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.
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 $1-\frac a2<0$.