x^2/(√e^x-1)[0,∞]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^{\infty} \frac{1}{\sqrt{e^x-1}}dx=π\\
&(2)  \displaystyle\int_0^{\infty} \frac{x}{\sqrt{e^x-1}}dx=2π \log 2\\
&(3)  \displaystyle\int_0^{\infty} \frac{x^2}{\sqrt{e^x-1}}dx=4π\left\{(\log 2)^2+\frac{π^2}{12}\right\}\\
&(4)  \displaystyle\int_0^{\infty} \frac{x^3}{\sqrt{e^x-1}}=2π\left\{4(\log 2)^3+π^2 \log 2+6ζ(3)\right\}
\end{alignat}










<証明>

次の定積分の結果を用います。[詳細はこちらです。(A)(B)]
\begin{alignat}{2}
&(A)  \displaystyle\int_0^{\frac{π}{2}}\left\{ \log (\cos x)\right\}^2dx=\frac{π^3}{24}+\frac{π}{2}( \log 2)^2\\
&(B)  \displaystyle\int_0^{\frac{π}{2}} \{\log (\cos x)\}^3 dx=-\frac{π}{2}(\log 2)^3-\frac{π^3}{8}\log 2-\frac{3}{4}πζ(3)\\
\end{alignat}



\((2)(3)(4)\) は次の順番で文字を置き換えます。
\((α)\) \(\sqrt{e^x-1}=t, x=\log (1+t^2)\) \(\displaystyle \left(dx=\frac{2t}{1+t^2}dt\right)\)
\((β)\) \(t=\tan r\) \(\displaystyle \left(dt=\frac{1}{\cos^2 r}dr\right)\)




\((1)\) \(e^x-1=t^2\) と置きます。\((e^xdx=2tdt)\)$$\displaystyle\int_0^{\infty} \frac{1}{\sqrt{e^x-1}}dx=\displaystyle\int_0^{\infty} \frac{1}{t} \cdot \frac{2t}{1+t^2}dt=2\displaystyle\int_0^{\infty} \frac{1}{1+t^2}dt=2 \cdot \frac{π}{2}=π$$








\begin{alignat}{2}
(2)  \displaystyle\int_0^{\infty} \frac{x}{\sqrt{e^x-1}}dx&=\displaystyle\int_0^{\infty} \frac{\log (1+t^2)}{t} \cdot \frac{2t}{1+t^2}dt=2 \displaystyle\int_0^{\infty} \frac{\log (1+t^2)}{1+t^2}dt\\
&=2 \displaystyle\int_0^{\frac{π}{2}}\frac{\log (1+\tan^2 r)}{1+\tan^2 r} \cdot \frac{1}{\cos^2 r}dr\\
&=2\displaystyle\int_0^{\frac{π}{2}} \log \frac{1}{\cos^2 r}dr=-4 \displaystyle\int_0^{\frac{π}{2}} \log (\cos r)dr\\
&=-4 \left(-\frac{π}{2}\log 2\right)=2π \log 2\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{x}{\sqrt{e^x-1}}dx=2π \log 2$$








\begin{alignat}{2}
(3)  \displaystyle\int_0^{\infty} \frac{x^2}{\sqrt{e^x-1}}dx&=\displaystyle\int_0^{\infty} \frac{\{\log (1+t^2)\}^2}{t} \cdot \frac{2t}{1+t^2}dt=2 \displaystyle\int_0^{\infty} \frac{\{\log (1+t^2)\}^2}{1+t^2}dt\\
&=2 \displaystyle\int_0^{\frac{π}{2}}\frac{\{\log (1+\tan^2 r)\}^2}{1+\tan^2 r} \cdot \frac{1}{\cos^2 r}dr\\
&=2\displaystyle\int_0^{\frac{π}{2}} \left(\log \frac{1}{\cos^2 r}\right)^2dr=2 \displaystyle\int_0^{\frac{π}{2}}\{-2 \log (\cos r)\}^2dr\\
&=8 \displaystyle\int_0^{\frac{π}{2}}\{\log (\cos r)\}^2dr=8\left\{\frac{π^3}{24}+\frac{π}{2}( \log 2)^2\right\}=4π\left\{(\log 2)^2+\frac{π^2}{12}\right\}
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{x^2}{\sqrt{e^x-1}}dx=4π\left\{(\log 2)^2+\frac{π^2}{12}\right\}$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^{\infty} \frac{x^3}{\sqrt{e^x-1}}dx&=\displaystyle\int_0^{\infty} \frac{\{\log (1+t^2)\}^3}{t} \cdot \frac{2t}{1+t^2}dt=2 \displaystyle\int_0^{\infty} \frac{\{\log (1+t^2)\}^3}{1+t^2}dt\\
&=2 \displaystyle\int_0^{\frac{π}{2}}\frac{\{\log (1+\tan^2 r)\}^3}{1+\tan^2 r} \cdot \frac{1}{\cos^2 r}dr\\
&=2\displaystyle\int_0^{\frac{π}{2}} \left(\log \frac{1}{\cos^2 r}\right)^3dr=2 \displaystyle\int_0^{\frac{π}{2}}\{-2 \log (\cos r)\}^3dr\\
&=-16 \displaystyle\int_0^{\frac{π}{2}}\{\log (\cos r)\}^3dr\\
&=-16\left\{-\frac{π}{2}(\log 2)^3-\frac{π^3}{8}\log 2-\frac{3}{4}πζ(3)\right\}\\
&=2π\left\{4(\log 2)^3+π^2 \log 2+6ζ(3)\right\}\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{x^3}{\sqrt{e^x-1}}=2π\left\{4(\log 2)^3+π^2 \log 2+6ζ(3)\right\}$$



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