# Integration by parts

• Jan 16th 2013, 08:44 AM
JDAWES
Integration by parts
Integrate x* 2*alpha*x*e^(-alpha*x^2) dx with limits infinity and zero?
• Jan 16th 2013, 09:36 PM
sbhatnagar
Re: Integration by parts
Quote:

Originally Posted by JDAWES
Integrate x* 2*alpha*x*e^(-alpha*x^2) dx with limits infinity and zero?

Hi JDAWES!
Your question is not clear. Is it

$\int_0^\infty 2\alpha x^{2 }e^{-\alpha x^2}dx$

or

$\int_0^\infty x^{2\alpha}e^{-\alpha x^2}dx$
?(Talking)
• Jan 17th 2013, 11:06 AM
zaidalyafey
Re: Integration by parts
$\int_0^\infty 2\alpha x^{2 }e^{-\alpha x^2}dx$

we can here use the substitution $t= x^2 \,\, \, dx = \frac{1}{2}t^{\frac{-1}{2}}\, dt$

so we get the following : $\alpha \int^{\infty}_0 \, t^{1/2}e^{-\alpha t }\, dt$

Now this is solved by the Laplace transform or gamma function which converges for $\alpha >0$

$\alpha \int^{\infty}_0 \, t^{{1/2}}e^{-\alpha t }\, dt = \alpha \frac{\Gamma {(\frac{3}{2})}}{\alpha^{\frac{3}{2}}}= \frac{\Gamma(\frac{1}{2})}{2\sqrt{\alpha }}= \frac{\sqrt{\pi}}{2\sqrt{\alpha }}$

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Now for the integrand : $\int_0^\infty x^{2\alpha}e^{-\alpha x^2}dx$

here if we use the same sub we get :

$\int^{\infty}_0 \, t^{\alpha -\frac{1}{2}}e^{-\alpha t }\, dt = \frac{\Gamma {(\alpha +\frac{1}{2})}}{2\alpha^{\alpha +\frac{1}{2}}}$