[mathjax]
積分問題34番
計算
\(\displaystyle\int_{0}^{\infty}\displaystyle\frac{x}{e^x-1}dx=\displaystyle\int_{0}^{\infty}\displaystyle\frac{x}{e^x(1-\frac{1}{e^x})}dx\)
\(=\displaystyle\int_{0}^{\infty}xe^{-x}\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{1}{e^{nx}}dx\) ※\(\displaystyle\frac{1}{e^x}<1\)より無限級数に展開
\(=\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{0}^{\infty}xe^{-(n+1)x}dx\)
\(=\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{0}^{\infty}\displaystyle\frac{t}{n+1}\cdot e^{-t}\cdot \displaystyle\frac{dt}{n+1}\) ※ \(t=(n+1)x\)と置換
\(=\displaystyle\sum_{n=0}^{\infty} \displaystyle\frac{1}{(n+1)^2}\displaystyle\int_{0}^{\infty}te^{-t}dt\) ※変数と関係ないところは外に出せる
\(=\displaystyle\sum_{n=0}^{\infty} \displaystyle\frac{1}{(n+1)^2}\) ※\(\displaystyle\int_{0}^{\infty}te^{-t}dt=1\)
\(=\displaystyle\frac{\pi^2}{6}\) ※バーゼル問題
答え
\(\displaystyle\frac{\pi^2}{6}\)
