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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.

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    $\begingroup$ It's $O(e^{-(a/2-1)x})$ as $x\to\infty$ and $O(x^{-1/2})$ as $x\to0$. $\endgroup$ – Lord Shark the Unknown Nov 4 '18 at 13:04
  • $\begingroup$ This integral can expressed by the Gamma function $\endgroup$ – Dr. Sonnhard Graubner Nov 4 '18 at 13:20
  • $\begingroup$ @LordSharktheUnknown I understand your first part, but why does the graph resemble $x^{-0.5}$ as x approaches 0? $\endgroup$ – Yip Jung Hon Nov 6 '18 at 3:57
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    $\begingroup$ @YipJungHon Since $\sinh ax\sim ax$ as $x\to 0$. $\endgroup$ – Lord Shark the Unknown Nov 6 '18 at 4:03
  • $\begingroup$ Okay I understand now, thanks $\endgroup$ – Yip Jung Hon Nov 6 '18 at 4:32
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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 $1-\frac a2<0$.

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