# Differentiation under the integral sign

1. Jan 21, 2004

### hliu8

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. Jan 21, 2004

### himanshu121

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

Last edited: Jan 21, 2004
3. Jan 22, 2004

### Tron3k

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

4. Jan 23, 2004