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Differentiation under the integral sign 
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#1
Jan2104, 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 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
Jan2104, 11:34 PM

P: 658

How it is done
Consider [tex]I(b)=\int_0^1 \frac{x^b1}{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 


#3
Jan2204, 09:00 PM

P: 45

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



#4
Jan2304, 09:01 AM

P: 3,243

Differentiation under the integral sign



#5
Jan2304, 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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