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.

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.
