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

\(\displaystyle v = e^{-x^2}\)

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

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