1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    thanks
     

    Attached Files:

  2. jcsd
  3. Dec 12, 2009 #2
    Let
    [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
    [tex]\frac{dP(x^2)}{dx}[/tex]
    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

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

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    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
    Thanks
    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

    User Avatar
    Staff Emeritus
    Science Advisor

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

    [tex]
    \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)
    [/tex]

    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:

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

    Integrating by parts,

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

    or

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

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

    [tex]\aligned
    \frac {dF(x)}{dx}
    &= 2x e^{-x^3} -
    \frac 1{2x}
    \left(
    \int_{\infty}^{x^2} e^{-xt^2}\,dt -
    \left(\left. t e^{-xt^2}\right|_{t=\infty}^{x^2}\right)
    \right)
    &= \frac 5 2\,x e^{-x^3} - \frac {F(x)}{2x}
    [/tex]
     
  8. Dec 13, 2009 #7
    Thanks makes a lot of sense
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook