convergence of $\int_0^{\infty}\frac{2x}{e^x-e^{-x}}dx $ I need help checking convergence of the integral:
$$\int_0^{\infty}\frac{2x}{e^x-e^{-x}}dx $$
I tried to change the integral to $\int_0^{\infty}\frac{x}{\sinh(x)}dx $ or comparing to other integrals but got stuck.
any suggestions?
 A: A convergence check is much easier than evaluation, since for small $x$ the integrand is $O(1)$ and for large $x$ it's $O(xe^{-x})$, giving respective antiderivative behaviours $O(x)$ and $O(x^2e^{-x})$. To evaluate, rewrite as$$\int_0^\infty\frac{2xe^{-x}dx}{1-e^{-2x}}=\sum_{k=0}^\infty\int_0^\infty 2xe^{-(2k+1)x}dx=\sum_k\frac{2}{(2k+1)^2}=\frac{\pi^2}{4}$$(the last $=$ is equivalent to the Basel problem).
A: As regards the convergence, you may note that
$$\lim_{x\to+\infty}\frac{2x\cdot x^2}{e^x-e^{-x}}=0$$
It follows that there is $R>0$ such that for any $x\geq R$,
$$0<\frac{2x}{e^x-e^{-x}}\leq \frac{1}{x^2}.$$
Hence
$$0<\int_R^{+\infty}\frac{2x}{e^x-e^{-x}}dx\leq \int_R^{+\infty}\frac{1}{x^2}dx=\frac{1}{R}.$$
P.S. Since $\frac{2x}{e^x-e^{-x}}$ is continuous in $(0,+\infty)$ with a finite limit at $0$ (it tends to $1$), the integral $\int_0^{R}\frac{2x}{e^x-e^{-x}}dx$ is finite for any $R>0$.
A: Note that the function 
$$ f \, : x \in \, ]0,+\infty[ \; \longmapsto \; \frac{2x}{e^{x} - e^{-x}} $$
is continuous on $]0,+\infty[$. 
When $x \to 0$:
$$ e^{x} - e^{-x} = \big( 1 + x + o(x) \big) - \big( 1 - x + o(x) \big) = 2x + o(x). $$
Therefore:
$$ \frac{2x}{e^{x} - e^{-x}} \; \mathop{\sim} \limits_{x \to 0} \; 1. $$
Since $x \mapsto 1$ is integrable, $f$ is integrable at $0$.
When $x \to +\infty$:
$$ \frac{2x}{e^{x} - e^{-x}} = \mathop{o} \limits_{x \to +\infty}\Big( \frac{1} {x^2} \Big) $$
It follows from Riemann's comparison theorem that $f$ is integrable at infinity.
A: You may also notice that since
$$ \frac{\sinh x}{x} = \prod_{k\geq 1}\left(1+\frac{x^2}{k^2\pi^2}\right) \tag{1}$$
we clearly have
$$ 0\leq \int_{0}^{+\infty}\frac{x}{\sinh x}\,dx \leq \int_{0}^{+\infty}\frac{dx}{1+\frac{x^2}{\pi^2}}=\frac{\pi^2}{2}.\tag{2} $$
Funny thing, the exact value of the given integral is just at the midpoint of $\left(0,\frac{\pi^2}{2}\right).$
