# Prove that $\int^{\infty}_0 \frac{e^x}{\sqrt{\sinh(ax)}}dx$ is convergent if $a>2$.

Prove that $$\int^{\infty}_0 \frac{e^x}{\sqrt{\sinh(ax)}}dx$$ is convergent if $$a>2$$.

I've simplified the expression to: $$\sqrt{\frac{2e^{2x}}{e^{ax}-e^{-ax}}}$$.

I'm thinking of finding an expression bigger that the above and showing convergence for that. I know that $$\int^{\infty}_k e^{-tx} dx$$ is convergent for $$t>0$$. The problem is that I can't seem to find an expression is guaranteed bigger than $$\sqrt{\frac{2e^{2x}}{e^{ax}-e^{-ax}}}$$ that seems to solve the problem elegantly. I've considered $$\sqrt{\frac{2e^{2x}}{e^{-ax}}}$$, but that only works for $$x>\frac{ln2}{4}$$ and it seems needlessly complicated.

Hints and suggestions appreciated! I've been stuck on this a while.

• It's $O(e^{-(a/2-1)x})$ as $x\to\infty$ and $O(x^{-1/2})$ as $x\to0$. – Lord Shark the Unknown Nov 4 '18 at 13:04
• This integral can expressed by the Gamma function – Dr. Sonnhard Graubner Nov 4 '18 at 13:20
• @LordSharktheUnknown I understand your first part, but why does the graph resemble $x^{-0.5}$ as x approaches 0? – Yip Jung Hon Nov 6 '18 at 3:57
• @YipJungHon Since $\sinh ax\sim ax$ as $x\to 0$. – Lord Shark the Unknown Nov 6 '18 at 4:03
• Okay I understand now, thanks – Yip Jung Hon Nov 6 '18 at 4:32

You can have a simpler solution using an asymptotic equivalent. Observe that $$\sinh(ax)\sim_{x\to+\infty}\begin{cases} \phantom{-}\frac12\mathrm e^{ax} &\text{if } a>0 \\[1ex] -\frac12\mathrm e^{-ax} &\text{if } a <0 \end{cases}$$ Now this integral is defined only if $$a>0$$, so the integrand is equivalent to $$\frac{\mathrm e^x}{\sqrt{\sinh(ax)}}\sim_{x\to+\infty}\sqrt 2\,\mathrm e^{\bigl(1-\tfrac a2\bigr)x},$$ and the integral of the latter converges if and only if $$1-\frac a2<0$$.