Improper integral x^(2) * e^(-x^2)

[TsAmE]

Homework Statement

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

Homework Equations

None.

The Attempt at a Solution

I used substitution:

$$t = x^{2}$$

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

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

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

 

[lanedance]

have you tried straight integartion by parts?

 

[gomunkul51]

Try:

x^2*e^(-x^2) = [-0.5*x]*[-2*x*e^(-x^2)dx]

[-0.5*x] = u
[-2*x*e^(-x^2)dx] = dv

Tell us if you got it !

Good Luck :)

 

[TsAmE]

I tried integration by parts but it didnt work out. I only learn how to integrate e^(-x^2) next year.

 

[HallsofIvy]

I tried integration by parts but it didnt work out. I only learn how to integrate e^(-x^2) next year.

How did you try? Did you try letting $u= x$ and $dv= xe^{-x^2}dx$.

 

[gomunkul51]

Try integration by parts with t=x, not t=x^2 !

I remind you:

integral(u*dv) = u*v - integral(v*du)

P.S.: If you learn in the next year how to get the anti-derivative of e^(-x^2), please tell us ;)