A mesmerizing result

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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[@stefanalecu9532]

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

[@orionspur]

Ohhhkaaay cool!

[@alielhajj7769]

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]

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]

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.

[@julioguilarte9438]

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 :)

[@MrWael1970]

Thank you very much.

[@slavinojunepri7648]

Cool result

[@yoav613]

Sport and math perfect!😊💯

[@jonsmith8579]

Logarithms

[@arkadelik]

Logarithms

[@maths_505]

Logarithms

[@nox_ell1549]

@@arkadelik Logarithms

[@Dedicate25]

Logarithms

[@stefanalecu9532]

Logarithms

[@txikitofandango]

Sir you are a poet

[@serdarakalin2209]

Perfect

[@PeterParker-gt3xl]

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.]

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

[@2thecartoonsimp1]

May I ask what app you use for your videos?

[@felipematus3021]

Greetings from the Czech Republic.

[@maths_505]

Greetings

[@felipematus3021]

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

[@archinsoni1254]

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

[@3manthing]

2:13 Have you seen the new Oppen-adder movie? "Now I have become the sigma, the summator of the series."😅😁

[@maths_505]

Nah haven't seen that one but damn that was cool 😂

[@xizar0rg]

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]

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

[@kgangadhar5389]

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

[@maths_505]

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

[@lizardwithahat4862]

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]

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

[@GearsScrewlose]

Sweet result.

[@giuseppemalaguti435]

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

[@insouciantFox]

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

[@maths_505]

Argentina

[@ayushrudra8600]

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

[@hizon525]

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 😂

[@7yamkr]

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