2}}}\\ &=&\frac{\pi}{\sin{\pi z}} \end{eqnarray}

[1]
\begin{eqnarray} \Gamma(-\omega m)\Gamma(-\omega^{2} m)&=&\Gamma(-\omega m)\Gamma((1+\omega) m)\\ &=&\Gamma(-\omega m)\Gamma(m+\omega m)\\ &=&(\omega m)_{m}\Gamma(-\omega m)\Gamma(\omega m)\\ &=&-(\omega m)_{m}\frac{\pi }{\omega m\sin{(\omega m\pi)}} \end{eqnarray}
[2]
\begin{eqnarray} \sin{(\omega m\pi)}&=&\sin{(\frac{1+i\sqrt{3}}{2}m \pi)}\\ &=&\sin{(\frac{1}{2}m\pi)}\cos{(i\frac{\sqrt{3}}{2}m\pi)}+\cos{(\frac{1}{2}m\pi)}\sin{(i\frac{\sqrt{3}}{2}m\pi)}\\ &=&\delta_{1,m(mod\ 2)}(-1)^{\frac{m-1}{2}}\cosh{(\frac{\sqrt{3}}{2} m\pi)}+i\delta_{0,m(mod\ 2)}(-1)^{\frac{m}{2}}\sinh{\frac{\sqrt{3}}{2}m\pi} \end{eqnarray}
[3]
\begin{eqnarray} \int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{3}}{k^{3}})}&=&\frac{2\pi}{\sqrt{3}}\sum_{m=1}^{\infty}(-1)^{m-1}\frac{m^{3}}{m!}\{-(\omega m)_{m}\frac{\pi}{\omega m\sin{(\omega m\pi)}}\}\\ &=&-\frac{2\pi^{2}}{\sqrt{3}}\sum_{m=1}^{\infty}(-1)^{m-1}\frac{m^{3}}{m!}\frac{(\omega m)_{m}}{\delta_{1,m(mod\ 2)}(-1)^{\frac{m-1}{2}}\cosh{(\frac{\sqrt{3}}{2} m\pi)}+i\delta_{0,m(mod\ 2)}(-1)^{\frac{m}{2}}\sinh{\frac{\sqrt{3}}{2}m\pi} } \end{eqnarray}

\begin{eqnarray} \Gamma(-im)\Gamma(-i^{2}m)\Gamma(-i^{3}m)&=&\Gamma(m)\Gamma(-im)\Gamma(im)\\ &=&\frac{1}{-im}\Gamma(m)\Gamma(im)\Gamma(1-im)\\ &=&\frac{(m-1)!\pi}{m\sinh{(\pi m)}} \end{eqnarray}
より
\begin{equation} \int_{0}^{\infty}\frac{dx}{\prod_{k=1}^{\infty}(1+\frac{x^{4}}{k^{4}})}=\pi^{2}\sqrt{2}\sum_{m=1}^{\infty}\frac{m^{2}}{\sinh{(\pi m)}} \end{equation}