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!

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.
