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Derivative of a function with variable also in the integrand

  1. May 3, 2012 #1
    F(y)=Integral(e^{-x^{2}y^{2}}dx between y and 0 (where y>0). I know you have to make substitutions with y, but then you have to make a further substitution along the way and that's where I'm a little lost. Can anyone help point me in the right direction? I know you can set u=-y^{2} and v=y in the integrand, but, as you'll see, you get stuck and you need to make another substitution somewhere.

    (The stupid symbol commands won't work for me, so sorry if it's hard to read the function)
     
    Last edited: May 3, 2012
  2. jcsd
  3. May 3, 2012 #2


    Did you mean [itex]\displaystyle{\int_0^y e^{-x^2y^2}dx}[/itex] ? I don't think there's an easy way to do this as it is non nomalized Gauss's bell function.

    If, for example, you'd have [tex]\int_0^\infty e^{-x^2y^2}dx=\frac{1}{y}\int_0^\infty e^{-u^2}du=\frac{\sqrt{\pi}}{2u}[/tex]

    DonAntonio
     
  4. May 3, 2012 #3

    mathman

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    Your discussion is very unclear. Based on the title it seems that you can take ey2 outside the integral, so you then need to get the derivative of a product.
     
  5. May 3, 2012 #4
    Yes, this is what I meant. I know you have to make a third substitution at some point, I'm just not sure how.
     
  6. May 3, 2012 #5


    Not really, only one: [tex]x^2y^2=u^2\Longleftrightarrow x=\frac{u}{y}\Longrightarrow dx=\frac{du}{y}[/tex] and voila. Of course, we're implicitly using the fact that the variable(s) is (are) all positive.

    DonAntonio
     
  7. May 4, 2012 #6

    HallsofIvy

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    The general Leibniz formula is
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)}\phi(x, t)dt= \beta'(x)\phi(x, \beta(x))- \alpha'(x)\phi(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial \phi}{\partial x}dx[/tex]
     
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