1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ at $0$, your integrand is $\sim 1/x^2$ $\endgroup$ – reuns Jan 15 '16 at 14:39
1
$\begingroup$

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 $a>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.