Prove that $ \int\limits_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $

Question

Is this conjecture true?

Conjecture: $$ \int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $$

I found it myself based on numerical evidence. Need help in analytical proof. Thanks.

Answer 1

It is easier than I exspected in the first place. However, first of all lets denote your integral as

$$\mathfrak I~=~\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x$$

We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}\frac{1-x}{1+x}$ thus write $\mathfrak I$ as

$$\begin{align*} \mathfrak I=\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x&=\int_0^\infty \frac1{1+\left(e^{2x}\frac{1-x}{1+x}\right)^2}\frac{x^2e^{2x}\mathrm d x}{(1+x)^2} \end{align*}$$

Now we can try the straightforward substitution $z=e^{2x}\frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to

$$\begin{align*} \mathfrak I &=\int_0^\infty \frac1{1+\left(e^{2x}\frac{1-x}{1+x}\right)^2}\frac{x^2e^{2x}\mathrm d x}{(1+x)^2}\\ &=\int_1^{-\infty} \frac1{1+z^2}\frac{-\mathrm dz}{2}\\ &=\frac12\int_{-\infty}^1 \frac{\mathrm d z}{1+z^2}\\ &=\frac12\left[\arctan{z}\right]_{-\infty}^1 \end{align*}$$

$$\therefore~\mathfrak I ~=~\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x~=~\frac{3\pi}8$$

Where we used the symmetry aswell as well-known values of the tangent function.

Answer 2

The integral is $$\int_0^x\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,\mathrm dx=\frac{1}{8} \left(\pi +4 \tan ^{-1}\left(\frac{(x-1) e^{2x}}{x+1}\right)\right)$$ as can be verified by direct differentiation.

You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.

Answer 3

Well, we can just rewrite the integral as: $$\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\int_0^\infty \frac{x^2}{e^{2x}\left(\frac{1-x}{1+x}\right)^2+e^{-2x}}\frac{1}{(1+x)^2}dx$$ $$=\int_0^\infty \frac{1}{e^{4x}\left(\frac{1-x}{1+x}\right)^2+1}\frac{1}{e^{-2x}}\frac{x^2}{(1+x)^2}dx =\int_0^\infty \frac{\color{blue}{e^{2x}}}{\left(\color{red}{e^{2x}\cdot\frac{1-x}{1+x}}\right)^2+1}\color{blue}{\frac{x^2}{(1+x)^2}dx}$$ Now set $\,\displaystyle{\color{red}{e^{2x} \frac{1-x}{1+x}}=t\Rightarrow \color{blue}{e^{2x}\frac{x^2}{(1+x)^2}dx}=-\frac12 dt}\,$ and the result follows.