Improper integration, for which values does integral converge

Question

We wish to know for which values of $\alpha \in \mathbb R$ does the following integral converge:

$$\int_{\mathbb R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^\alpha} \, dx\,dy$$

What I did:

Move to polar coordinates -

$$\int_{\mathbb R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^\alpha} \, dx \, dy = \int_0^\infty \int_0^{2\pi}r \frac{\sin(r^2)}{(r^2+1)^\alpha} \, d\theta \, dr = 2\pi \int_0^\infty r\frac{\sin(r^2)}{(r^2+1)^\alpha} \, dr$$

We can further simplify it by performing the substitution $u = r^2$:

$$2\pi \int_0^\infty r\frac{\sin(r^2)}{(r^2+1)^\alpha} \, dr = \pi \int_0^\infty \frac{\sin u}{(u+1)^\alpha} \, du$$

So essentially, the question now boils down to "to which values of $\alpha$ does $\int_0^\infty \frac{\sin u}{(u+1)^\alpha} \, du$ converge"

But I can't seem to find an anti-derivative. Comparison test won't do us much good here either I think, as we wish to find the boundary of $\alpha$ for which it will or wont converge. What to do? Is my simplification the right direction?

Answer 1

For all $L>0$, $\int_0^L \sin(u)\,du\le 2$ and $\frac{1}{(1+u)^\alpha}$ monotonically decreases to $0$ as $u\to\infty$ for $\alpha>0$. Now, apply Dirichlet's test.

Answer 2

As stated above in the comments, this can also be done with manipulations, let $b\in\mathbb{R}_{>0}$ be given, we obtain by integration by parts $$ \int_{0}^{b}\frac{\sin(u)}{(1+u)^{\alpha}}\mathrm{d}u=\left[\tfrac{-\cos(u)}{(u+1)^{\alpha}} \right]_{u=0}^{u=b}+\alpha\int_{0}^{b}\tfrac{\cos(u)}{(u+1)^{1+\alpha}}\mathrm{d} u $$ For $\alpha>0$ the first term goes to zero as $b\rightarrow\infty$, in the second term, one can estimate $$ \left|\int_{0}^{b}\tfrac{\cos(u)}{(u+1)^{1+\alpha}}\mathrm{d} u\right|\leq \int_{0}^{b} \tfrac{1}{(u+1)^{1+\alpha}}\mathrm{d}u=-\alpha\left[\frac{1}{(1+u)^{\alpha}}\right]_{u=0}^{u=b}. $$ Clearly, also the second term is bounded as $b\rightarrow\infty$ and one obtains the existence of the integral for every $\alpha>0$.