A mesmerizing result

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A beautiful iterated integral with full solution development leading to a closed form with this awesome linear combination of zeta functions.
My complex analysis lectures:
• Complex Analysis Lectures
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Опубликовано:

27 сен 2024

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Комментарии : 52

@stefanalecu9532 3 месяца назад

Bro had a football fan moment before proceeding to jump to an integral You're something special

@alielhajj7769 3 месяца назад

If we make it a triple integral with the same structure you will get two additional zetas and and additional -2, the generalization is pretty cool and I think the limit as we increase the dimension of integration is convergent and it is actually equal to -1 + the sum from n=2 to infinity of zeta(n)-1 !!! Really cool and worth investigating

@CM63_France 3 месяца назад

Hi, "ok, cool" : 1:55 , 5:24 , 8:03 , 8:43 , "terribly sorry about that" : 2:05 , 2:44 , 4:34 , 8:20 , 9:35 .

@leroyzack265 3 месяца назад

The final results with ascending coefficients of zeta is really cool. "We enjoyed the video. Thank you see you next time". Oops I'm using some one's words.

@jonsmith8579 3 месяца назад

Logarithms

@arkadelik 3 месяца назад

Logarithms

@maths_505 3 месяца назад

Logarithms

@nox_ell1549 3 месяца назад

@@arkadelik Logarithms

@Dedicate25 3 месяца назад

Logarithms

@stefanalecu9532 3 месяца назад

Logarithms

@julioguilarte9438 3 месяца назад

No wonder you are a sigma with this talent you have when it comes to solving these kinds of integrals. I hope I will someday become as good as you, meanwhile, keep up this amazing work :)

@txikitofandango 3 месяца назад

Sir you are a poet

@MrWael1970 3 месяца назад

Thank you very much.

@felipematus3021 3 месяца назад

Greetings from the Czech Republic.

@maths_505 3 месяца назад

Greetings

@felipematus3021 3 месяца назад

@@maths_505 I knew there was something weird with my comment 🤣🤣🤣

@داریوشسروری Месяц назад

‏‬‏505

@داریوشسروری Месяц назад

اَلسَلامُ عَلَيْكُم وَرَحْمَةُ اَللهِ وَبَرَكاتُهُ‎

@داریوشسروری Месяц назад

اَلسَلامُ عَلَيْكُم وَرَحْمَةُ اَللهِ وَبَرَكاتُهُ‎

@داریوشسروری Месяц назад

اَلسَلامُ عَلَيْكُم وَرَحْمَةُ اَللهِ وَبَرَكاتُهُ‎

@serdarakalin2209 3 месяца назад

Perfect

@giuseppemalaguti435 3 месяца назад

I=-Σ1/((k+1)(k+2)^4)...poi bisognerebbe sintetizzare con zeta function..

@Nottherealbegula4 3 месяца назад

May I ask what app you use for your videos?

@hizon525 3 месяца назад

2:13 Kamaal your sigma dad joke literally sent me hysterically laughing across the house. I have to thank the Gods i am home alone right now. Ive never been more comedically t-boned before in my life 😂

@ayushrudra8600 3 месяца назад

portugal played well but they porbbaly won't win the whole thing - my guess is a quarter final exit

@insouciantFox 3 месяца назад

Who you got for the Copa? Also I dare you to use ξ as a variable

@maths_505 3 месяца назад

Argentina

@satyam-isical 3 месяца назад

I thought we're gonna find the value of zeta2 zeta3 zeta 4😂

@داریوشسروری Месяц назад

لوتللت

@orionspur 3 месяца назад

Ohhhkaaay cool!

@archinsoni1254 3 месяца назад

Please integrate x^a*cosnx*e^(-kx)

@slavinojunepri7648 3 месяца назад

Cool result

@PeterParker-gt3xl 3 месяца назад

Hope to learn about Zeta fn. of odd powers apart from Euler's even powers. He loved using ln. to "lessen the labor".

@aravindakannank.s. 3 месяца назад

bro rest well bro i think you pulled an all nighter to watch football.

@xizar0rg 3 месяца назад

Can this be extrapolated out with more integrals? Like integral over unit cube of lnx*lny*lnz*ln(1-xyz) being (guessing result if it continues as before) eta2 +eta3 + eta4+ eta5 - 5? Or generally N Integrals_01 of Product(ln(x_n), 1, N) = Sum(eta(n), 2, n+1) - (n+1)? (I probably fucked up some notation there, but hopefully what I mean is understood.)

@BridgeBum 3 месяца назад

On notation: zeta, not eta. On the concept: seems worth exploring!

@kgangadhar5389 3 месяца назад

I was wondering the same. If its the case then Its can give some insight on Zeta and its distribution.

@maths_505 3 месяца назад

Oh yeah it's definitely gonna carry over thanks to symmetry.

@lizardwithahat4862 3 месяца назад

I tried to find a solution to the general integral: ∫...∫ln(x1)•...•ln(xN)•ln(1-x1x2...xN)dxN... dx1 (with all the integrals going from 0 to 1) and got this: I = -2N + sum{j=2 to 2N}Zeta(j)

@username-ur6dq 3 месяца назад

I looked at numerical answers for these integrals, and it seems there is a factor of -1 multiplied for each n for the solution you provided@@lizardwithahat4862