# 2 answers to same integral?

1. May 15, 2008

### siresmith

2 answers to same integral??!!

1. The problem statement, all variables and given/known data

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))

2. Relevant equations

What is the right answer?

3. 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!

2. May 15, 2008

### Staff: Mentor

3. May 15, 2008

### 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?

4. May 15, 2008

### Hootenanny

Staff Emeritus
If you show us your working, perhaps we could point out where you've gone wrong.

5. May 15, 2008

### 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?

6. May 15, 2008

### 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.

7. May 15, 2008

### HallsofIvy

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

8. May 15, 2008

### DavidWhitbeck

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

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