Improper integral doubt

Question

I have to study the convergence of the following improper integral $$\int_0^{+\infty} \frac{x^\alpha \sinh(\beta x)}{(\sinh(x))^\gamma}dx$$

So I split it $$\int_0^a \frac{x^\alpha \sinh(\beta x)}{(\sinh(x))^\gamma}dx + \int_a^{+\infty} \frac{x^\alpha \sinh(\beta x)}{(\sinh(x))^\gamma}dx$$ the first is asymptotic to $$\frac{\beta x^{\alpha+1}}{x^\gamma}=\frac{\beta}{x^{\gamma - \alpha -1}} $$ so it for $\beta=0$ converges, for $\gamma \ge \alpha +2$ diverges and for $\gamma< \alpha +2$ converges.

The second is asymptotic to $$x^\alpha\frac{e^{x(\beta - \gamma)}}{2^{1-\gamma}} $$ Is that correct?

My doubt is for which $\alpha, \beta, \gamma$ the second integral converges — Richy65, Jan 18 '21 at 14:20

score 1 · Accepted Answer · answered Jan 18 '21 at 14:32

The second integral is equal to $0$ if $\beta = 0$, otherwise it is asymptotic to \begin{equation} \text{sgn}(\beta)\frac{x^\alpha e^{(|\beta| -\gamma) x}}{2^{1-\gamma}} \end{equation} Hence it converges if $|\beta| -\gamma < 0$ or if $|\beta|-\gamma=0$ and $\alpha < -1$.

Improper integral doubt

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