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3 Challenging Integration problems

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data
    1. Show that [tex]\int_0^\infty x^{n}e^{-ax}dx = \frac{n!} {a^{n+1}}[/tex]
    for n = 0, 1, 2, 3...

    2. Show that [tex]\int_{-\infty}^\infty x^{2n}e^{-ax^{2}}dx =\frac{{\surd \pi} (2n-1)!!} {2^{n}a^{(2n+1)/2}}[/tex]
    for n = 0, 1, 2, 3...

    Assumption: [tex]\int_{-\infty}^\infty e^{-ax^{2}}dx =\surd \frac{\pi} a [/tex]

    3. Evaluate [tex]\int {\frac{1} {A^{x^2}+Bx+C}} dx [/tex]
    For all possible real values of A, B, C.

    For #1 and #2, you may use mathematical induction, if you like.

    Notation: 7!! = 7 * 5 * 3 * 1

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Jun 11, 2008
  2. jcsd
  3. Jun 11, 2008 #2
    1) Use gamma function or integration by parts
    2) Use gamma function (or induction)
    3) Seperate the problem in certain cases.
  4. Jun 11, 2008 #3


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    Since you post them here, I assume that you want to solve these yourself and haven't posted them as challenge for us :)

    For the first one, try partial integration.
    For the second one, try differentiating [tex]\int e^{-a x^2}[/tex] (this is a familiar integral in physics).
    For the third one, I have no idea yet (but solve 1 and 2 first :biggrin:)
  5. Jun 11, 2008 #4
    i actually would appreciate if you can finish the challenge
    gosh, I've spend so much time figuring these questions out.
    but i just can't get thru 'em
  6. Jun 11, 2008 #5
    I would have appreciated it if someone would have finished my questions....especially without trying it myself...
  7. Jun 12, 2008 #6

    Gib Z

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    Any one can figure out challenge questions to give someone but not be able to get through them. So don't "gosh" at us and take some advice.
  8. Jun 12, 2008 #7


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    It's not hard, we almost gave you the answer, you just have to do the algebra now.
    For example, let's look at the first one:

    \int_0^\infty x^{n}e^{-ax}dx = \frac{n!} {a^{n+1}}

    You can work from here and check what happens if you integrate by parts all the time (just write it out 2 or 3 times and you will see the pattern).

    But let's just set up a nice formal proof by induction.
    First check that it works for n = 1 (that's kinda trivial, just don't forget the boundary term!)
    Then suppose that
    \int_0^\infty x^{n}e^{-ax}dx = \frac{n!} {a^{n+1}}
    and try to integrate

    \int_0^\infty x^{n + 1}e^{-ax}dx
    by parts (hint: write [tex]x^{n+1}e^{-ax} = x \cdot (x^{n} e^{-ax})[/tex])
  9. Jun 12, 2008 #8


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    For the first one I much prefer differentiation to integration by parts: start with

    [tex]\int_{0}^{\infty} dx~e^{-ax},[/tex]

    evaluate that, and start differentiating both sides with respect to a to see the pattern arise. If you actually need a formal proof, just use induction as suggested above, but I would differentiate the expression again instead of doing it by integration by parts. Doing it this way avoids the annoying boundary terms from integrating. =)

    Of course, if you're allowed to just assume the answer straight away and then use induction, just use the differentiating method straight away in the induction proof without doing the differentiation to 'discover' the pattern first.
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