# Differentiation under the integral sign

by hliu8
Tags: differentiation, integral, sign
 P: 3 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.
 P: 662 How it is done Consider $$I(b)=\int_0^1 \frac{x^b-1}{lnx} dx$$ 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 $$I'(b)=\int_0^1 \frac{x^b lnx}{lnx} dx$$ $$I'(b)=\int_0^1 x^b dx=\frac{1}{b+1}$$ $$=> I(b)= \int \frac{1}{b+1} db +c$$ 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
 P: 45 Funny, I was just reading Surely You're Joking, Mr. Feynman and I was wondering about that also.
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## Differentiation under the integral sign

 Originally posted by Tron3k Funny, I was just reading Surely You're Joking, Mr. Feynman and I was wondering about that also.