Differentiation under the integral sign

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.

himanshu121

How it is done

Consider
[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

Tron3k

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

MathematicalPhysicist

what is written about this in the book, is there a technical explanation about it?

Muzza

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...