@@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.
@@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.
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.
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.
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)
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
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.
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
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
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.
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.
To get early access to content: www.patreon.com/Maths505 You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA==
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?
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.
"terribly sorry about that"
"ouhhkayy cool"
When I took Complex Analysis from Nahari, I couldn't tell his ζ's from his ξ's. Like yours. But great integral!
What is this cult of Nahari you speak of? How might I seek out membership?
@@xleph2525 en.wikipedia.org/wiki/Zeev_Nehari
Other than that, nihari is also the national dish of Pakistan
@@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.
@@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.
Very cool. Also appreciating that your Zeta looks just like the Tasmanian Devil. 🌪️
@@orionspur I've never felt more proud of the way I write zeta 😂
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.
you didn’t need to take the derivative, you already had the laurent series so you just find the coefficient of z^-1
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.
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)
Failed my driving exam today. These videos really help me cope. Thank you Kamal.
You'll get it next time!
@@mcalkis5771 you'll get it next time bro
WHAT A BEAUTY
Solving integrals can be a hobby
Beautiful result , you've made amazing video
Thank you so much man!
This is awesome. Thank you.
The Stietjes constants are evaluated by replacing I*theta by I*n*theta and multiply by n!/(2*pi)
Beautiful solution!
@@michaelihill3745 thank you my friend
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
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.
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
you already have the Laurent series, the coefficient of 1/z is the residue. why are you going through all that derivative crap
yeah came here to say the same thing
Crazy integral bro
You should start the project of writing a book about integration,it will be a great work.
This one's going down on the history books
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
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.
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.
To get early access to content:
www.patreon.com/Maths505
You can follow me on Instagram for write ups that come in handy for my videos:
instagram.com/maths.505?igshid=MzRlODBiNWFlZA==
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?
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.
@@dnsfsn oh i get it now. thanks for responding!
you could have ued tau instead of 2pi and would be even more cool xd
Very nice!! Can you please solve the integral ((i)^tanx )(tanx)^i from 0 to pi,the result is pi e^(-pi).thanks💯
Gamma subscription k. What does it mean or what is their representing?
How do you even find those integrals in tbe first place?
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.
thank you for this fantastic integral.Mario more and more ...Ok cool!!!
What is yours scientific degree you have got skills Man congrats take care
@@michallichmira8458 well in about a year and a half I'll have a masters degree
@@maths_505 congrats
damn that one is cool
"math people of youtube" lmao
Amazing... 9nce again
Nice way to build your Balrog audience.
Math is inclusive after all
The result is even more beautiful if you replace 2π by τ.
Cope
The 25th "one of the most beautiful integral i've solved on this channel" this year 😁
In hindsight, I guess I shouldn't be surprised by the result, given how connected ɣ is to ζ(1).