I Solving an Integral using Feyman's trick

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The integral from 0 to 1 of (sin(ln(x))/ln(x) dx was initially approached using Feynman's technique, but the user encountered a discrepancy in results. They suspected an error occurred during the calculation of a constant in step 5, where they took the limit as alpha approached negative infinity. Other participants clarified that the limit should actually be taken as alpha approaches positive infinity to avoid divergence and correctly calculate the constant. Ultimately, this adjustment led to the correct result of pi/4 for the integral. The discussion emphasized the importance of proper limit handling in integral calculations.
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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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