# $\int_0^\infty\exp(-x^2)\sin(x)~dx$ Evaluate Integral

I try to show that

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

using

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

• I would replace the sine by a complex exponential, and try to complete the square and see if that leads anywhere. – 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.
• 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 – brinki1602 Dec 26 '13 at 13:14