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Derivative of an integral function

  1. Dec 4, 2012 #1
    Hello there,

    I have the function

    $$f(t) = \int_0 ^{t} \frac{1}{\sqrt{t-\tau}} \mathrm{d}\tau$$

    The integral can be found, yielding an expression as

    $$f(t) = 2 \sqrt{t} $$

    The question is, if I tried to calculate the derivative of $$f$$, using the "derivation under the integral " rule I get

    $$\frac {\mathrm{d} f }{\mathrm{d}t} = \int_0 ^{t} \frac{\mathrm{d}}{\mathrm{d}t} \frac{1}{\sqrt{t-\tau}} \mathrm{d}\tau + \frac{1}{\sqrt{t-t}} $$

    which is a problem I never run across.
    I went through the derivation of the "derivation under the integral " but I am not getting anywhere, for the moment.
    Anybody's help would be appreciated, thanks
  2. jcsd
  3. Dec 4, 2012 #2


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    Homework Helper

    Loosely speaking, the first term with the derivative inside the integral will generate a "##-1/\sqrt{t-t}##" term, as ##d(t-\tau)^{-1/2}/dt = -d(t-\tau)^{-1/2}/d\tau##, so the diverging terms cancel out.

    If you change variables to ##y = t - \tau## before taking the t derivative you also avoid the problem altogether.
  4. Dec 4, 2012 #3

    as always, much appreciated.

  5. Dec 4, 2012 #4


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    Science Advisor
    Gold Member
    2017 Award

    you can write out the Newton quotient in terms of t then write

    (t + Δt-[itex]\tau[/itex])[itex]^{-1/2}[/itex] in a Taylor series
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