Is this limitation correct? 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!
 A: 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
$$4 \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} = \frac{\pi^2}{3}$$
ADDENDUM
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.
