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Taking the derivative of a definite integral

  1. Aug 21, 2008 #1
    How do I do it? For example, if I have:

    [tex]\int_{0}^{\infty}sf(x) dx[/tex]

    How do I take the derivative with respect to x?

    I was trying to derive the formula for an inverse laplace transform when I realized that I didn't know how to take the derivative of a definite integral.
  2. jcsd
  3. Aug 21, 2008 #2


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    [tex]\frac{d}{dx} \left( \int_{0}^{\infty}sf(x) dx \right) = 0[/tex]
    because the bracketed thing does not depend on x (it is just a number). It doesn't matter whether the integral is definite or not by the way.

    Note however, that
    [tex]\frac{d}{da} \left( \int_{0}^{a}sf(x) dx \right) \neq 0[/tex]
    [tex]\frac{d}{db} \left( \int_{0}^{\infty}s f(x, b) dx \right) \neq 0[/tex]
    [tex]\frac{d}{ds} \left( \int_{0}^{a}sf(x) dx \right) \neq 0[/tex]
    in general!
  4. Aug 21, 2008 #3

    Interesting, but I’m not so sure about the last part of your statement. Since the indefinite integral does not result in just a number but an algebraic. If it is an indefinite integral, then the derivative of the integral of f(x) should be f(x), as the two operations cancel each other.

    As far as the derivative of the definite integral, once you reduce the definite integral to just a number, then certainly the derivative is zero. However, what if you kept it in algebraic form? Then the question would be at what value do you evaluate the derivative? Can you not use the MVT to determine where to evaluate the derivative?
  5. Aug 21, 2008 #4


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    You are right, I don't know what I was thinking. Of course, [itex]\int f(x) dx[/itex] still depends on x. Never mind that part.

    You mean something like this?
    [tex]\frac{d}{dx} \left( \int_a^b f'(x) \, dx \right) = \frac{d}{dx}( f(b) - f(a) ) = 0[/tex]

    Otherwise, it's not quite clear to me what you mean.
  6. Aug 21, 2008 #5
    To be candid, I am not 100% sure of what I mean either,:confused: but something is tugging at my brain concerning taking the derivative of a definite integral.

    Maybe if I try an example that will at least clarify what I am thinking of.
    Suppose we have a simple f(x) = x^2 Now we take the definite integral of that within the limits x = 3 and x = 4 . The numerical result is 12.333…
    And of course if we differentiate that number the result is zero. Nothing earth shaking about any of that.

    But let’s leave the integral of x^2 in its algebraic format so we have this:
    (x^3/3 where x = 4) – (x^3/3 where x =3) and Not evaluate that but differentiate it as it is which results in:
    (x^2 where x = 4) – (x^2 where x = 3) and then we evaluate that we have
    16 – 9 = 7.

    The significance of this comes in when we consider the MVT which states
    ( f(4) – f(3) ) / 4 – 3 = Df(c) where c is the mean value on the interval from 3 to 4.
    For f(x) = x^2, Df(x) = 2X
    Putting it all together then for f(x) = x^2, 7 = 2x and x = 3.5 which is the mean value on the interval.

    So, I come to my very tentative conclusion that the derivative of a definite integral of f(x) should be f(x) evaluated at the mean value on the interval defined by the limits of the definite integral. At least, this interpretation gives some meaning to the limits of integration of the definite integral which carries over into the derivative.

    I am interested in getting your reaction to this idea no matter how critical.:smile:
  7. Aug 21, 2008 #6


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    You might be thinking of the very general "Leibniz' formula":
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta(x)}{dx}f(x,\beta(x))- \frac{d\alpha(x)}{dx}f(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x}dt[/tex]
    One result of that is that
    [tex]\frac{d }{dx} \int_a^b f(t)dt= 0[/tex]
    because all of those derivatives are 0.

    (Edited to correct sign error. Thanks, NoMoreExams.)
    Last edited by a moderator: Aug 21, 2008
  8. Aug 21, 2008 #7
    Are your limits correct?
  9. Aug 21, 2008 #8

    D H

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    This algebraic form?

    [tex]g(s) = \left( \int_{0}^{\infty}sf(x) dx \right) = s\int_{0}^{\infty}f(x) dx[/tex]

    This is just a number times s; it doesn't depend on x at all, so the derivative wrt x is zero.

    Halls is correct. You can google "Liebniz integral rule" for more info.
  10. Aug 21, 2008 #9
    I did :)
  11. Aug 21, 2008 #10
    Well, the actual problem I want to solve is:

    [tex]F(s) = \int_{0}^{\infty} e^{-st}f(t) dt[/tex]

    for f(t). Now I want to take the derivative of both sides, and then multiply by [tex]e^{-st}[/tex], and then get f(t).

    Now I know from looking at Laplace transforms that the infinity part always evaluates to 0, so in reality the laplace transform is just that integral evaluated at 0.

    So I have:

    [tex]F(s) = \int e^{-st}f(t) dt \left|t = 0[/tex]

    But the problem is that I can't just plug in the infinity or 0 because I want to take the derivative of both sides, and I can't just take the derivative because the integral is definite. This is going to sound really stupid but is there a way to cancel the "plug in" operation?

    And just to let you guys know my most advanced math class is differential equations.
  12. Aug 21, 2008 #11


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    I don't think that's going to work. Probably you have to pull some trick like
    [tex]\int_{\cdots}^{\cdots} e^{-a x} dx = \delta(a)[/tex]
    (sorry haven't done math in like 6 weeks, bit rusty here) and use that to write some integral over F(s) which in the end gives you [tex]\int f(t) \delta(t - s) dt = f(s)[/tex].
  13. Aug 21, 2008 #12


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    I am now. I edited what I had before because of NoMoreExams' question!
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