# Checking convergence of improper integral

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!

• In the first part $(e^x-1)^{-1}>e^{-x}$ so your estimate is wrong. Concerning the second part the crucial is behavior at $0.$ The integrand behaves approximately as $x^2/(xx^a)= x^{1-a}$ The convergence holds for $a<2.$ Aug 24, 2022 at 13:14

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.