calculus - Prove $int_{0}^infty mathrm{d}yint_{0}^infty sin(x^2+y^2)mathrm{d}x=int_{0}^infty mathrm{d}xint_{0}^infty sin(x^2+y^2)mathrm{d}y=pi/4$

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?

Answer

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, $$\frac{1}{2} \sqrt{\frac{\pi }{2}}\int_0^t \left(\sin \left(y^2\right)+\cos \left(y^2\right)\right)dy=\frac{\pi}{4} \left(C\left(\sqrt{\frac{2}{\pi }} t\right)+S\left(\sqrt{\frac{2}{\pi }} t\right)\right)$$ $$\frac{1}{2} \sqrt{\frac{\pi }{2}}\int_0^\infty \left(\sin \left(y^2\right)+\cos \left(y^2\right)\right)dy=\frac{\pi}{4}$$