A mesmerizing result

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  • Опубліковано 17 чер 2024
  • 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:
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КОМЕНТАРІ • 47

  • @stefanalecu9532
    @stefanalecu9532 Місяць тому +31

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

  • @orionspur
    @orionspur Місяць тому +36

    Ohhhkaaay cool!

  • @alielhajj7769
    @alielhajj7769 Місяць тому +3

    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
    @CM63_France Місяць тому +3

    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
    @leroyzack265 Місяць тому +6

    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
    @julioguilarte9438 Місяць тому +3

    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
    @MrWael1970 Місяць тому

    Thank you very much.

  • @slavinojunepri7648
    @slavinojunepri7648 Місяць тому

    Cool result

  • @yoav613
    @yoav613 Місяць тому

    Sport and math perfect!😊💯

  • @jonsmith8579
    @jonsmith8579 Місяць тому +17

    Logarithms

  • @txikitofandango
    @txikitofandango Місяць тому +1

    Sir you are a poet

  • @serdarakalin2209
    @serdarakalin2209 Місяць тому

    Perfect

  • @PeterParker-gt3xl
    @PeterParker-gt3xl Місяць тому

    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.
    @aravindakannank.s. Місяць тому

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

  • @3manthing
    @3manthing Місяць тому

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

    • @maths_505
      @maths_505  Місяць тому +1

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

  • @2thecartoonsimp1
    @2thecartoonsimp1 Місяць тому

    May I ask what app you use for your videos?

  • @archinsoni1254
    @archinsoni1254 Місяць тому

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

  • @xizar0rg
    @xizar0rg Місяць тому +1

    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
      @BridgeBum Місяць тому

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

    • @kgangadhar5389
      @kgangadhar5389 Місяць тому

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

    • @maths_505
      @maths_505  Місяць тому

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

    • @lizardwithahat4862
      @lizardwithahat4862 Місяць тому +1

      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
      @username-ur6dq Місяць тому

      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

  • @felipematus3021
    @felipematus3021 Місяць тому

    Greetings from the Czech Republic.

    • @maths_505
      @maths_505  Місяць тому

      Greetings

    • @felipematus3021
      @felipematus3021 Місяць тому

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

  • @GearsScrewlose
    @GearsScrewlose Місяць тому

    Sweet result.

  • @giuseppemalaguti435
    @giuseppemalaguti435 Місяць тому

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

  • @insouciantFox
    @insouciantFox Місяць тому

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

  • @hizon525
    @hizon525 Місяць тому +1

    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
    @ayushrudra8600 Місяць тому +1

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

  • @Ghostwriter_zone
    @Ghostwriter_zone Місяць тому

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