積分解説26

まず、$\log \mathrm{\Gamma }(x)$のフーリエ級数展開を考えます。

${\displaystyle a_n=2{\int }_0^1\log \mathrm{\Gamma }(x)\cos 2n\pi xdx}$

${\displaystyle b_n=2{\int }_0^1\log \mathrm{\Gamma }(x)\sin 2n\pi xdx}$ とおくと、

${\displaystyle \log \mathrm{\Gamma }(x)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}(a_n\cos 2n\pi x+b_n\sin 2n\pi x)}$

が成立します。また、

$\begin{array}{rcl}& & a_0\\ & =& 2{\int }_0^1\log \mathrm{\Gamma }(x)dx\\ & =& {\int }_0^1\log (\mathrm{\Gamma }(x)\mathrm{\Gamma }(1-x))dx\\ & =& {\int }_0^1\log \frac{\pi }{\sin \pi x}dx\\ & =& \log 2\pi \end{array}$

より、$a_0=\log 2\pi$

さらに、

$\begin{array}{rcl}& & a_n\\ & =& 2{\int }_0^1\log \mathrm{\Gamma }(x)\cos 2n\pi xdx\\ & =& {\int }_0^1\log (\mathrm{\Gamma }(x)\mathrm{\Gamma }(1-x))\cos 2n\pi xdx\\ & =& {\int }_0^1\log \frac{\pi }{\sin \pi x}\cos 2n\pi xdx\\ & =& \log \pi {\int }_0^1\cos 2n\pi xdx-{\int }_0^1\log \sin \pi x\cos 2n\pi xdx\\ & =& -{\left[\frac{\log \sin \pi x\sin 2n\pi x}{2n\pi }\right]}_0^1+\frac{1}{2n}{\int }_0^1\frac{\sin 2n\pi x\cos \pi x}{\sin \pi x}dx\\ & =& \frac{1}{2n}{\int }_0^1\frac{\sin 2\pi x\cos \pi x}{\sin \pi x}dx\\ & =& \frac{1}{n}{\int }_0^1{\cos }^2\pi xdx\\ & =& \frac{1}{n}{[\frac{x}{2}+\frac{\sin 2\pi x}{4\pi }]}_0^1\\ & =& \frac{1}{2n}\end{array}$

より、${\displaystyle a_n=\frac{1}{2n}}$

また、ワイエルシュトラスの乗積表示より、

${\displaystyle \log \mathrm{\Gamma }(x)=-\gamma x-\log x+\sum _{k=1}^{\mathrm{\infty }}(\frac{x}{k}-\log (1+\frac{x}{k}))}$

であるから、

$\begin{array}{rcl}& & b_n\\ & =& 2{\int }_0^1(-\gamma x-\log x+\sum _{k=1}^{\mathrm{\infty }}(-\frac{x}{k}+\log (1+\frac{x}{k})))\sin 2n\pi xdx\\ & =& 2\gamma {\left[\frac{x\cos 2n\pi x}{2n\pi }\right]}_0^1+\frac{1}{n\pi }{[\cos 2n\pi x\log x]}_0^1-\frac{1}{n\pi }{\int }_0^1\frac{\cos 2n\pi x}{x}dx+2\sum _{k=1}^{\mathrm{\infty }}{\int }_0^1(\frac{1}{k}x\sin 2n\pi x-\log (1+\frac{x}{k})\sin 2n\pi x)dx\\ & =& \frac{\gamma }{n\pi }-\frac{1}{n\pi }\underset{t\to 0}{lim}\cos 2n\pi t\log t-\frac{1}{n\pi }{[\text{Ci}(2n\pi x)]}_0^1+\frac{1}{n\pi }\sum _{k=1}^{\mathrm{\infty }}(-\frac{1}{k}+\log (1+\frac{1}{k})-{[\text{Ci}(2n\pi (k+x))]}_0^1)\\ & =& \frac{\gamma -\text{Ci}(2n\pi )}{n\pi }-\frac{1}{n\pi }\underset{t\to 0}{lim}(\cos 2n\pi t\log t-\text{Ci}(2n\pi t))+\frac{1}{n\pi }\sum _{k=1}^{\mathrm{\infty }}(\log (1+\frac{1}{k})-\frac{1}{k}+\text{Ci}(2nk\pi )-\text{Ci}(2n(k+1)\pi ))\\ & =& \frac{\gamma -\text{Ci}(2n\pi )}{n\pi }-\frac{1}{n\pi }\underset{t\to 0}{lim}(\cos 2n\pi t\log t-\gamma -\log 2n\pi t-{\int }_0^{2n\pi t}\frac{1-\cos u}{u}du)+\frac{-\gamma +\text{Ci}(2n\pi )}{n\pi }\\ & =& \frac{\gamma -\text{Ci}(2n\pi )+\gamma +\log 2n\pi -\gamma +\text{Ci}(2n\pi )}{n\pi }\\ & =& \frac{\gamma +\log 2n\pi }{n\pi }\end{array}$

より、${\displaystyle b_n=\frac{\gamma +\log 2n\pi }{n\pi }}$

よって、

${\displaystyle \log \mathrm{\Gamma }(x)=\frac{1}{2}\log 2\pi +\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2n}\cos 2n\pi x+\frac{\gamma +\log 2n\pi }{n\pi }\sin 2n\pi x)}$

がわかりました。この式に${\displaystyle x=\frac{3}{4}}$を代入すると、

$\begin{array}{rcl}& & \log \mathrm{\Gamma }\left(\frac{3}{4}\right)\\ & =& \frac{1}{2}\log 2\pi +\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2n}\cos \frac{3}{2}n\pi +\frac{\gamma +\log 2n\pi }{n\pi }\sin \frac{3}{2}n\pi )\\ & =& \frac{1}{2}\log 2\pi +\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n}+\frac{\gamma +\log 2\pi }{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{2n+1}+\frac{1}{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}\log (2n+1)}{2n+1}\\ & =& \frac{1}{2}\log 2\pi -\frac{1}{4}\log 2-\frac{1}{4}\gamma -\frac{1}{4}\log 2\pi +\frac{1}{\pi }{\beta }^{\prime }(1)\\ & =& -\frac{1}{4}(\gamma -\log \pi )+\frac{1}{\pi }{\beta }^{\prime }(1)\end{array}$

よって、

${\displaystyle {\beta }^{\prime }(1)=\frac{\pi }{4}(\gamma -\log \pi )+\pi \log \mathrm{\Gamma }\left(\frac{3}{4}\right)}$

が証明されました。