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Differentiating Under Integral Sign

  1. Feb 8, 2016 #1
    I(α) = 0∫e-(x2+α/x2) dx

    Differentiating under the integral sign leads to:

    I(α) = 0∫-e-(x2+α/x2)/x2 dx

    Here I am supposed to let u = sqrt(a)/x, but the -x2 doesn't cancel out,

    Wolfram-Alpha tells me the answer is: e(-2 sqrt(α) sqrt(π))/(2 sqrt(α)). I understand where the sqrt(π))/(2)sqrt(α) comes from, but not the 2 sqrt(α)) in the numerator.

    Thanks!
     
    Last edited: Feb 8, 2016
  2. jcsd
  3. Feb 8, 2016 #2

    mfb

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    Differentiating what with respect to what, and why?
    Your integral is missing something ("dx"?).
     
  4. Feb 8, 2016 #3

    HallsofIvy

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    First, please, please, please, include a "dx" in the integral!
    Yes, the derivative, with respect to "a", of [itex]\int_0^\infty e^{-x^2- a/x^2}dx[/itex] is [itex]-\int_0^\infty \frac{e^{-x^2- a/x^3}}{x^2}dx[/itex].

    Is your question about actually doing that integration? Are you required to?
     
  5. Feb 8, 2016 #4
    Yes, with dx. Sorry! (Also, how do you format equations?)

    For this problem, I am supposed to find I(1) using u = sqrt(x)/a and the answer is e-2sqrt(π)/2 .
     
  6. Feb 9, 2016 #5

    Samy_A

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    You have ##\displaystyle l(a)=\int_0^\infty e^{-x^2- a/x^2}dx##
    Then ##\displaystyle l'(a)=-\int_0^\infty \frac{e^{-x^2- a/x^2}}{x^2}dx##
    Apply the substitution ##u=\frac{\sqrt a}{x}## to this last integral, and compare what you get with the expression for ##l(a)##.
    This will give you an easy differential equation for ##l##.

    You will have to use the value of the Gaussian integral at some point:
    ##\displaystyle l(0)=\int_0^\infty e^{-x^2}dx=\frac{\sqrt \pi}{2}##.

    See the LaTeX link at the bottom left of the edit box.
     
    Last edited: Feb 9, 2016
  7. Feb 9, 2016 #6
    Right, so I have [itex] I'(α) = 1/\sqrt{α}\int_0^∞ e^{-x^2 - u^2} du = 1/\sqrt{α}\int_0^∞ e^{-a/u^2 - u^2} du [/itex], which looks just like the [itex]l(a)=\int_0^\infty e^{-x^2- a/x^2}dx[/itex]

    I don't understand what to do next.
     
  8. Feb 9, 2016 #7

    Samy_A

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    I also have a minus sign, giving:

    ##\displaystyle I'(\alpha)=-\frac{1}{\sqrt \alpha}\int_0^\infty e^{-x^2- \alpha /x^2}dx=-\frac{1}{\sqrt \alpha}I(\alpha)##

    Now solve the differential equation ##I'(\alpha)=-\frac{1}{\sqrt \alpha}I(\alpha)##.
     
    Last edited: Feb 9, 2016
  9. Feb 9, 2016 #8
    Got it, thanks so much!
     
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