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Derivative of integral

  1. Jul 3, 2007 #1
    what is
    d/dx of INT[ exp(-x^2/2*sin(t)^2) dt ] from 0 to pi/2
  2. jcsd
  3. Jul 3, 2007 #2


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    Staff: Mentor

    Welcome to the PF, mnaeem. Homework and coursework questions need to be posted in the Homework Help section of the PF (where I have moved it), and not in the general technical forums. Also, we require that you show your own work on the problem, before we can offer our tutorial help. We do not give out answers here on the PF.

    So what do you think is the way to approach this question? And what does it mean to take a derivative of a definite integral?
  4. Jul 3, 2007 #3


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    Staff: Mentor

    Is this what you mean (in LaTex)?

    [tex]\frac{d}{dx} \int^{\frac{\pi}{2}}_{0} e^{\frac{-x^2}{2}} sin^2(t) dt[/tex]

    Or do you mean?

    [tex]\frac{d}{dx} \int^{\frac{\pi}{2}}_{0} e^{\frac{-x^2}{2} sin^2(t)} dt[/tex]
  5. Jul 3, 2007 #4
    it should be this INT[ exp[-x^2/(2*sin(t)^2)] dt ] from 0 to pi/2

    this is the alternate form of classical Q function. I am trying to findout the derivative of this alternate function.
  6. Jul 4, 2007 #5
    I guess you could substitute u=-x^2/(2*sin(t)^2), and solve the integral wrt t and then differentiate that wrt x.
  7. Jul 4, 2007 #6
    I want to clarify what berkeman mentioned, are you actually taking the derivative of the DEFINITE integral of this function? If so then what is ANY definite integral, I mean what type of thing is it?
  8. Jul 4, 2007 #7
    That's not a definite integral. It has an x in it, hence it will be a function of x.
  9. Jul 4, 2007 #8

    D H

    Staff: Mentor

    It's still a definite integral. The integration is being performed with respect to t, not x.

    There are two ways to solve this problem: integrate first and then differentiate, or differentiate via Leibniz's integral rule and then integrate.
  10. Jul 4, 2007 #9


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    Staff Emeritus
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    For this "simple case", Leibniz integral rule reduces to
    [tex]\frac{d}{dx} \int_a^b f(x,t)dt= \int_a^b\frac{\partial f}{\partial x} dt[/tex]
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