Math Help - Integration Help

1. Integration Help

Hello, this is my first post. I was wondering if somebody could shed some light on how to solve the following integral:

$\int_0^{\infty} \sqrt{x}e^{-(1/2)x^2} dx$

I have tried all sorts: parts, reduction formulae and cannot seem to find a suitable substitution. Please Help!

2. Originally Posted by Electric
Hello, this is my first post. I was wondering if somebody could shed some light on how to solve the following integral:

int_0^infinity x^(1/2)e^[-(1/2)x^2]dx

I have tried all sorts: parts, reduction formulae and cannot seem to find a suitable substitution. Please Help!
Wolfram Mathematica Online Integrator

It looks as if this has no "closed form" solution. You can solve it using the exponential integral.

I'm not sure of this though, and there are a few amazing integrators on this forum. Maybe one of them can find a solution.

3. Yeah, it is a tricky one! I have a feeling there is a sneaky substitution and then the error or gamma function has to be used somewhere however, I still cannot find a solution

4. Originally Posted by Electric
Yeah, it is a tricky one! I have a feeling there is a sneaky substitution and then the error or gamma function has to be used somewhere however, I still cannot find a solution
Maybe pm Krizalid and see if he can help.

5. Originally Posted by Electric
Hello, this is my first post. I was wondering if somebody could shed some light on how to solve the following integral:

$\int_0^{\infty} x^{1/2}e^{-(1/2)x^2} dx$

I have tried all sorts: parts, reduction formulae and cannot seem to find a suitable substitution. Please Help!
Yeah \infty is weird, right? When I'm typing teX I say in my head "integral from zero to inf-T"

6. put $x=\sqrt2\sqrt t$ and that'd turn the integral into a Gamma function one.

7. Ah yes, thanks Krizalid. Overlooked that substitution For anyone interested to know the solution, here it is:

$x=\sqrt{2t}, dx=\frac{1}{\sqrt{2t}} dt$

$=\int_0^{\infty} {2t}^{-(1/4)}e^{-t} dt$

$since\;\;\; \Gamma{(n)}=\int_0^{\infty} t^{n-1}e^{-t} dt,$

$=>\frac{1}{2^{1/4}} \int_0^{\infty} t^{-(1/4)} e^{-t} dt$

$=\frac{1}{2^{1/4}}*\Gamma{(3/4)}$

$=0.841*1.225$

$=1.030$