Show that $\int^\infty_0$ $\int^\infty_0$ sin($x^2$+$y^2$) dxdy value is $\frac{\pi}{4}$ I am trying to show that the value of $\int^\infty_0$$\int^\infty_0$ sin($x^2$+$y^2$) dxdy is $\frac{\pi}{4}$ using Fresnel integrals. I'm having trouble splitting apart the integrand in order to actually be able to use the Fresnel integrals. Any help is appreciated. 
Answer:
 $\int^\infty_0$ $\int^\infty_0$ sin($x^2$+$y^2$) =  $\int^\infty_0$ $\int^\infty_0$ sin($x^2$)cos($y^2$)+cos($x^2$)sin($y^2$)dxdy
= $\int^\infty_0$ $\frac{\sqrt2\pi}{4}$ cos$(y^2)$+$\frac{\sqrt2\pi}{4}$ sin$(y^2)$ dy (the $\frac{\sqrt2\pi}{4}$ comes from established Fresnel integrals values)
= do the same thing for dy and then you get $\frac{\pi}{4}$ as desired.
Now I'm working on doing this with polar coords.
 A: In real-world situation, often it is too ideal to consider all the interactions from arbitrarily long distance. So we may first suppress long-distance interactions and then let the suppression disappear. In mathematical terms, it means that we may understand the integral
$$ \int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2}) \,dxdy \tag{1} $$
in an appropriate summability sense. Here, let us consider the Gaussian summability, i.e., we understand (1) by
$$ \lim_{\epsilon \downarrow 0} \int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2})e^{-\epsilon(x^{2}+y^{2})} \,dxdy $$
Then it follows that
\begin{align*}
\int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2})e^{-\epsilon(x^{2}+y^{2})} \,dxdy
&= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} r \sin(r^{2}) e^{-\epsilon r^{2}} \, dr d\theta \\
&= \frac{\pi}{4} \int_{0}^{\infty} \sin u \, e^{-\epsilon u} \, du \qquad (u = r^{2}) \\
&= \frac{\pi}{4} \frac{1}{1+\epsilon^{2}}.
\end{align*}
Taking $\epsilon \to 0$ we get the desired answer.
A: Let's recall the definition of the convergence of an improper multiple integral (of the first kind). Assume that a function $f \colon \mathbb{R}^m \to \mathbb{R}$ is continuous almost everywhere. Consider the following sequence of sets $\{E_n\}_{n = 1}^\infty$:


*

*Each $E_n$ is an open Jordan-measurable subset of $\mathbb{R}^m$.

*$\overline{E_n} \subset E_{n + 1}$ and $\cup_{n = 1}^\infty E_n = \mathbb{R}^m.$


Consider the corresponding sequence of Riemann integrals:
$$ I_n = \int\limits_{E_n} f(x) dx, \quad n = 1, 2, \dots$$
If for every sequence $\{E_n\}_{n = 1}^\infty$, satisfying 1 and 2, there exists a finite limit $I = \lim_{n \rightarrow \infty} I_n$ independent of the choice of $\{E_n\}_{n = 1}^\infty$, then the improper multiple integral  $$\int_\limits{\mathbb{R}^m}f(x)dx$$ converges (exists) and is equal to $I$. Otherwise (if this limit is infinite or does not exist), this integral diverges.
To apply this definition to other domains $E \subset \mathbb{R}^m$ (in your case $E$ is the first quadrant) instead of $f$ we consider the following function
$$F(x) = \begin{cases}f(x), &x \in E,\\ 0, &x \in \mathbb{R}^m \setminus E.\end{cases}$$
Then
$$\int_\limits{E}f(x)dx \triangleq \int_\limits{\mathbb{R}^m}F(x)dx.$$
In your case, consider two sequences $\{E_n\}_{n = 1}^\infty$ and $\{E'_n\}_{n = 1}^\infty$ satisfying 1 and 2:
$$ E_n = \{(x, y) \in \mathbb{R}^2 \mid |x| + |y| < n\},\ E'_n = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 2 \pi n\}, n \in \mathbb{N}.$$
For the first sequence we have:
$$I_n = \iint\limits_{E_n}F(x, y)dxdy = \int_0^n dx\int_0^n\sin(x^2+y^2)dy = 2\int_0^n \sin{x^2} dx\int_0^n\cos{y^2}dy,$$
and hence (using Fresnel integrals) we have $\lim_{n \rightarrow \infty} I_n = \frac{\pi}{4}.$
For the second sequence (using polar coordinates) we have:
$$I_n = \iint\limits_{E'_n}F(x, y)dxdy = \int_0^\frac{\pi}{2} d\varphi\int_0^{\sqrt{2\pi n}} r\sin(r^2)dr = \frac{\pi}{4}(1 - \cos 2\pi n) = 0,$$
and hence $\lim_{n \rightarrow \infty} I_n = 0$, which means that this limit depends on the choice of a sequence and hence by definition this integral diverges.
Also notice that for improper multple integrals there is no notion of conditional convergence because of the following theorem:

Assume that a function $f \colon \mathbb{R}^m \to \mathbb{R}$ $(m \geq 2)$ is continuous almost everywhere. Then the following two integrals 
  $$1. \int_\limits{\mathbb{R}^m}f(x)dx \qquad 2. \int_\limits{\mathbb{R}^m}|f(x)|dx $$
  converge or diverge simultaneously.

A: Does it work like that? $$\int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2}) \,dx\ dy=\Im\left\{\int_{0}^{\infty}\int_{0}^{\infty} e^{i(x^2+y^2)} \ dx \ dy\right\}=\Im\left\{\int_{0}^{\infty}e^{i x^2} \ dx\int_{0}^{\infty} e^{i y^2} \ dy\right\}=\frac{\pi}{4}$$
