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Differentiation under the integral sign

  1. Jan 21, 2004 #1
    Hello everyone,
    This is my first post. I would like to understand better the idea of differentiation under the integral sign. I read about it in
    http://mathworld.wolfram.com/LeibnizIntegralRule.html and Feynman's autobiography, about evaluating an integral by differentiation under the integral sign, but how exactly it is done.

    Thank to everyone.
  2. jcsd
  3. Jan 21, 2004 #2
    How it is done

    [tex]I(b)=\int_0^1 \frac{x^b-1}{lnx} dx[/tex]

    now u can see clearly that after plugging the limits the variable x will vanish the only variable remains is b so the integration will be a function with b

    While integrating w.r.t x u consider b as a constant similarly when differentiating w.r.t b u consider x as a constant
    So , u have

    [tex]I'(b)=\int_0^1 \frac{x^b lnx}{lnx} dx[/tex]
    [tex]I'(b)=\int_0^1 x^b dx=\frac{1}{b+1}[/tex]
    [tex]=> I(b)= \int \frac{1}{b+1} db +c[/tex]

    If b=0 I(b)=0 => c=0

    Therefore I(b)=ln(b+1)

    So clearly it is afunction of b now with no x
    Last edited: Jan 21, 2004
  4. Jan 22, 2004 #3
    Funny, I was just reading Surely You're Joking, Mr. Feynman and I was wondering about that also.
  5. Jan 23, 2004 #4
    what is written about this in the book, is there a technical explanation about it?
  6. Jan 23, 2004 #5
    No, Feynman basically says that his "mathematical toolbox" (which included differentiation under the integral sign) was different from others', so he could solve problems others couldn't...
    Last edited: Jan 23, 2004
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