x^(2) * e^(-x^2) (improper integrals)

[TsAmE]

Show that $\displaystyle \int_{0}^{\infty}x^{2}e^{-x^{2}}dx = \frac{1}{2}\int_{0}^{\infty}e^{-x^{2}}dx.$

I used substitution:

$\displaystyle t = x^{2}$

$\displaystyle dx = \frac{dt}{2x}$

$\displaystyle \frac{1}{2}\int_{0}^{\infty}\sqrt{t}e^{-t}dx$

Then tried using integration by parts but then I didnt get an answer and got stuck.

 

[Deleted member 4993]

Show that $\displaystyle \int_{0}^{\infty}x^{2}e^{-x^{2}}dx = \frac{1}{2}\int_{0}^{\infty}e^{-x^{2}}dx.$

I used substitution:

$\displaystyle t = x^{2}$

$\displaystyle dx = \frac{dt}{2x}$

$\displaystyle \frac{1}{2}\int_{0}^{\infty}\sqrt{t}e^{-t}dx$

Then tried using integration by parts but then I didnt get an answer and got stuck.

$\displaystyle \int_{0}^{\infty}x^{2}e^{-x^{2}}dx \ = \ \frac{1}{2}\int_{0}^{\infty}x\cdot 2x e^{-x^{2}}dx \ = \ -\frac{1}{2}\int_{0}^{\infty}x\cdot d[e^{-x^{2}}] \ = \ -\frac{1}{2}\left[ x\cdot e^{-x^2}\right ]_0^{\infty} + \frac{1}{2}\int_{0}^{\infty}e^{-x^{2}}dx$

 

[TsAmE]

$\displaystyle \int_{0}^{\infty}x^{2}e^{-x^{2}}dx \ = \ \frac{1}{2}\int_{0}^{\infty}x\cdot 2x e^{-x^{2}}dx \ = \ -\frac{1}{2}\int_{0}^{\infty}x\cdot d[e^{-x^{2}}] \ = \ -\frac{1}{2}\left[ x\cdot e^{-x^2}\right ]_0^{\infty} + \frac{1}{2}\int_{0}^{\infty}e^{-x^{2}}dx$

Where did the d come from in the 3rd equation?

 

[Deleted member 4993]

$\displaystyle \int_{0}^{\infty}x^{2}e^{-x^{2}}dx \ = \ \frac{1}{2}\int_{0}^{\infty}x\cdot 2x e^{-x^{2}}dx \ =$
Another way

Integration by parts

u = x

$\displaystyle v = e^{-x^2}$

$\displaystyle dv = -2xe^{-x^2}dx$

$\displaystyle \ = \ -\frac{1}{2}\left[ x\cdot e^{-x^2}\right ]_0^{\infty} + \frac{1}{2}\int_{0}^{\infty}e^{-x^{2}}dx$