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2 answers to same integral?

  • Thread starter siresmith
  • Start date
  • #1
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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 cant 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!
 

Answers and Replies

  • #2
Borek
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  • #3
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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
Hootenanny
Staff Emeritus
Science Advisor
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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.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
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Hi siresmith! :smile:

If you integrate [tex]\int x^2 e^{-2Bx^2}[/tex] by parts,

you still end up with [tex]\int e^{-2Bx^2}[/tex] … didn't you? :confused:
 
  • #6
Feynman trick-- take the derivative of I with respect to B.

[tex]
I= \int e^{-2Bx^2}
[/tex]

If you know that integral already you will already be done. If not integrate your integral [tex]\frac{dI}{dB}[/tex] 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
HallsofIvy
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Homework Helper
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I am right that in integration by parts you let u= x and [itex]dv= xe^{-Bx^2}[/itex]?
 
  • #8
Yup and the boundary term vanishes, the odd integral vanishes and you're left with something like I/2B.
 

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