Find the following integral:

$\displaystyle \int{\frac{e^{-x}}{\sqrt{x}}~dx}$

Unable to use integration by parts I believe for this problem..

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- Apr 17th 2008, 10:47 PMIdeasmanHard Integral
Find the following integral:

$\displaystyle \int{\frac{e^{-x}}{\sqrt{x}}~dx}$

Unable to use integration by parts I believe for this problem.. - Apr 17th 2008, 11:08 PMJhevon
- Apr 17th 2008, 11:22 PMlllll
not sure if this will be of any help, but you could also express [tex] e^{-x} as an infinite series then integrate it out.

so you would have:

$\displaystyle \int \frac{\sum^{\infty}_{n=0} \frac{-x^n}{n!}}{\sqrt{x}} dx$

=$\displaystyle \int \sum^{\infty}_{n=0} \frac{-x^{n-0.5}}{n!} dx$

=$\displaystyle \sum^{\infty}_{n=0} \frac{-x^{n+0.5}}{(n+0.5)n!} +C $ - Apr 18th 2008, 04:17 AMtopsquark
If this is supposed to be

$\displaystyle \int_0^{\infty} {\frac{e^{-x}}{\sqrt{x}}~dx}$

then this is a gamma function.

-Dan - Apr 18th 2008, 04:21 AMmr fantastic
- Apr 19th 2008, 06:44 PMJhevon
- Apr 19th 2008, 06:54 PMPaulRS
Yes

Let $\displaystyle u=\sqrt[]{x}$ then $\displaystyle \int_0^{\infty} {\frac{e^{-x}}{\sqrt{x}}dx}=2\int_0^{\infty} {e^{-u^2}du}$$\displaystyle =\int_{-\infty}^{\infty} {e^{-u^2}du}=\sqrt[]{\pi}$

Gaussian integral - Wikipedia, the free encyclopedia - Apr 19th 2008, 06:58 PMKrizalid
The well known gaussian integral. There're lots of proofs of its result. :D