# Help with a tricky improper integration problem

1. Aug 23, 2012

### stonecoldgen

Show that:

∫(x^2)e^(-x^2)=(1/2)∫(e^(-x^2)

(both integrals are from 0 to ∞)

I don't even know if it's possible to integrate these functions, so it makes me think that what's interesting about the problem is that it only works with this integration limits.

I know improper integration, but don't know if prooving this "numerically" is the right path.

Thanks, I would appreciate any advice

2. Aug 23, 2012

### Curious3141

Hint:

$$\int_0^{\infty} x^2e^{-x^2}dx = \int_0^{\infty} x.(xe^{-x^2})dx$$

Integrate that by parts (use $u = x, dv = xe^{-x^2}dx$), then use L'Hopital's Rule to calculate the limit of one of the expressions.

Last edited: Aug 23, 2012
3. Aug 23, 2012

### stonecoldgen

is this correct?

I integrated by pats (differentiating x and integrating x^2e^{-x^2} )

until I got to the point where (e^{-x^2})/2 (evaluated from o to infinity)=0

Using L'Hopital's Rule, I effectively got 0=0

Is this the way?

4. Aug 23, 2012

### Curious3141

No, I think you messed up the integration by parts. You should end up with:

$$\int_0^{\infty} x^2e^{-x^2}dx = xe^{-x^2}|_0^{\infty} +\frac{1}{2}\int_0^{\infty}e^{-x^2}dx$$

and the first expression on the RHS vanishes when you take the limit by L' Hopital's.

5. Aug 23, 2012

### stonecoldgen

It's the same thing (maybe my english isn't that good, I don't know) but Thanks!!!

6. Aug 23, 2012

### Curious3141

How is it the same thing? You said you got "0 = 0" which means nothing. You also said you got $\frac{1}{2}\int_0^{\infty}e^{-x^2}dx = 0$ which is clearly wrong.

Anyway, if you're confident you have the right solution, then it's fine.

7. Aug 23, 2012

### stonecoldgen

∫e-x2=∫√(-lny)
(from 0 to ∞) (from 0 to 1)

I tried this one by solving for y in the right side of the equation, getting y=-ex2

However, when checking the solution graphically, it's not right.

So what the hell am I supposed to do? I also wonder if the integration limits changed when expressed in y or x.

Thanks

8. Aug 24, 2012

### Millennial

You should do a substitution to change the limits and get the log into play, then integrate by parts. Your previous result might help here.

Last edited: Aug 24, 2012
9. Aug 24, 2012

### Curious3141

$$\int_0^{\infty} x^2e^{-x^2}dx = \frac{1}{2}\int_0^{\infty}e^{-x^2}dx$$
and make the substitution $e^{-x^2} = y$ on the LHS. The result follows very quickly.