Convergence of an improper integral with the hyperbolic sine

Question

Let us consider the following improper integral:

$$ \int_{0}^\infty \frac{x^\alpha\sinh (\beta x)}{(\sinh (x))^\gamma} dx $$

I have to understand for which parameters $\alpha,\beta,\gamma$ the above integral converges.

Then, we must study the limit for $M \to \infty$ and that for $\epsilon \to 0$ of the integral

$$ \int_{\epsilon}^M \frac{x^\alpha\sinh (\beta x)}{(\sinh (x))^\gamma} dx $$

I know that $\sinh (x)=\sum\limits_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$.

If $x \to 0$, I have that $\sinh(x) \sim x$, therefore I treat the hyperbolic function as $x$ and hence obtain $\beta x^{\alpha+1-\gamma}$ and, the convergence depends on the exponent $\alpha+1-\gamma$ as in the standard examples given in any book. So, I compute the integral and then I study the two limits.

1) Is my argument correct? \ 2) Are there other possible ways to solve the exercise?

Answer 1

Hints:

Your argument is not quite correct: it is an improper integral, because of both bounds, so you have to split the integral in two, say as $$\int_{0}^\infty \frac{x^\alpha\sinh (\beta x)}{(\sinh (x))^\gamma} \mathrm dx=\int_{0}^1 \frac{x^\alpha\sinh (\beta x)}{(\sinh (x))^\gamma} \mathrm dx + \int_{1}^\infty \frac{x^\alpha\sinh (\beta x)}{(\sinh (x))^\gamma} \mathrm dx.$$ As the integrand has a constant sign, we indeed can use equivalents.

• On the $0$ side, we do have $\;\dfrac{x^\alpha\sinh \beta x }{(\sinh x)^\gamma}\sim_0\dfrac{x^\alpha \beta x}{x^\gamma}=\beta x^{\alpha-\gamma+1} $, and it converges if and only if $\;\alpha-\gamma+1>-1$, i.e. $\;\gamma<\alpha+2$.
• On the $+\infty$ side, the equivalents are different: $\;\sinh x\sim_{+\infty}\frac12\mathrm e^{x}$ and $\;\sinh x\sim_{-\infty}-\frac12\mathrm e^{-x}$, so you'll have to examine the cases $\beta>0$, $\;\beta<0$. If $\beta>0$, $$\dfrac{x^\alpha\sinh \beta x }{(\sinh x)^\gamma}\sim_\infty \begin{cases} \dfrac{x^\alpha\mathrm e^{\beta x}}{2\mathrm e^{\gamma x}}=\dfrac12x^\alpha\mathrm e^{(\beta-\gamma)x}&\text{if }\beta >0, \\ -\dfrac{x^\alpha\mathrm e^{-\beta x}}{2\mathrm e^{\gamma x}} = -\dfrac12x^\alpha\mathrm e^{-(\beta+\gamma)x}&\text{if }\beta <0, \end{cases}$$ and these latter integrals converge if and only if $\beta<\gamma$ or $\beta+\gamma>0$ respectively.