一道积分题

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\[\begin{aligned} \int_{0}^{+\infty}\frac{xe^{-x}}{(1+e^{-x})^2}\mathrm{d}x & =\int_{0}^{+\infty}\frac{xe^{x}}{(1+e^{x})^2}\mathrm{d}x \cr & = \int_{1}^{+\infty}\frac{ln(u)}{(1+u)^2}\mathrm{d}u \cr & = -\int_{1}^{+\infty}ln(u) \mathrm{d}\frac{1}{(1+u)} \cr & = -\frac{ln(u)}{1+u}|_{1}^{+\infty}+\int_{1}^{+\infty}\frac{1}{u(1+u)}\mathrm{d}u \cr & = ln(\frac{u}{u+1})|_{1}^{+\infty} \cr & = ln(2) \end{aligned} \]

这道题还有别的解法，比如这张图（来自微信群

方法一： 柯西主值^[1]

\[\begin{aligned} \int_{0}^{+\infty} x d \frac{1}{1+e^{-x}}& = \left.\frac{x}{1+e^{-x}}\right|_{0} ^{+\infty}-\int_{0}^{+\infty} \frac{e^{x}}{1+e^{x}} \mathrm{d} x \cr & = \lim _{x \rightarrow+\infty}\left(\frac{x}{1+e^{-x}}-\ln \left(1+e^{x}\right)\right)+\ln 2 \cr & =\lim _{x \rightarrow+\infty}\left(\frac{x\left(e^{x}+1\right)-x}{e^{x}+1}-\ln \left(1+e^{x}\right)\right)+\ln 2 \cr & = \lim _{x \rightarrow+\infty}\left(\ln e^{x}-\ln \left(1+e^{x}\right)-\frac{x}{e^{x}+1}\right)+\ln 2 \cr & =\lim _{x \rightarrow+\infty}\left(\ln \frac{e^{x}}{1+e^{x}}-\frac{x}{e^{x}+1}\right)+\ln 2=\ln 2 \end{aligned} \]

方法三： 无穷级数

因为

\[\begin{aligned} \ln (1+x) & =\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{n};\quad (-1<x \leqslant 1 ) \cr \Gamma (n) & =(n-1)! \end{aligned} \]

\[\begin{aligned} \int_{0}^{+\infty} \frac{x e^{-x}}{\left(1+e^{-x}\right)^{2}} \mathrm{d} x &=\sum_{n=1}^{\infty}(-1)^{n+1} \int_{0}^{+\infty}x \cdot n e^{-n x} \mathrm{d}x;\quad|e^{-x}|<1, x \in(0,+\infty) \cr & = \sum_{n=1}^{\infty}(-1)^{n+1} \cdot \frac{1}{n} \cdot \Gamma(2) \cr & =\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cr & =\ln 2. \end{aligned} \]

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1. 柯西主值 ↩︎