For which positive a this integral is convergent?

Question

For which positive $a$ this integral is convergent?

$$\int_0^\infty \frac{\sin x}{x^{a}+x^{2a}}\mathrm dx$$

I tried splitting it to two integrals (one from $0$ to $1$). I am not sure, but it seems to be convergent for all $a$.

Answer 1

You are on the right track, splitting the domain is fine. As $x\to 0^+$, $$\frac{\sin x}{x^{a}+x^{2a}}\sim \frac{x}{x^{a}}=\frac{1}{x^{a-1}}$$ and therefore the integral ove r $$(0,1]$ is convergent if and only if $a-1<1$.

What about the convergence of the integral $[1,+\infty)$?

Note that, by integration by parts, $$\int_1^\infty \frac{\sin x}{x^{a}+x^{2a}} dx=\left[\frac{-\cos x}{x^{a}+x^{2a}}\right]_1^\infty -\int_1^\infty \frac{\cos x(ax^{a-1}+2ax^{2a-1})}{(x^{a}+x^{2a})^2} dx$$ and, for $a>0$, $$\frac{|\cos x|(ax^{a-1}+2ax^{2a-1})}{(x^{a}+x^{2a})^2} \leq \frac{ax^{a-1}+2ax^{2a-1}}{(x^{a}+x^{2a})^2}\sim \frac{2a}{x^{4a-(2a-1)}}=\frac{2a}{x^{2a+1}}.$$ What may we conclude?