個人的に大変だった積分1

この記事を書いている人がまだ一様収束とかをあまり理解していないため、シグマとインテグラルを無言で入れ替えます。ごめんなさい。(一応、$\overset{!}{=}$で知らせています。)
\begin{align*} \int_{0}^{\infty}\frac{x^{s-1}e^{-x}}{1+e^{-2x}}dx&=\int_{0}^{\infty}x^{s-1}e^{-x}\sum_{k=0}^{\infty}\left(-e^{-2x}\right)^kdx\\ &\overset{!}{=}\sum_{k=0}^{\infty}(-1)^k\int_{0}^{\infty}x^{s-1}e^{-x(2k+1)}dx\\ &=\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{\infty}u^{s-1}e^{-u}\frac{1}{(2k+1)^s}du\\ &=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^s}\int_{0}^{\infty}u^{s-1}e^{-u}du\\ &=\Gamma(s)\beta(s)\\ \therefore \beta(s)&=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-x}}{1+e^{-2x}}dx \end{align*}

\begin{align*} J_1&=\int_{0}^{\frac{\pi}{2}}\ln(\sin{x})dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln(\cos{x})dx\quad \because \mathrm{king\:property}\\ 2J_1&=\int_{0}^{\frac{\pi}{2}}\ln(\sin{x})dx+\int_{0}^{\frac{\pi}{2}}\ln(\cos{x})dx\\ J_1&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln(\sin{x}\cos{x})dx\\ &=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln\left(\frac{1}{2}\sin{2x}\right)dx\\ &=-\frac{1}{2}\ln{2}\int_{0}^{\frac{\pi}{2}}dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln(\sin{2x})dx\\ &=-\frac{\pi}{4}\ln{2}+\frac{1}{4}\int_{0}^{\pi}\ln(\sin{t})dt\quad\because 2x=t\\ &=-\frac{\pi}{4}\ln{2}+\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\ln(\sin{t})dt+\frac{1}{4}\int_{\frac{\pi}{2}}^{\pi}\ln(\sin{t})dt\\ &=-\frac{\pi}{4}\ln{2}+\frac{1}{4}J_1+\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\ln(\cos{u})du\quad\because t-\frac{\pi}{2}=u\\ &=-\frac{\pi}{4}\ln{2}+\frac{1}{2}J_1\\ \therefore J_1&=-\frac{\pi}{2}\ln{2} \end{align*}

\begin{align*} J_2&=\int_{0}^{\frac{\pi}{2}}\ln(\tan{x})dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln(\sin{x})dx-\int_{0}^{\frac{\pi}{2}}\ln(\cos{x})dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln(\sin{x})dx-\int_{0}^{\frac{\pi}{2}}\ln(\sin{x})dx\quad\because \mathrm{king\:property}\\ &=0 \end{align*}

\begin{align*} I&=\int_{0}^{\frac{\pi}{4}}x\tan{x}dx\\ &=\left[-x\ln(\cos{x})\right]_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}}\ln(\cos{x})dx\\ &=\frac{\pi}{8}\ln{2}+I'\\ I'&=\int_{0}^{\frac{\pi}{4}}\ln(\cos{x})dx\\ &=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln\left(\cos{\frac{t}{2}}\right)dt\quad\because 2x=t\\ &=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\ln\left(\frac{\cos{t}+1}{2}\right)dt\\ &=-\frac{1}{4}\ln{2}\int_{0}^{\frac{\pi}{2}}dt+\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\ln(\cos{t}+1)dt\\ &=-\frac{\pi}{8}\ln{2}+\frac{1}{4}\int_{0}^{1}\ln\left(\frac{1-y^2}{1+y^2}+1\right)\frac{2}{1+y^2}dy\quad \because y=\tan\frac{t}{2}\\ &=-\frac{\pi}{8}\ln{2}+\frac{1}{2}\int_{0}^{1}\ln\left(\frac{2}{1+y^2}\right)\frac{1}{1+y^2}dy\\ &=-\frac{\pi}{8}\ln{2}+\frac{1}{2}\ln{2}\int_{0}^{1}\frac{1}{1+y^2}dy-\frac{1}{2}\int_{0}^{1}\frac{1}{1+y^2}\ln(1+y^2)dy\\ &=-\frac{1}{2}\int_{1}^{\infty}\frac{1}{1+u^2}\ln\left(\frac{1+u^2}{u^2}\right)du\quad\because y=\frac{1}{u}\\ &=-\frac{1}{2}\int_{1}^{\infty}\frac{1}{1+u^2}\ln(1+u^2)du+\int_{1}^{\infty}\frac{1}{1+u^2}\ln{u}du \end{align*}
すこし紛らわしくなるので、積分を分けます。
\begin{align*} I_1&=\int_{1}^{\infty}\frac{1}{1+u^2}\ln(1+u^2)du\\ I_2&=\int_{1}^{\infty}\frac{1}{1+u^2}\ln{u}du \end{align*}
つまり、
\begin{align*} I'&=-\frac{1}{2}I_{1}+I_{2} \end{align*}
では$I_1$から計算しましょう。
\begin{align*} I_1&=\int_{1}^{\infty}\frac{1}{1+x^2}\ln(1+x^2)dx\\ &=\int_{0}^{\infty}\frac{1}{1+x^2}\ln(1+x^2)dx-\int_{0}^{1}\frac{1}{1+x^2}\ln(1+x^2)dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln\left(\frac{1}{\cos^2{t}}\right)dt+2\int_{0}^{\frac{\pi}{4}}\ln(\cos{t})dt\quad \because x=\tan{t}\\ &=-2\int_{0}^{\frac{\pi}{2}}\ln(\cos{t})dt+2I'\\ &=-2J_1+2I'\\ &=\pi\ln{2}+2I'\\ 　\\ I_2&=\int_{1}^{\infty}\frac{1}{1+x^2}\ln{x}dx\\ &=\int_{0}^{\infty}\frac{1}{1+x^2}\ln{x}dx-\int_{0}^{1}\frac{1}{1+x^2}\ln{x}dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln(\tan{u})du+\int_{0}^{\infty}\frac{ue^{-u}}{1+e^{-2u}}du\quad\because x=\tan{u}　と　\ln{x}=-u\\ &=J_2+\int_{0}^{\infty}\frac{u^{2-1}e^{-u}}{1+e^{-2u}}du\\ &=\Gamma(2)\beta(2)\\ &=\beta(2) \end{align*}
以上より、(必死)
\begin{align*} I'&=-\frac{1}{2}I_1+I_2\\ &=-\frac{\pi}{2}\ln{2}-I'+\beta(2)\\ I'&=-\frac{\pi}{4}\ln{2}+\frac{1}{2}\beta(2) \end{align*}
よって(大声)
\begin{align*} I&=\frac{\pi}{8}\ln{2}+I'\\ &=\frac{1}{2}\beta(2)-\frac{\pi}{8}\ln{2} \end{align*}