This is one of the coolest integrals ever solved

My complex analysis lectures:

Support my work and get early access to content via Patreon:

You can follow me on Instagram for write ups that come in handy for my videos and DM me in case you need math help:

My LinkedIn:

Advanced MathWear:

[Calcprof]

When I took Complex Analysis from Nahari, I couldn't tell his ζ's from his ξ's. Like yours. But great integral!

[sammtanX]

"terribly sorry about that"
"ouhhkayy cool"

[sbares]

4:36 No need to calculate the residue like that. The residue is the -1st term of the Laurent expansion (this can be, and often is, taken as the *difinition* of the residue), and you already helpfully wrote that up.

[orionspur]

Very cool. Also appreciating that your Zeta looks just like the Tasmanian Devil. 🌪️

[theelk]

you didn’t need to take the derivative, you already had the laurent series so you just find the coefficient of z^-1

[mcalkis]

Failed my driving exam today. These videos really help me cope. Thank you Kamal.

[xanterrx]

Beautiful result, you've made amazing video

[vascomanteigas]

The Stietjes constants are evaluated by replacing I*theta by I*n*theta and multiply by n!/(2*pi)

[threepointone]

This one's going down on the history books

[NurBiswas-fcty]

You should start the project of writing a book about integration, it will be a great work.

[taterpun]

4:03 I believe there is a slight mistake, you forgot the γ/z pole for k=0 in the sum. When contouring, the non-1/z terms would cancel, leading to I = 2πγ. But your calculations after are equivalent, so the same answer is still achieved

[yoav]

Very nice!! Can you please solve the integral ((i)^tanx )(tanx)^i from 0 to pi, the result is pi e^(-pi).thanks💯

[zyklos]

So actually the Laurent Series does the trick. Since if you wouldnt know, you would have to "invent" / calculate this constant(s), which is probably just another sum with some limit, hard to calculate.

[amirb]

you already have the Laurent series, the coefficient of 1/z is the residue. why are you going through all that derivative crap

[serdarakalin]

Kamal, z hast a simple Pole at 1 with the residue of Gamma, cauchy integral Theorem says the contour integral ist sum of residues Times 2pi IS the value of the integral, 1+ e**(i * Theta ) ist the circle centered at 1, you can calculate directly from cauchy

[danielc.martin]

you could have ued tau instead of 2pi and would be even more cool xd

[marioyard]

thank you for this fantastic integral.Mario more and more ...Ok cool!!!

[lagnugg]

i have one question: on a unit circle |z| = 1 there are 2 points where Re(1+z) = 1, and these are the points where ζ(1+z) is not defined. why can we just ignore these points when we apply cauchy's residue theorem?

[michallichmira]

What is yours scientific degree you have got skills Man congrats take care

[maths_]

To get early access to content:
You can follow me on Instagram for write ups that come in handy for my videos: