Two Answers to Same Integral: Which One is Correct?

[siresmith]

2 answers to same integral??!

Homework Statement

Ive been giong mental over this. There seems to be two different answers to the same integral. I can't work out which is correct.

Integral from 0 to infinity of:

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

Homework Equations

What is the right answer?

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?

Please help, i have an important exam tomorrow!

 

[Borek]

Try http://integrals.wolfram.com

 

[siresmith]

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

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

 

[Hootenanny]

But why ia the integration by parts method wrong?? I am sure i did it right. does integration by parts not work on certain functions or something weird 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.

 

[tiny-tim]

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.

<br /> I= \int e^{-2Bx^2}<br />

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.

 

[HallsofIvy]

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

 

[DavidWhitbeck]

Yup and the boundary term vanishes, the odd integral vanishes and you're left with something like I/2B.