求积分：1／π∫(-∞,+∞) x^2／[e^(-x)+e^x]dx

xfhaoym

求解一个积分：

杨协成

引入一个特殊函数：双伽玛函数 \(\psi(x)\)，其定义为对数伽玛函数的导数，
\[\psi(x):={\frac{\mathrm{d}}{\mathrm{d}x}}\ln{\Gamma(x)}\]
双伽玛函数 \(\psi(x)\) 有以下积分表达和关系：
\[\psi(x)=\int_0^\infty\left({\frac{e^{-t}}{t}}-{\frac{e^{-xt}}{1-e^{-t}}}\right)\,\mathrm{d}t\]
\[\psi(1-x)-\psi(x)=\pi\cot{(\pi x)}\]

先计算一个中间步骤需要使用的积分：
\[\begin{aligned}
&\int_{-\infty}^\infty\frac{x^2}{\cosh{x}}\,\mathrm{d}x \\
=&4\int_0^\infty\frac{\sinh{x}}{\sinh{(2x)}}\,x^{2}\,\mathrm{d}x \\
\stackrel{x=t/4}{=}&16\int_0^\infty\frac{\sinh{(t/4)}}{\sinh{(t/2)}}\,t^{2}\,\mathrm{d}t \\
=&\frac{1}{4}\frac{\partial^2}{\partial\alpha^2}\left[\int_0^\infty\frac{\sinh{(\alpha t/2)}}{\sinh{(t/2)}}\,\mathrm{d}t\right]_{\alpha=1/2} \\
=&\frac{1}{4}\frac{\partial^2}{\partial\alpha^2}\left[\int_0^\infty\frac{e^{-(1-\alpha)t/2}-e^{-(1+\alpha)t/2}}{1-e^{-t}}\,\mathrm{d}t\right]_{\alpha=1/2} \\
=&\frac{1}{4}\frac{\partial^2}{\partial\alpha^2}\left[\left(\int_0^\infty\frac{e^{-t}}{t}-\frac{e^{-(1+\alpha)t/2}}{1-e^{-t}}\,\mathrm{d}t\right)-\left(\int_0^\infty\frac{e^{-t}}{t}-\frac{e^{-(1-\alpha)t/2}}{1-e^{-t}}\,\mathrm{d}t\right)\right]_{\alpha=1/2} \\
=&\frac{1}{4}\frac{\partial^2}{\partial\alpha^2}\left[\psi\left(\frac{1+\alpha}{2}\right)-\psi\left(\frac{1-\alpha}{2}\right)\right]_{\alpha=1/2} \\
=&\frac{1}{4}\frac{\partial^2}{\partial\alpha^2}\left[\pi\cot{\left(\pi\frac{1-\alpha}{2}\right)}\right]_{\alpha=1/2} \\
=&\frac{\pi}{4}\left[\frac{\pi^2}{4}\sec^2{\left(\frac{\pi\alpha}{2}\right)}{\tan\left(\frac{\pi\alpha}{2}\right)}\right]_{\alpha=1/2} \\
=&\frac{\pi^3}{8}\sec^2{\left(\frac{\pi}{4}\right)}{\tan\left(\frac{\pi}{4}\right)} \\
=&\frac{\pi^3}{4}
\end{aligned}\]

所以原积分为
\[\frac{1}{\pi}\int_{-\infty}^\infty\frac{x^2}{e^{-x}+e^x}\,\mathrm{d}x
=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{x^2}{\cosh{x}}\,\mathrm{d}x
=\frac{1}{2\pi}\cdot\frac{\pi^3}{4}
=\frac{\pi^2}{8}\]

xfhaoym

谢谢LSA的。有一位是这样做的：