I was trying to compute the integral
$$\ I = \int_0^{\infty} \frac{x \sinh(x)}{(1+\cosh^2(x))^2} dx$$
So, I tried defining:
$$\ I(a) = \int_0^{\infty} \frac{ x \sinh(ax)}{(1+\cosh^2(x))^2}dx$$
Then
$$\ I'(a) = \int_0^{\infty} \frac{x\cosh(ax)}{(1+\cosh^2x)^2} dx$$
Then I substituted $\cosh(x) = t$, but I was stuck with this method and could not proceed any further.
So, I tried another approach.
I used the fact that we can express $\sinh(x)$ as:
$$ \sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$
Then swapping the integral and summation signs, we get:
$$\ I = \sum_{n=0}^{\infty}\frac{1}{(2n+1)!} \cdot \int_0^{\infty} \frac{x^{2n+2}}{(1+\cosh^2(x))^2} dx$$
But I am unable to proceed with the evaluation of the inner integral.
Can someone help me evaluate the value of this integral?
Any help is appreciated. Thank you for reading!
Best Answer
Here is a way using the Feynman technique:
$$I=\int_0^{\infty} \frac{x \sinh(x)}{(1+\cosh^2(x))^2} dx=\int_1^\infty\frac{\cosh^{-1}(x)}{(x^2+1)^2}dx$$
Define:
$$I(a)= \int_1^\infty\frac{\cosh^{-1}(ax)}{(x^2+1)^2}dx\implies I’(a)=\int_1^\infty\frac x{\sqrt{(ax)^2-1}(x^2+1)^2}dx$$
It can be shown that:
$$I’(a)=\frac{a^2\pi}{4(a^2+1)^\frac32}\implies I(a)=\frac\pi4\left(\sinh^{-1}(a)-\frac a{\sqrt{a^2+1}}\right)+C$$
Clearly $I(0)=0$ giving $C=0$. Therefore, our integral is $I(1)$:
$$\boxed{\int_0^{\infty} \frac{x \sinh(x)}{(1+\cosh^2(x))^2} dx =\frac\pi8\left(2\sinh^{-1}(1)-\sqrt2\right)}$$