Prove $\int_{0}^\infty \mathrm{d}y\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}x=\int_{0}^\infty \mathrm{d}x\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}y=\pi/4$ How can we prove that
\begin{aligned}
&\int_{0}^\infty \mathrm{d}y\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}x\\
=&\int_{0}^\infty \mathrm{d}x\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}y\\=&\cfrac{\pi}{4}
\end{aligned}
I can prove these two are integrable but how can we calculate the exact value?
 A: I do not know if you are supposed to know this. So, if I am off-topic, please forgive me.
All the problem is around Fresnel integrals. So, using the basic definitions,$$\int_{0}^t \sin(x^2+y^2)dx=\sqrt{\frac{\pi }{2}} \left(C\left(\sqrt{\frac{2}{\pi }} t\right) \sin
   \left(y^2\right)+S\left(\sqrt{\frac{2}{\pi }} t\right) \cos
   \left(y^2\right)\right)$$ where appear sine and cosine Fresnel integrals. $$\int_{0}^\infty \sin(x^2+y^2)dx=\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\sin \left(y^2\right)+\cos
   \left(y^2\right)\right)$$ Integrating a second time,$$\frac{1}{2} \sqrt{\frac{\pi }{2}}\int_0^t \left(\sin \left(y^2\right)+\cos
   \left(y^2\right)\right)dy=\frac{\pi}{4}   \left(C\left(\sqrt{\frac{2}{\pi }}
   t\right)+S\left(\sqrt{\frac{2}{\pi }} t\right)\right)$$ $$\frac{1}{2} \sqrt{\frac{\pi }{2}}\int_0^\infty \left(\sin \left(y^2\right)+\cos
   \left(y^2\right)\right)dy=\frac{\pi}{4}  $$
A: Although this problem relates to Fresnel integration, if you take the following for granted then it is quite easy to show.
Given that (Fresnel integrals):
$$ \int_0^\infty \cos (x^2)dx = \int_0^\infty \sin (x^2)dx 
= \sqrt{\frac{\pi}{8}} $$
Using trigonometry:
$$ \sin(x^2 + y^2) = \sin(x^2)\cos(y^2) + \cos(x^2)\sin(y^2)$$
And since both are symmetric, the original integral becomes:
$$\int_0^\infty dx \int_0^\infty dy \sin(x^2 + y^2) 
= 2\int_0^\infty \sin(x^2)dx \int_0^\infty \cos(y^2)dy 
= 2 \sqrt{\frac{\pi}{8}} \sqrt{\frac{\pi}{8}} = \frac{\pi}{4}$$
