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

Question

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 1

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

Answer 2

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: $$\int_0^\infty dx \int_0^\infty dy \sin(x^2 + y^2) = 2\int_0^\infty \sin(x^2)dx \int_0^\infty \cos(y^2)dy = 2 \sqrt{\frac{\pi}{8}} \sqrt{\frac{\pi}{8}} = \frac{\pi}{4}$$