I Solving an Integral using Feyman's trick

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Hey guys ! I just need a little help on a integral I was trying to solve using feyman's technique.

This is the integral from 0 to 1 of (sin(ln(x))/ln(x) dx, which has been solved in one of the videos of bprp, but I'm trying to solve it using a different technique, and I end up with a different result, which is false of course, but the thing is I want to know where I messed up (not sure, but I have a feeling it's on point 5 where I calculate the constant)

Those are my steps :

-step 1 : I do a u sub and define my function I(alpha)

-step 2 : I do the partial derivative with respect to alpha of I(alpha)

-step 3 : I find the antiderivative of sin(u)exp(alpha*u)

-step 4 : I plug my antiderivative into my integral, and calculate it. Then I integrate back I'(alpha) to find back I(alpha)

-step 5 : I use the limit as alpha goes to -inf to make the integral equals to 0 to be able to calculate the constant

-step 6 : I use I(1) to calculate the integral, but in the end I find -3pi/4 instead of pi/4.Do you guys know where I messed up ? That would help me a lot, thanks !View attachment a27d24_721afbacfbfb4ce0905c9097e193e4e0_mv2.webp
 
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Bprp is an acronym for Youtuber BlackPenRedPen right?
 
Yes !
jedishrfu said:
Bprp is an acronym for Youtuber BlackPenRedPen right?
 
Because the limits of the integral are from ## x=-\infty## to ##x=0##, you want to take the limit as ##\alpha \to +\infty## to make it vanish. I think that'll fix your sign problem on the constant.
 
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But if I take the limit as alpha goes to +inf, wouldn't the integral diverges ? Because the limit as alpha goes to +inf of exp(alpha*x) is equal to +inf right ? But if I take the limit as alpha goes to -inf, then exp(alpha*x) will be equal to zero, and the integral would vanish right ?
 
The integral only converges when ##\alpha>0## because you're integrating over the negative ##x##-axis. The same issue applies when you take the limit.
 
Flamitique said:
Hey guys ! I just need a little help on a integral I was trying to solve using feyman's technique.

This is the integral from 0 to 1 of (sin(ln(x))/ln(x) dx, which has been solved in one of the videos of bprp, but I'm trying to solve it using a different technique, and I end up with a different result, which is false of course, but the thing is I want to know where I messed up (not sure, but I have a feeling it's on point 5 where I calculate the constant)

Those are my steps :

-step 1 : I do a u sub and define my function I(alpha)

-step 2 : I do the partial derivative with respect to alpha of I(alpha)

-step 3 : I find the antiderivative of sin(u)exp(alpha*u)

-step 4 : I plug my antiderivative into my integral, and calculate it. Then I integrate back I'(alpha) to find back I(alpha)

-step 5 : I use the limit as alpha goes to -inf to make the integral equals to 0 to be able to calculate the constant

-step 6 : I use I(1) to calculate the integral, but in the end I find -3pi/4 instead of pi/4.Do you guys know where I messed up ? That would help me a lot, thanks !View attachment 284090
When I went through it, I calculated the constant ##C## to be ##\pi/2## not ##-\pi/2##
 
Thank you vela for your help, now I understand ! So yes stevendaryl, now if I take the limit as alpha goes to +inf of the inverse tangent of alpha, C will be equal to pi/2, and the result of the integral is indeed pi/4 ! Thanks for your help!
 
Take some time to learn latex. It will really be worth it and will help us answer your questions better.
 
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