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I found the integral in the Fermi gas theory. There is an approximate formula for specific integrals: $$\int_{-\infty}^{+\infty}F(\epsilon)\frac{\partial f(\epsilon)}{\partial \epsilon}d\epsilon\approx-F(\mu)-\frac{\pi^2T^2}{6}F''(\mu)$$ where $f(\epsilon)=\left(\exp\left( \frac{\epsilon-\mu}{T} \right)+1\right)^{-1}$ is the Fermi-Dirac distribution.

The formula can be obtained after substitution of $F(\epsilon)$ in form of Taylor series: $$F(\epsilon)\approx F(\mu)+F'(\mu)(\epsilon -\mu)+\frac{F''(\mu)}{2}(\epsilon -\mu)^2$$ Because $\frac{\partial f(\epsilon)}{\partial \epsilon}=-\left(4T\cosh^2\left( \frac{\epsilon-\mu}{2T} \right)\right)^{-1}$, it can be proved that: $$\int_{-\infty}^{+\infty}F(\mu)\frac{\partial f(\epsilon)}{\partial \epsilon}d\epsilon=-F(\mu)$$ $$\int_{-\infty}^{+\infty}F'(\mu)(\epsilon -\mu)\frac{\partial f(\epsilon)}{\partial \epsilon}d\epsilon=0$$ But I have a problem with the last integral: $$\int_{-\infty}^{+\infty}\frac{F''(\mu)}{2}(\epsilon -\mu)^2\frac{\partial f(\epsilon)}{\partial \epsilon}d\epsilon=-F''(\mu)T^2\int_{-\infty}^{+\infty} \frac{x^2}{\cosh(x)^2} dx$$

Olexot
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1 Answers1

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$$\int_{-\infty}^\infty\frac{x^2}{\cosh^2 x}dx\\ =2\int_0^\infty\frac{x^2}{\cosh^2 x}dx\\ =2\int_0^\infty2x\frac{2}{1+e^{2x}}dx\text{ (I.B.P.)}\\ =2\int_0^\infty\frac{x}{1+e^{x}}dx\text{ (Sub $2x\mapsto x$)}\\ =2\int_0^\infty\frac{xe^{-x}}{1+e^{-x}}dx\\ =2\int_0^\infty x\sum_{n=1}^\infty (-1)^{n+1}e^{-nx}dx\\ =2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} =2\eta(2)=\frac{\pi^2}6$$

Kemono Chen
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    Nice. And if we don't know the eta function we can show that the sum is $ (1−\frac{1}{2})\zeta(2)$ by writing out even/odd terms. – Migos Dec 15 '18 at 00:56
  • Impressive, thank you very much! I wouldn't figure out how to reduce it to the Basel problem. Could you please explain why you choosed $(1-\tanh(x))$ instead of $\tanh(x)$ in integrating by parts? Is it just for zeroing at infinity or did you see the series in advance? – Olexot Dec 15 '18 at 01:58