In this video I integrate this horrible looking integral using series expansions and many other tricks. It has a very satisfying solution. Subscribe for more maths videos
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Use Feynman's trick let I(a)=int_0^(oo) [1-exp(-ax)]*exp(-x)/x dx I’(a)= int_0^(infity) exp[-(a+1)x] dx=1/(1+a) I(a)=ln(1+a)+c, I(0)=c=0 I(a)=ln(1+a) # The general form is int_0^(oo) [exp(-ax)-exp(-bx)]/x dx =ln(b/a) Original question can rewrite int_0^(oo) [exp(-x)-exp(-(1+a)x)]/x dx =ln(1+a)
this is great! Thank you!
You're very welcome!
Hey there, i looked up your name on insta, but only found a profile with a profile picture showing a siloutte of a guy standing in what seems to be some sort of concrete hallway. is that you?
No it is not haha, my insta is @jagoalexander_
Use Feynman's trick
let I(a)=int_0^(oo) [1-exp(-ax)]*exp(-x)/x dx
I’(a)= int_0^(infity) exp[-(a+1)x] dx=1/(1+a)
I(a)=ln(1+a)+c, I(0)=c=0
I(a)=ln(1+a) #
The general form is
int_0^(oo) [exp(-ax)-exp(-bx)]/x dx
=ln(b/a)
Original question can rewrite
int_0^(oo) [exp(-x)-exp(-(1+a)x)]/x dx
=ln(1+a)
Fabulous!!!
The Feynman "trick" is an embarrassment of riches