This is one of the coolest integrals ever solved

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[@sammtanX]

"terribly sorry about that"
"ouhhkayy cool"

[@Calcprof]

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

[@xleph2525]

What is this cult of Nahari you speak of? How might I seek out membership?

[@Calcprof]

@@xleph2525 en.wikipedia.org/wiki/Zeev_Nehari

[@maths_505]

Other than that, nihari is also the national dish of Pakistan

[@tomasstride9590]

@@maths_505 One thing that confuses me is why when finding the residue you did not just pick out the 1/z term from the series for zeta(1+z)/z . The expression was already in the form of a Laurent series so there was no need to do any work at all.

[@maths_505]

@@tomasstride9590 better for teaching purposes to be more detailed. Perhaps someone watching the video isn't familiar with contour integration so doing a bit of extra work will encourage them to do a bit of research without feeling too lost.

[@orionspur]

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

[@maths_505]

@@orionspur I've never felt more proud of the way I write zeta 😂

[@mcalkis5771]

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

[@jamiepianist]

You'll get it next time!

[@maths_505]

@@mcalkis5771 you'll get it next time bro

[@theelk801]

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

[@CliffHanger-fg6uy]

Exactly. I wanted to see if someone else had pointed it out.
Way over-complicated the process of finding the residue.
At least the way I was taught, the residue is originally just the coefficient of z^-1 in the Laurent series. From that, the formula the video gives using derivatives and multiplication by z is easily derived.

[@Klemvar]

agreed, once a Laurent expansion is given there is no need to do further computations (so this integral depends only on the Laurent expansion of the zeta function)

[@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.

[@kingzenoiii]

WHAT A BEAUTY

[@xanterrx9741]

Beautiful result , you've made amazing video

[@jaycash4381]

Thank you so much man!

[@MathsScienceandHinduism]

Solving integrals can be a hobby

[@MrWael1970]

This is awesome. Thank you.

[@vascomanteigas9433]

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

[@michaelihill3745]

Beautiful solution!

[@maths_505]

@@michaelihill3745 thank you my friend

[@amirb715]

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

[@theelk801]

yeah came here to say the same thing

[@dannymooren9160]

Crazy integral bro

[@taterpun6211]

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

[@CliffHanger-fg6uy]

I don’t think he says anything technically incorrect. The only pole encircled by the contour is z = 0, and it is a repeated pole of order 2. That’s what he says.
The only complaint should be that his presentation is slightly misleading by boxing just the 1/z^2 term without explicitly mentioning the 1/z contribution. Still, I don’t think anything was incorrect.
The missed opportunity was not just directly observing the residue from the first Laurent series given. By definition, the residue is just the coefficient of the 1/z term. The whole derivative procedure was needless.

[@asparkdeity8717]

The definition of a pole of order 2 is that your Laurent expansion has a non-zero z^-2 coefficient as the one with lowest power, so the z^-1 term is not to be worried about. In fact even better if u know the z^-1 coefficient, which is just the definition of residue and u rightly point out is γ straight away

[@threepointone415]

This one's going down on the history books

[@zyklos229]

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.

[@yoav613]

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

[@NurBiswas-fc6ty]

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

[@serdarakalin2209]

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

[@nickkrempel5888]

That's not correct. The initial integral is just a standard integral of a function ℝ→ℂ, which is equivalent to two independent integrals ℝ→ℝ (the real and imaginary parts). This is very different from a contour integral (which in parametrized form involves an additional complex multiplication by the derivative of the parametrizing function - incidentally these are also both different from a 2d line integral, which involves an additional real multiplication by the norm of the derivative). ζ has a simple pole at z=1 with residue 1, not γ, and also the residue theorem has a factor of 2πi not 2π.

[@danielc.martin]

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

[@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?

[@dnsfsn]

Actually since its 1+z, none of the points of the contour are poles of the zeta function, which only has a pole at z=1.

[@lagnugg]

@@dnsfsn oh i get it now. thanks for responding!

[@marioyard]

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

[@montreearmy]

Gamma subscription k. What does it mean or what is their representing?

[@antoine2571]

"math people of youtube" lmao

[@paulmenard3232]

How do you even find those integrals in tbe first place?

[@maths_505]

I play around with functions, integrals and series quite alot. There's another write-up on my patreon showcasing such playfulness leading to fascinating results.

[@dariuszpanchyrz2784]

Amazing... 9nce again

[@unturnedd]

damn that one is cool

[@maths_505]

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[@michallichmira8458]

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

[@maths_505]

@@michallichmira8458 well in about a year and a half I'll have a masters degree

[@michallichmira8458]

@@maths_505 congrats

[@mikeoffthebox]

Nice way to build your Balrog audience.

[@maths_505]

Math is inclusive after all

[@jeremyjedynak]

The result is even more beautiful if you replace 2π by τ.

[@maalikserebryakov]

Cope

[@lextratherese7277]

The 25th "one of the most beautiful integral i've solved on this channel" this year 😁

[@SkorjOlafsen]

In hindsight, I guess I shouldn't be surprised by the result, given how connected ɣ is to ζ(1).