積分メモ①
この記事は、この積分の導出過程をメモしたものです。
$\displaystyle\int_0^1\int_0^1\left(\frac{\ln\dfrac{1}{x}-\ln\dfrac{1}{y}}{\ln\ln\dfrac{1}{x}-\ln\ln\dfrac{1}{y}}\right)^2\,dxdy=\frac{31\zeta(5)}{15\zeta(4)}-\frac{7\zeta(3)}{6\zeta(2)}$
\begin{align}
\displaystyle\int_0^1\int_0^1\left(\frac{\ln\dfrac{1}{x}-\ln\dfrac{1}{y}}{\ln\ln\dfrac{1}{x}-\ln\ln\dfrac{1}{y}}\right)^2\,dxdy&=\int_0^1\int_0^1\int_0^1\int_0^1\ln^{t+u}\dfrac{1}{x}\ln^{2-t-u}\dfrac{1}{y}\,dtdudxdy
\\&=\int_0^1\int_0^1\int_0^1\int_0^1\ln^{t+u}\dfrac{1}{x}\ln^{2-t-u}\dfrac{1}{y}\,dxdydtdu
\\&=\int_0^1\int_0^1\int_0^1s^{2-t-u}e^{-s}\left(\int_0^1v^{t+u}e^{-v}\,dv\right)\,dsdtdu
\\&=\int_0^1\int_0^1\Gamma(3-t-u)\Gamma(1+t+u)\,dtdu
\\&\overset{\text{u+t=z}}{=}\int_0^1\int_u^{u+1}\Gamma(3-z)\Gamma(1+z)\,dzdu
\end{align}
部分積分によって
\begin{align}
\displaystyle\left.\int_0^1u'\int_0^1\Gamma(3-z)\Gamma(1+z)\,dzdu=u\int_u^{u+1}\Gamma(3-z)\Gamma(1+z)\,dz\right|^{u=1}_{u=0}-\int_0^1 \Gamma(2-u)\Gamma(2+u)du+\int_0^1u\Gamma(3-u)\Gamma(1+u)du
\end{align}
より,もとの計算は次の3つの計算の和であることが分かり,その値は各々以下の通りである.
(1)$\displaystyle\int_1^2\Gamma(3-z)\Gamma(1+z)dz=\frac{7\zeta(3)}{4\zeta(2)}$
(2)$\displaystyle\int_0^1u\Gamma(2-u)\Gamma(2+u)du=\frac{7\zeta(3)}{3\zeta(2)}-\frac{31\zeta(5)}{30\zeta(4)}$
(3)$\displaystyle\int_0^1u\Gamma(3-u)\Gamma(1+u)du=\frac{31\zeta(5)}{30\zeta(4)}-\frac{7\zeta(3)}{12\zeta(2)}$
1つずつ示す.
\begin{align} \displaystyle\int_1^2\Gamma(3-z)\Gamma(1+z)dz&\overset{\text{z-1=s}}{=}\int_0^1\Gamma(2-s)\Gamma(2+s)ds \\&=\int_0^1s(1-s^2)\Gamma(1-s)\Gamma(s)ds\,\,\,\,\,\,\,\,\,※\,\Gamma(x+1)=x\Gamma(x) \\&=\frac{1}{\pi}\int_0^\pi\frac{s(1-s^2)}{\sin\pi s}ds\,\,\,\,\,\,\,\,\,\,\,※\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x} \\&\overset{s\pi=w}{=}\int_0^\pi\frac{\dfrac{w}{\pi}\left(1-\dfrac{w^2}{\pi^2}\right)}{\sin w}dw \\&=\frac{1}{\pi}\int_0^\pi\left(\ln\tan\frac{w}{2}\right)'w\left(1-\frac{w^2}{\pi^2}\right)dw \\&=\left.\frac{1}{\pi}\ln\tan\frac{w}{2}\cdot w\left(\ln\tan\frac{w}{2}\right)'w\left(1-\frac{w^2}{\pi^2}\right)\right|^{w=\pi}_{w=0}-\frac{1}{\pi}\int_0^\pi\ln\tan\frac{w}{2}dw+\frac{3}{\pi^3}\int_0^\pi w^2\ln\tan\frac{w}{2}dw \\&=\frac{3}{\pi^3}\int_0^\pi w^2\ln\tan\frac{w}{2}dw \\&\overset{{\frac{w}{2}=r}}{=}\frac{24}{\pi^3}\int_0^\frac{\pi}{2}r^2\ln\tan r dr \\&=-\frac{48}{\pi^3}\int_0^\frac{\pi}{2}\sum_{k=1}^\infty\frac{r^2\cos2(2k-1)r}{2k-1}dr \\&=\frac{6}{\pi^3}\sum_{k=1}^\infty\frac{(\pi^2(2k-1)^2-2)\sin2\pi k}{(2k-1)^4}+\frac{12}{\pi^2}\sum_{k=1}^\infty\frac{\cos2\pi k}{(2k-1)^3} \\&=\frac{7\zeta(3)}{4\zeta(2)} \end{align}
\begin{align} \displaystyle\int_0^1u\Gamma(2-u)\Gamma(2+u)du&=\int_0^1u^2(1-u^2)\Gamma(1-u)\Gamma(u)du \\&=\pi\int_0^1\frac{u^2(1-u^2)}{\sin\pi u}du \\&\overset{{\pi u=s}}{=}\int_0^\pi\frac{\dfrac{s^2}{\pi^2}\left(1-\dfrac{s^2}{\pi^2}\right)}{\sin s}ds \\&=\int_0^\pi\left(\ln\tan\frac{s}{2}\right)'\cdot\frac{s^2}{\pi^2}\left(1-\frac{s^2}{\pi^2}\right)ds\ \\&=\left.\ln\tan\frac{s}{2}\cdot\frac{s^2}{\pi^2}\left(1-\frac{s^2}{\pi^2}\right)\right|^{s=\pi}_{s=0}-\frac{2}{\pi^4}\int_0^\pi\ln\tan\frac{s}{2}\cdot s(\pi^2-2s^2)ds \\&=-\frac{2}{\pi^2}\int_0^\pi s\ln\tan\frac{s}{2}ds+\frac{4}{\pi^4}\int_0^\pi s^3\ln\tan\frac{s}{2}ds \\&=-\frac{7\zeta(3)}{6\zeta(2)}+\frac{4}{\pi^4}\int_0^\pi s^3\ln\tan\frac{s}{2}ds \\&\overset{{\frac{s}{2}=t}}{=}-\frac{7\zeta(3)}{6\zeta(2)}+\frac{64}{\pi^4}\int_0^\frac{\pi}{2} t^3\ln\tan t\,dt \\&=-\frac{7\zeta(3)}{6\zeta(2)}-\frac{128}{\pi^4}\int_0^\frac{\pi}{2}\sum_{k=1}^\infty\frac{t^3\cos2(2k-1)t}{2k-1}dt \\&=-\frac{7\zeta(3)}{6\zeta(2)}-\frac{128}{\pi^4}\sum_{k=1}^\infty\frac{1}{2k-1}\int_0^\frac{\pi}{2}t^3\cos2(2k-1)t\,dt \\&=-\frac{7\zeta(3)}{6\zeta(2)}+\frac{8}{\pi^3}\sum_{k=1}^\infty\frac{(2k-1)^2\pi^2-6}{(2k-1)^4}\sin 2k\pi-\frac{48}{\pi^4}\sum_{k=1}^\infty\frac{1}{(2k-1)^5}-\frac{48}{\pi^4}\sum_{k=1}^\infty\frac{\cos2k\pi}{(2k-1)^5}+\frac{48}{\pi^4}\sum_{k=1}^\infty\frac{\cos2k\pi}{(2k-1)^3} \\&=-\frac{7\zeta(3)}{6\zeta(2)}-\frac{96}{\pi^4}\sum_{k=1}^\infty\frac{1}{(2k-1)^5}+\frac{24}{\pi^2}\sum_{k=1}^\infty\frac{1}{(2k-1)^3}\\ &=\frac{7\zeta(3)}{3\zeta(2)}-\frac{31\zeta(5)}{30\zeta(4)} \end{align}
\begin{align} \displaystyle\int_0^1u\Gamma(3-u)\Gamma(1+u)du&=\int_0^1(1-t)\Gamma(2+t)\Gamma(2-t)dt \\&\overset{{1-u=t}}{=}\int_0^1\Gamma(2+t)\Gamma(2-t)dt-\int_0^1t\Gamma(2+t)\Gamma(2-t)dt \\&=\frac{31\zeta(5)}{30\zeta(4)}-\frac{7\zeta(3)}{12\zeta(2)} \end{align}
よって,
$\displaystyle\int_0^1\int_0^1\left(\frac{\ln\dfrac{1}{x}-\ln\dfrac{1}{y}}{\ln\ln\dfrac{1}{x}-\ln\ln\dfrac{1}{y}}\right)^2\,dxdy=\frac{31\zeta(5)}{15\zeta(4)}-\frac{7\zeta(3)}{6\zeta(2)}$
が示された.