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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?

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  • $\begingroup$ Try an integration by parts on the last derived result and recall that $|\cos(x)|\leq 1$ to obtain that it converges for every $\alpha>0$. For $\alpha\leq 0$ you should easily find that it diverges. $\endgroup$ – Alex Jun 16 '18 at 22:30
  • $\begingroup$ calculating the anti-derivative seems to require trigonometric integrals and gamma function which we were not taught. Maybe there is another way that I am missing? One that doesn't require actually calculating the integral $\endgroup$ – Oria Gruber Jun 16 '18 at 22:46
  • $\begingroup$ Finding an antiderivative doesn't seem very promising. If $\alpha>1$ then this converges, by comparison with $\displaystyle \int_0^\infty \frac{du}{(1+u)^\alpha}.$ If $\alpha=1,$ then I expect that this is "properly improper," i.e. the integral of the absolute value is $+\infty,$ whereas $\displaystyle\lim_{a\,\to\,+\infty} \int_0^a$ is a finite number. $\qquad$ $\endgroup$ – Michael Hardy Jun 16 '18 at 23:06
  • $\begingroup$ Apply Abel's (aka Dirchelt's) Test and find that the integral converges for $\alpha>0$ . $\endgroup$ – Mark Viola Jun 16 '18 at 23:08
  • $\begingroup$ I think that the that $$\int_{\Bbb R^2}\frac{\sin(x^2+y^2)^{\alpha}}{x^2+y^2+1} \ dx \ dy,$$ should be $$\int_{\Bbb R^2}\frac{\sin(x^2+y^2)}{(x^2+y^2+1)^{\alpha}} \ dx \ dy,$$ right under “What I did: $\endgroup$ – Crosby Jun 16 '18 at 23:54
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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.

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  • $\begingroup$ That’s great! First time I hear of this test, didn’t know if it before. Thank you for showing this to me. $\endgroup$ – Oria Gruber Jun 16 '18 at 23:22
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jun 16 '18 at 23:23
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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 $\alpha>0$.

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