Differentiation under the integral sign

  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. 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. Funny, I was just reading Surely You're Joking, Mr. Feynman and I was wondering about that also.
  5. what is written about this in the book, is there a technical explanation about it?
  6. 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
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?