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Derivative of an integral

  1. Dec 12, 2009 #1
    hey i'm having problems trying to find the derivative of the function

    i'm not sure how the seperate the x out of the integral

    i want to find the derivative of F(x) where
    [tex] F(x) = \int_{+\infty}^{x^2} e^{-xt^2} dt [/tex]


    Attached Files:

  2. jcsd
  3. Dec 12, 2009 #2
    [tex]P(x) = \int e^{-xt^2} \textrm{ d}t[/tex]
    (with arbitrary constant)
    Then you probably know how to find P'(x), but the hard part is finding
    Let s be the squaring function (i.e. let s(x) = x^2), then what you want to find is:
    [tex]\frac{d(P\circ s)(x)}{dx}[[/tex]
    which you can find using the chain-rule, and then you just do:
    [tex]F'(x) = \frac{dP(x^2)}{dx} - \left(\lim_{t \to +\infty} P(t)\right)'[/tex]
  4. Dec 12, 2009 #3
    Thanks i did do it the way you suggested but what confused me is that there is an e^-x

    under the integral. I was trying to find a way to it in a form so that only t was under the integral.
    For example when
    i had
    [tex] F(x) = \int e^{-(x +t)} dt [/tex]
    i got it into the following form

    [tex] F(x) = e^-x \int e^-t dt [/tex]
    from here i use the product rule to find the derivative of F(x).

    I thought i would have had to do something similar to move
    [tex] e^{-x} [/tex]
    outside the integral. or doesnt it matter in this case since i can treat as a constant in the power

    Last edited: Dec 12, 2009
  5. Dec 12, 2009 #4

    D H

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    The general rule for taking the derivative of a definite integral is the Leibiz integral rule:

    [tex]\frac d{dx}\int_{a(x)}^{b(x)} f(x,t) dt =
    \frac{db(x)}{dx} f(x,b(x)) \,-\, \frac{da(x)}{dx} f(x,a(x)) \,+\,
    \int_{a(x)}^{b(x)} \frac{\partial f(x,t)}{\partial x} dt[/tex]

    Applying that rule to this specific problem is not all that tough. (Is this homework?)

    BTW, did you really mean to use +infinity for the lower limit on your integral?
  6. Dec 12, 2009 #5
    no this not a homework question, its been many years since my first degree, i'm basically going through basic calculus notes to get ready for my post graduate course in mathematics in finance.
    i did mean to use +infinity, since the exponential is to the - ve power, this shouldnt be a problem since this will tend to 0 as t-> infinity. does this make sense or am i missing something
  7. Dec 12, 2009 #6

    D H

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    Since this is not homework, applying the Leibniz integral rule to this particular integral yields

    \frac {dF(x)}{dx} = \frac d{dx}\int_{\infty}^{x^2} e^{-xt^2}\,dt
    = 2x e^{-x^3} - \int_{\infty}^{x^2} t^2e^{-xt^2}\,dt \qquad\qqaud (1)

    That integral on the right-hand side can be re-expressed in terms of the original integral. The easiest way is to integrate the original integral, [itex]F(x)=\int \exp(-xt^2)\,dt[/itex], by parts:

    u &= e^{-xt^2} & du &= -2xte^{-xt^2}\,dt \\
    dv &= dt & v &= t\endaligned[/tex]

    Integrating by parts,

    \int e^{-xt^2}\,dt = te^{-xt^2} + 2x\int t^2 e^{-xt^2}\,dt


    \int t^2 e^{-xt^2}\,dt = \frac 1{2x}\left(\int e^{-xt^2}\,dt - te^{-xt^2}\right)

    Applying this in (1) yields (assuming I didn't make a dumb mistake somewhere),

    \frac {dF(x)}{dx}
    &= 2x e^{-x^3} -
    \frac 1{2x}
    \int_{\infty}^{x^2} e^{-xt^2}\,dt -
    \left(\left. t e^{-xt^2}\right|_{t=\infty}^{x^2}\right)
    &= \frac 5 2\,x e^{-x^3} - \frac {F(x)}{2x}
  8. Dec 13, 2009 #7
    Thanks makes a lot of sense
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