# 2 answers to same integral?

siresmith

## Homework Statement

Ive been giong mental over this. There seems to be two different answers to the same integral. I cant work out which is correct.

Integral from 0 to infinity of:

x(squared) exp(-2Bx(squared))

## The Attempt at a Solution

Integration by parts gives the answer as: -1/2B

My course lecturer gives the answer as: (1/8B) x route of (pie/2B)

I think she must have used the general formula:

Integral from -infinity to + infinity of x(power2n)exp(-Ax(squared)) = [1x3x5...(2n-1)]x(pie/A)/2(powern)A(powern)

and divided it by 2 as we only want the integral from 0 to infinity.

Whos right?? Can anyone work this out?
Please! write it down and see if you get the answer i did by integration by parts. If you do, do you think the lecturer is wrong?

Mentor
siresmith
Ok, used the integrator and.... It came out as the standard integral suggests.

But why ia the integration by parts method wrong?? Im sure i did it right. does integration by parts not work on certain functions or something wierd like that? Or did i just get it wrong- what do you get?

Staff Emeritus
Gold Member
But why ia the integration by parts method wrong?? Im sure i did it right. does integration by parts not work on certain functions or something wierd like that? Or did i just get it wrong- what do you get?
If you show us your working, perhaps we could point out where you've gone wrong.

Homework Helper
Hi siresmith!

If you integrate $$\int x^2 e^{-2Bx^2}$$ by parts,

you still end up with $$\int e^{-2Bx^2}$$ … didn't you?

DavidWhitbeck
Feynman trick-- take the derivative of I with respect to B.

$$I= \int e^{-2Bx^2}$$

If you know that integral already you will already be done. If not integrate your integral $$\frac{dI}{dB}$$ by parts and to find it as an expression in terms of I. Solve the de by separating and integrating.

But you should already know what I is, it's a classic result and it must have been given to you.

I am right that in integration by parts you let u= x and $dv= xe^{-Bx^2}$?