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Homework Help: 2 answers to same integral?

  1. May 15, 2008 #1
    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. jcsd
  3. May 15, 2008 #2


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    Staff: Mentor

  4. May 15, 2008 #3
    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?
  5. May 15, 2008 #4


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    If you show us your working, perhaps we could point out where you've gone wrong.
  6. May 15, 2008 #5


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    Homework Helper

    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:
  7. May 15, 2008 #6
    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 [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.
  8. May 15, 2008 #7


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    I am right that in integration by parts you let u= x and [itex]dv= xe^{-Bx^2}[/itex]?
  9. May 15, 2008 #8
    Yup and the boundary term vanishes, the odd integral vanishes and you're left with something like I/2B.
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