Checking convergence of improper integral

Question

I have an exercise that I'm not sure if I got right.

I'm asked to check convergence of $$ \int_0^\infty \frac{\sin^2(x)}{(e^x-1)x^\alpha} $$

depending on parameter $\alpha$.

I want to divide it into two parts, one that is on interval $(0, 1)$, and another on $(1, \infty)$.

First the one on $(1, \infty)$. $$ \int_1^\infty \frac{\sin^2(x)}{(e^x-1)x^\alpha} \leq \int_1^\infty \frac{1}{(e^x-1)x^\alpha} $$ since $\sin^2(x) \leq 1$, then also $$\int_1^\infty \frac{1}{(e^x-1)x^\alpha} \leq \int_1^\infty \frac{1}{e^x x^\alpha}$$ since the denominator is bigger on the RHS, because $x$ is a positive number. Since $e^x$ is bigger than any $x^\alpha$ we know that $\int_1^\infty \frac{1}{e^x x^\alpha}$ converges even for $\alpha \le 0$.

So thanks to comparison test we know that $\int_1^\infty \frac{\sin^2(x)}{(e^x-1)x^\alpha}$ converges.

Now onto the second part - $x \in (0, 1)$. We can rewrite $$ \int_0^1 \frac{\sin^2(x)}{(e^x-1)x^\alpha} $$ as $$ \int_0^1 \frac{x\sin^2(x)}{(e^x-1)x^2x^{\alpha-1}} $$

Thanks to known limits $\lim_{x\to0}\frac{\sin x}{x} = 1$ and $\lim_{x \to 0}\frac{x}{e^x-1} = 1$ we see that our integrand should behave like $\frac{1}{x^{\alpha-1}}$, which converges when $2 \le \alpha$.

This way we've shown that $$ \int_0^\infty \frac{\sin^2(x)}{(e^x-1)x^\alpha} $$ is convergent when $2 \le \alpha$.

I see two problems with that solution: First and foremost I'm not sure if this second part is ok. Can I use those limits in such way? Also, it doesn't prove that it is divergent when $\alpha \le 2$, does it?

Cheers!

Answer 1

As you noticed, $$ \int_{1}^{+\infty}\frac{\sin^2(x)}{(e^x-1)x^\alpha}\,dx \leq \int_{1}^{+\infty}\frac{1}{(e^x-1)x^\alpha}\,dx \sim \int_{1}^{+\infty}\frac{dx}{x^\alpha e^x} $$ so over the sub-interval $[1,+\infty)$ we have convergence for any $\alpha\in\mathbb{R}$.
Over $(0,1)$ we have to consider the behaviour of the integrand function in a right neighborhood of the origin: $$ \frac{\sin^2(x)}{(e^x-1)x^\alpha}\sim \frac{x^2}{x\cdot x^\alpha}=x^{1-\alpha}$$ and we have convergence provided that $1-\alpha > -1$, i.e. $\color{blue}{\alpha < 2}$.
If $\alpha=2$ we have a simple pole at the origin (which is not integrable) and if $\alpha > 2$ then $$ \int_{0}^{1}\frac{\sin^2(x)}{(e^x-1)x^{\alpha}}\,dx > \int_{0}^{+\infty}\frac{\sin^2(x)}{(e^x-1)x^2}\,dx = +\infty $$ so the LHS is also divergent.