# Hard Integral

• Apr 17th 2008, 10:47 PM
Ideasman
Hard 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 PM
Jhevon
Quote:

Originally Posted by Ideasman
Find the following integral:

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

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

is the integral indefinite? if so, there is no solution in elementary functions, you would need the erf function
• Apr 17th 2008, 11:22 PM
lllll
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 AM
topsquark
Quote:

Originally Posted by Ideasman
Find the following integral:

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

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

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 AM
mr fantastic
Quote:

Originally Posted by Jhevon
is the integral indefinite? if so, there is no solution in elementary functions, you would need the erf function

..
• Apr 19th 2008, 06:44 PM
Jhevon
Quote:

Originally Posted by topsquark
If this is supposed to be
$\displaystyle \int_0^{\infty} {\frac{e^{-x}}{\sqrt{x}}~dx}$
then this is a gamma function.

-Dan

yes, if we had limits like that, then a substitution would turn the integrand into $\displaystyle 2e^{-u^2}$ which is fun to do...
• Apr 19th 2008, 06:54 PM
PaulRS
Quote:

Originally Posted by topsquark
If this is supposed to be
$\displaystyle \int_0^{\infty} {\frac{e^{-x}}{\sqrt{x}}~dx}$
then this is a gamma function.

-Dan

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 PM
Krizalid
The well known gaussian integral. There're lots of proofs of its result. :D