Limit of a double integral I have to analyse the following limit
\begin{equation}
\lim_{n\rightarrow \infty}\int_{-n}^{n}\int_{-n}^{n}\sin(x^2+y^2)dxdy
\end{equation}
I've tried to see it in polar coordinates but it didn't work. Do you have any idea how to work this problem? I put this problem in Wolfram Alpha and I get the result is $\pi$ and the double integral it's some special
 A: What class is that for? I don't imagine you're taking any average ordinary Calc II class.
Following Ryszard's hint, we write
$$
\lim_{n \to \infty} \int_{-n}^n\int_{-n}^n \sin(x^2 + y^2) \ \mathrm{d}x\mathrm{d}y = \lim_{n \to \infty} \int_{-n}^n\int_{-n}^n\sin(x^2)\cos(y^2)\ + \cos(x^2)\sin(y^2)\ \mathrm{d}x\mathrm{d}y.
$$
We can break that sum under the double integral into a sum of integrals and, assuming the individual limits exist, we can write
$$
\lim_{n\to\infty} \int_{-n}^n\int_{-n}^n\sin(x^2)\cos(y^2)\ \mathrm{d}x\mathrm{d}y + \lim_{n\to\infty}\int_{-n}^n\int_{-n}^n\cos(x^2)\sin(y^2)\ \mathrm{d}x\mathrm{d}y.
$$
From here we note that the integrands are products of single variable functions that are independent of each other, so we can rewrite the double integrals as products of simple integrals. Also, all functions are even, so
$$
4\lim_{n\to\infty}\int_0^n\sin(x^2)\ \mathrm{d}x\int_0^n \cos(y^2)\ \mathrm{d}y + 4\lim_{n\to\infty} \int_0^n\cos(x^2)\ \mathrm{d}x \int_0^n \sin(y^2)\ \mathrm{d}y.
$$
Finally, we notice that both terms of the sum, despite having flipped variable names, will evaluate to the same thing, therefore
$$
\lim_{n\to\infty} \int_{-n}^n\int_{-n}^n\sin(x^2 + y^2) \ \mathrm{d}x\mathrm{d}y = 8\lim_{n\to\infty}\int_0^n\sin(x^2)\ \mathrm{d}x\int_0^n\cos(x^2)\ \mathrm{d}x. \tag{1} \label{1}
$$
Now here is why it matters what class this task is for. If you're allowed to quote well-known results, we're done. Those integrals in the right hand side of \eqref{1} are called Fresnel integrals $S(x)$ and $C(x)$ (can you guess which is which?) and their limit at infinity is $S(x\to\infty) = C(x\to\infty) = \sqrt{\pi/8}$. If you're taking a complex analysis class, you can setup a contour integral to evaluate them. If it's a real analysis class, there are options too.
