Integration Problem

  • Thread starter qspeechc
  • Start date
  • #1
840
14
I feel really dumb for asking this, because I know it's something simple I'm just not seeing. Ok, given that

[tex]\int _{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi } [/tex]

how to I find

[tex]\int _{-\infty}^{\infty} x^2e^{-x^2}dx = ? [/tex]

I have tried the substitution u=x^2, and integration by parts, but nothing is working. Any help? Thanks
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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619
The easy way to do this problem is to generalize your first integral. Can you show the integral of exp(-ax^2) is sqrt(pi/a) for a>0? (Use a substitution x=sqrt(a)*u). Now differentiate that with respect to a. Finally put a=1 again.
 
  • #3
412
3
This might help:

d/dx (x*[e^-x^2]) = ...
Solve it and then integrate!
 
  • #4
statdad
Homework Helper
1,495
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Integration by parts should work. You have

[tex]
\int_{-\infty}^\infty x^2 e^{-x^2} \, dx = \sqrt \pi
[/tex]

Set

[tex]
u = x, \quad dv =x e^{-x^2} dx
[/tex]

Then

[tex]
\int u \, dv = uv - \int v \, du
[/tex]

should, with careful attention to the [tex] uv [tex] term at the infinities, work fine.
 
  • #5
840
14
Thanks everyone.
 

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