Very helpful, thank you! Bounding the integral on the upper arc of the semicircle to show that it goes to 0 in the limit takes some extra work, though.
What about the convergence or real integral in the disk where we convert and integrate? If real integral doesn't convergence in the disk, it's impossible to do next steps.
This is way late but answering is helping me understand. The difference is that the upper arc of the semicircle contributes to the value of the integral. You are correct that the actual contour integral in the complex plane has the same value for both sets of bounds, but in the infinite case the value of the integral is entirely from the real line, as the upper arc contributes 0. The Wikipedia article on Residue Theorem was helpful for understanding the semicircle case.
WOW YOU JUST HELPED ME TO PREPARE FOR MY EXAMS.EVERYTHING NOW MAKES SENSE
Glad I could help you prepare
Love your applications, commentary and intuition. Thanks for the great videos!
Very helpful, thank you! Bounding the integral on the upper arc of the semicircle to show that it goes to 0 in the limit takes some extra work, though.
It was helpful
Thank You❤
thank you very much for the videos, really helpful
You're very welcome!
At 8.39 how do you get to those functions in the limit?
this integrals so nasty when my mom get in to the room i had to close the computer
lol
What about the convergence or real integral in the disk where we convert and integrate? If real integral doesn't convergence in the disk, it's impossible to do next steps.
Why does this only work for -∞ and ∞ ? If the boundaries is -10 and 10 we will get the same singularity in ∂R not the same answer…
This is way late but answering is helping me understand. The difference is that the upper arc of the semicircle contributes to the value of the integral. You are correct that the actual contour integral in the complex plane has the same value for both sets of bounds, but in the infinite case the value of the integral is entirely from the real line, as the upper arc contributes 0. The Wikipedia article on Residue Theorem was helpful for understanding the semicircle case.
Saved my day :D
Glad I could help!
why can you say cosz=real part of e^iz? if you solve directly for cos z, the answer will be the same?
might be a bit late but its because of the relation
e^iz = cos(z) + isin(z)
Where cos(z) is the real part, and sin(z) is the imaginary part
Real is dual to imaginary -- complex numbers are dual.
"Always two there are" -- Yoda.