How to prove that: $\int_{0}^{\infty} $$ \frac{1}{\sinh(x)\cdot x }dx=\int_{0}^{\infty} \frac {1}{\frac 1 2(e^x-e^{-x})\cdot x} dx$ Diverges.
SOLUTION ATTEMPT: I thought about separating this integral from $0$ to $a$ and from $a$ to $\infty$. and tried to using Comparasion test, but didn't know with what integral to compare. also thought about Taylor's Expansion and the relation between Series of functions and Integrals, but also couldn't find any starting point.
Near $x = 0$, we have $$\dfrac1{\sinh(x)\cdot x} = \dfrac2{(e^x-e^{-x})\cdot x} \sim \dfrac1{x^2}$$ The integral of $\dfrac1{x^2}$ diverges at $x=0$ and hence so does $$\int_0^a \dfrac{dx}{\sinh(x)\cdot x}$$ for all $a>0$.