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Substitution in integral

  1. Feb 5, 2006 #1

    I've got a problem I've been working on for hours.

    I get a clue;

    If the integral (from zero to infinity) of e^(-x^2) is sqrt(pi)/2, what is
    the integral (from zero to infinity) of e^(-bx^2)?

    I've tried substitution, but I kind of got it wrong. If x = y/sqrt(b), I get the same integral as in the clue. But then I'm stuck with a 1/sqrt(b) which I cant get rid of. Anyone up for the challenge? Thanks..
  2. jcsd
  3. Feb 5, 2006 #2


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    Homework Helper

    What do you mean get rid of? sqrt(b) is a constant, and it will appear in the final answer.
  4. Feb 5, 2006 #3
    But is sqrt(b) a constant when it is the same thing as y/x? Can I move it outside the integral?
  5. Feb 5, 2006 #4
    I'll take the whole problem:

    The integral (from zero to +infinity) of c*x^2*e^(-bx^2) dx = 1

    I get the clue: integral (from zero to +infinity) of e^(-x^2) = sqrt(pi) / 2

    What is c?
  6. Feb 5, 2006 #5

    George Jones

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    Staff Emeritus
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    Gold Member

    You know

    [tex]\frac{\sqrt{\pi}}{2} = \int_{0}^{\infty} e^{-x^{2}} dx.[/tex]

    Use the substitution [itex]u = \sqrt{b} x[/itex] to calculate

    [tex]I \left( b \right) = \int_{0}^{\infty} e^{-bx^{2}} dx[/tex]

    for any [itex]b[/itex]. Then differentiate with respect to [itex]b[/itex] both sides of

    [tex]I \left( b \right) = \int_{0}^{\infty} e^{-bx^{2}} dx[/tex]

    to find the integral that you want.

  7. Feb 5, 2006 #6
    Thanks alot George and StatusX. Appreciate you taking your time. :!!)
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