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Stimulated by some physics backgrounds, I want to know whether the following limitation is correct or not?

$$ \large\lim_{\beta\rightarrow+\infty}\frac{\int_{-\infty}^{\infty}f(x)\frac{(\beta x)^2}{(1+e^{-\beta x})(1+e^{\beta x})}dx}{f(0)\int_{-\infty}^{\infty}\frac{(\beta x)^2}{(1+e^{-\beta x})(1+e^{\beta x})}dx}=1 $$

where $\beta>0$ is a real parameter and the real function $f(x)$ ( $\geqslant 0$ over $\mathbb{R}$) is an analytical function of $\mathbb{R}$ and $0<f(0)<\infty$. Here we assume that the integral $$ \large\int_{-\infty}^{\infty}f(x)\frac{(\beta x)^2}{(1+e^{-\beta x})(1+e^{\beta x})}dx $$ is well defined and finite.

1.If the above limitation is true, how to prove it ?

2.How to calculate the integral $$ \large\int_{-\infty}^{\infty}\frac{x^2}{(1+e^{- x})(1+e^{x})}dx $$ which is relevant to this problem?

Thank you very much!

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I'm just going to answer your second question. You may manipulate the denominator to find that the integral is equal to

$$\frac14 \int_{-\infty}^{\infty} dx \frac{x^2}{\cosh^2{(x/2)}} = 2 \int_0^{\infty} dx \frac{x^2}{\left ( e^{x/2}+e^{-x/2}\right)^2} = $$

We may Taylor expand the denominator to get

$$2 \int_0^{\infty} dx \, x^2 \, e^{-x} \left ( 1+e^{-x}\right)^{-2} = 2 \sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^{\infty} dx \, x^2 \, e^{-(k+1) x}$$

The integral is recognized as related to a factorial and the integral is therefore equal to

$$4 \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} = \frac{\pi^2}{3}$$


I will show that

$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} = \frac{\pi^2}{12}$$

Write the sum as

$$\begin{align}1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots &= 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots - 2 \left (\frac{1}{2^2} + \frac{1}{4^2}+\cdots \right ) \end{align}$$

This (rearranging terms in the sum) is justified because the sum is absolutely convergent. Now, we may write this as

$$ \sum_{k=0}^{\infty} \frac{1}{(k+1)^2} - \frac{2}{2^2} \sum_{k=0}^{\infty} \frac{1}{(k+1)^2} = \frac12 \sum_{k=0}^{\infty} \frac{1}{(k+1)^2} $$

I assume that you are familiar with the latter sum as being equal to $\zeta(2) = \pi^2/6$, and we have what was to be shown.

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@ Ron Gordon, thank you very much – Kai Li Aug 2 '13 at 17:22
@ Ron Gordon, I am sorry that I also don't to know how to evaluate the series in your last equation? Thanks. – Kai Li Aug 3 '13 at 12:24
@K-boy Multiply this series by $(-1)^2$ to get $4 \sum_{k=0}^{\infty} \frac{(-1)^{k+2}}{(k+1)^2} $ then let $n=k+1$ the series becomes $4 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} $. Now find the even parts and odd parts of $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$. – user10444 Aug 3 '13 at 13:13
@ Ron Gordon, Thank you both. – Kai Li Aug 3 '13 at 15:16
@ user10444, Thank you both. – Kai Li Aug 3 '13 at 15:17

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