I try to show that

$$\int_0^\infty\exp(-x^2)\sin(x) \, \mathrm dx=\frac{1}{2}\sum_{k=0}^\infty(-1)^k\frac{k!}{(2k+1)!}$$


$$\frac{d}{dt}\int_0^\infty\exp(-tx) \, \mathrm dx=\int_0^\infty(-x)\exp(-tx) \, \mathrm dx$$ somehow. Tried several approaches to get the $\exp(-x^2)$ away, but none of it worked. Always fall back to an integral with $\exp(-x^2)$ .

  • $\begingroup$ I would replace the sine by a complex exponential, and try to complete the square and see if that leads anywhere. $\endgroup$ – Spine Feast Dec 26 '13 at 12:14

$\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \ldots$ So look at $\displaystyle\int_0^\infty e^{-x^2}x^{2k+1}dx = \frac12\int_0^\infty e^{-t}t^kdt = \dfrac{k!}2$ and you are done. Note: this is more elementary than the proposed method. Let me try the other way.

  • $\begingroup$ t tried this and I thinks that's way I am supposed to show it, since we had an example where we showed that $\int_0^\infty x^n\exp(-x)dx=n!$ using $\int_0^\infty \exp(-tx)dx=1/t$ and the method above... so that's the answer. Thanks $\endgroup$ – brinki1602 Dec 26 '13 at 13:14

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.