2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.