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Differentiation under the integral sign

by hliu8
Tags: differentiation, integral, sign
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Jan21-04, 05:20 PM
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 and Feynman's autobiography, about evaluating an integral by differentiation under the integral sign, but how exactly it is done.

Thank to everyone.
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Jan21-04, 11:34 PM
himanshu121's Avatar
P: 658
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
Jan22-04, 09:00 PM
P: 45
Funny, I was just reading Surely You're Joking, Mr. Feynman and I was wondering about that also.

Jan23-04, 09:01 AM
P: 3,243
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.
what is written about this in the book, is there a technical explanation about it?
Jan23-04, 09:09 AM
P: 695
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...

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