Contour Integrals: A Complex Proof of Reflection Formula

Hope everyone enjoyed! Longer style with this one but very rewarding complex proof - please comment with any questions or suggestions for new topics, and as always, subscribe to stay updated.
~ Thanks for watching!
Lets keep going to 2k!
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[@pokemil5705]

I'm so glad you turned the volume down slightly when the marker was squeaky. It's so nice, thanks :) And overall, it is a really nice video

[@ADDiOUMAARIR]

Very nice ❤, keep going my friend ❤❤❤

[@OscgrMaths]

@@ADDiOUMAARIR Thank you!

[@حسينالقطري-ب8ص]

I have encountered many proofs of this before, but struggled to grasp them fully.
Your explanation has finally clarified it for me. It remains rigorous yet you have made it remarkably engaging. I truly appreciate your clear and enjoyable explanation. Thank you very much.
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We know that
|a+b| ≤ |a| + |b|
Replacing b with -b, we get
|a-b| ≤ |a| + |-b|, but |'b|=|b|, so
|a-b| ≤ |a| + |b|. Rearranging:
|a-b| - |b| ≤ |a|. [*]
With the same argument, by replacing a with -a, we get
|a-b| - |a| ≤ |b|. [**]
Combining * and **, we get
||a|-|b|| ≤ |a-b|
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Well, I am not good at complex analysis, but I think the answer for (why) at 22:12, is because otherwise the function will no longer be analytic, and then we cannot integrate.
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Thanks again for this great video.

[@OscgrMaths]

@@حسينالقطري-ب8ص Thanks for the comment!

[@franolich3]

Very nice indeed! Well worth the wait!!!

[@OscgrMaths]

Thanks so much! Glad you enjoyed.

[@DavidMFChapman]

Good job!

[@OscgrMaths]

@@DavidMFChapman Thank you!!

[@Georgeclassified]

Awesome and rigorous Video as always! By the way I would like to make a suggestion about a future video you can consider making on the relationship between the zeta function and the dirichlet eta function OR on the analetic continuation of the zeta function and its contribution to the studying of primes and number theory(euler product formula for the zeta function)❤!

[@OscgrMaths]

Thanks so much! I'll definitely take a look at these thanks for the suggestion.

[@Georgeclassified]

@@OscgrMaths My pleasure😄

[@dean532]

Do you have huge whiteboards over there? Let me know! We can put one to use! 😅 May be on the street where people begin to think we’re mad scientists. Just kidding; it’s Just a whim*

[@dakcom-mk6mp]

Nice

[@OscgrMaths]

@@dakcom-mk6mp Thanks!

[@leofoxpro2841]

Hello, great videos!!

[@OscgrMaths]

@@leofoxpro2841 Thank you!

[@girianshiido]

You act as if the two horizontal line segments of the keyhole contour are on the real axis: they’re not.

[@OscgrMaths]

@@girianshiido Great question! Thats the why I parametrised with 0 and 2pi. Also, as epsilon reaches 0, they do approach the real axis given that the circle they are coming from essentially approaches a point at the origin. Hope that clarifies!

[@bahiihab-y2r]

hi sir i really enjoyed the video but i have a question concerning the branch cut: why the pole -1 (is a real number ) isn't branch cut i mean for example in the integral proposed by Math 505 int from 0 to inf cos(x))/(π^2 - 4x^2) the poles when are real he made a branch cut i hope that you understand what i mean

[@OscgrMaths]

@@bahiihab-y2r I know what you mean - my branch cut was only positive real axis so since the pole was negative it's okay. I could have chosen imaginary which would have changed the contour but I find real easier normslly

[@gregoriuswillson4153]

such a nice applications of contour integrations , btw may i know what book you use to learn this method

[@OscgrMaths]

To learn complex analysis and contour integration, Introduction To Complex Analysis by Priestley is a classic! Thanks for the comment.

[@gregoriuswillson4153]

@@OscgrMaths Thanks for the reply bro , i would love to try

[@niom-nx7kb]

13:44 is always going to be...negative :)

[@ishantkhanchi5539]

Your profile picture is really interesting. Is it an atom?

[@OscgrMaths]

I believe it is electron shells!

[@christoffel840]

Why is x between 0 and 1?

[@OscgrMaths]

@@christoffel840 Because 1-x and x have to be greater than 0 as inputs to the beta function. Hope that helps!

[@sis_sos]

Flammablemaths clones are rampant

[@skyblue4558]

How to reach you? Could you share an email?

[@OscgrMaths]

@@skyblue4558 Sure, you can email me at [e-mail]