# Differentiation under the integral sign

by hliu8
Tags: differentiation, integral, sign
 P: 658 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