Technically, the integral of 1/sinh does not diverge, precisely because in your parametrization, you chose a symmetric interval with a single parameter ε, and the integrand is odd, so the integral is always 0, for all ε and R. So in the limit as ε -> 0, the integral is 0. It would have been a problem if the intervals were not symmetric, but then the contour integral over γ would not have been doable. So actually, everything is okay here, and there is no divergence.
🤓🤓 Honestly, I didn't expect that! These integral is super cool and this video is actually so great (definitely I'll back and watch it again). I learned and enjoyed (especially when you use two methods). Thank you so much dear *QN³* ❤️
I took x to lnx and took partial fractions and simplified to get 2lnx*(1/(x-1) + 1/(x+1)) and the bounds now go from 0 to infinity. I tried integration by parts and it started to look really circular and also divergent, edit: corrected some mistakes. Interestingly from zero to 1 the first integral ln(x)/(x-1) equals pi^2/6 exactly! And both integrals combined equal pi^2/4 ! Wow! After everything including the factor of 2 from Sinh you wind up with pi^2 again, and the combined integral from 0 to 1 is equal to the integral from 1 to infinity, and the indefinite integral is in terms of two dilogarithms and the product of two functions
In the calculation,there was a pole at the origin too, why you didn't avoid it by making a semicircular detour around it, like the one you have made at around z = +iπ ?
Technically, the integral of 1/sinh does not diverge, precisely because in your parametrization, you chose a symmetric interval with a single parameter ε, and the integrand is odd, so the integral is always 0, for all ε and R. So in the limit as ε -> 0, the integral is 0.
It would have been a problem if the intervals were not symmetric, but then the contour integral over γ would not have been doable. So actually, everything is okay here, and there is no divergence.
Superior teaching, with this, you made something I was unsure about into child’s play (almost). The added bonus at the end is sincerely appreciated.
🤓🤓
Honestly, I didn't expect that! These integral is super cool and this video is actually so great (definitely I'll back and watch it again).
I learned and enjoyed (especially when you use two methods).
Thank you so much dear *QN³* ❤️
Your channel is the best.
:D
I must be too lucky to have discovered your channel
🤓 I love this video.
Thank you dear *QN³* for bringing it to us 💗
Thank you Sir.
Happy New Year
I took x to lnx and took partial fractions and simplified to get 2lnx*(1/(x-1) + 1/(x+1)) and the bounds now go from 0 to infinity. I tried integration by parts and it started to look really circular and also divergent, edit: corrected some mistakes.
Interestingly from zero to 1 the first integral ln(x)/(x-1) equals pi^2/6 exactly! And both integrals combined equal pi^2/4 ! Wow! After everything including the factor of 2 from Sinh you wind up with pi^2 again, and the combined integral from 0 to 1 is equal to the integral from 1 to infinity, and the indefinite integral is in terms of two dilogarithms and the product of two functions
In the calculation,there was a pole at the origin too, why you didn't avoid it by making a semicircular detour around it, like the one you have made at around z = +iπ ?
It is a removable singularity
Dear QN^3, I wish you a happy new year!
Could you please do the same integral over exp(i*a*x)*cosh(x)/sinh(x).
👍
Incredible !!! Can you recommend any good books for self study in complex analysis?
Happy New Year
Hello Dear *QN^3*
Happy New Year
U should have made the height of this rectangle 3π/2
It makes the integral much simpler