one seriously crazy integral!

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  • Опубліковано 28 лис 2024

КОМЕНТАРІ • 63

  • @CM63_France
    @CM63_France Місяць тому +12

    Hi,
    "ok, cool" : 1:14 , 2:16 , 3:25 , 4:25 , 6:17 ,
    "terribly sorry about that" : 5:01 , 6:49 , 8:35 .

    • @RohitPanda08
      @RohitPanda08 Місяць тому +2

      woah! he's here

    • @stefanalecu9532
      @stefanalecu9532 26 днів тому +1

      I think you meant to add 8:35 to the ok cool category

    • @CM63_France
      @CM63_France 26 днів тому

      @@stefanalecu9532 ok, I noted a "TSAT" at 8:35, maybe I confused, thanks a lot.

  • @SamLindskougMBGY
    @SamLindskougMBGY Місяць тому +16

    can already tell its gonna be a banger

  • @stefanalecu9532
    @stefanalecu9532 Місяць тому +13

    I am not the French guy, so I'll take his place this time around
    "ok cool": 1:14, 2:16 (pen scratches ASMR), 3:24, 6:16, 8:34
    "terribly sorry (about that)": 0:03, 6:47

    • @jpf119
      @jpf119 Місяць тому

      I do need a UA-cam plugin to un-OKcool the video, could someone help? 🤣

    • @alphazero339
      @alphazero339 Місяць тому +1

      Hello, thanks for your attempt. You did a good job, you can be him while he's gone. I also enjoyed that you included ASMR pen scratch in the list. Overall, the experience was positive as I can now click any of the timestamps you provided and hear the author speak the phrase. I do not know what happened to French guy, but we can hope he is okay and safe. For now, I'm voting you to take his place.

    • @CM63_France
      @CM63_France Місяць тому +1

      @@alphazero339 wait a minute 🤣

    • @stefanalecu9532
      @stefanalecu9532 26 днів тому

      ​@@alphazero339 I can't guarantee I'll be on time, but I'll try if the guy above me can't do it

  • @Nottherealbegula4
    @Nottherealbegula4 Місяць тому +23

    Waiting for the french guy to list the "ok cool" and "terribly sorry about that" (:

    • @alphazero339
      @alphazero339 Місяць тому

      He must have retired

    • @jpf119
      @jpf119 Місяць тому +1

      I might unsubscribe because of that🤣

    • @CM63_France
      @CM63_France Місяць тому +2

      Terribly sorry about that, I am back 🗼💈

    • @alphazero339
      @alphazero339 Місяць тому

      @CM63_France okaay, cool

    • @Haxislive766
      @Haxislive766 28 днів тому +1

      😂

  • @ByeUnderGround24
    @ByeUnderGround24 Місяць тому +17

    Do a x=1/t sub on original integral that gives I = -I + (alpha-beta) int 0 to ∞ dx/1+x² so I = π(alpha-beta)/4
    Thanks a lot for sharing new methods and manipulations . Each and every video of yours is a gem!

    • @venkatamarutiramtarigoppul2078
      @venkatamarutiramtarigoppul2078 Місяць тому +1

      Ya i also did that i am way too lasy . What to do . But the solution develepment in the video is kinda cool

    • @rishabhhappy
      @rishabhhappy Місяць тому +1

      awesome approach bhaiya

    • @Grecks75
      @Grecks75 Місяць тому +3

      Wow, can it be that simple? I just checked it. Incredible, awesome! What a symmetry! 🎉

    • @ByeUnderGround24
      @ByeUnderGround24 Місяць тому +1

      ​@@rishabhhappy rishabh bro 🌚🗿

    • @zachariastsampasidis8880
      @zachariastsampasidis8880 Місяць тому

      Well it's an equivalent transformation since he did x to arctanx to π/2 - χ

  • @ddg-norysq1464
    @ddg-norysq1464 23 дні тому

    I think this is the cleanest solution I've ever seen 😮

  • @MrWael1970
    @MrWael1970 Місяць тому

    Very nice idea. Thank you

  • @trelosyiaellinika
    @trelosyiaellinika 29 днів тому

    Yep, you have a talent in choosing really elegant integrals.

    • @maths_505
      @maths_505  29 днів тому +1

      @@trelosyiaellinika thanks mate

  • @michaelihill3745
    @michaelihill3745 Місяць тому

    That was pretty awesome!

  • @xanterrx9741
    @xanterrx9741 Місяць тому

    Great video i hope for more videos.

  • @farfa2937
    @farfa2937 Місяць тому +1

    Expectation: Some combination of gamma, beta and zeta. Reality: A linear function with delusions of grandeur.

  • @NeofeaxFeax
    @NeofeaxFeax Місяць тому +2

    sub x-> 1/x gives you the answer on a single line

  • @Aplicapitagoras
    @Aplicapitagoras Місяць тому +2

    @Math 505 any tips on how to identify which substituition I should make to solve more difficult integrals?

  • @SabaSa6a
    @SabaSa6a Місяць тому

    The expansion of tan is not necessary here
    I remember that there is a similar question giving α=sqrt(2)

  • @ademnasri6557
    @ademnasri6557 19 днів тому

    Can I1 be solved by doing a simple fraction decomposition from the begining ? If yes please tell me how i've trying and i found no solution

  • @giuseppemalaguti435
    @giuseppemalaguti435 Місяць тому

    X=tgθ..poi uso Feymann I(α)..I'(α)..I(β)..I'(β).2I'=π/2....I=(π/4)(α-β)

  • @chancia8990
    @chancia8990 Місяць тому

    okay this is iis awesome

  • @raghavendraPi
    @raghavendraPi 20 днів тому

    Kamaal bhai, no video for 12 days 🥲, missing the awesome problems you solve

  • @shadowreaper-gq6bz
    @shadowreaper-gq6bz 28 днів тому

    can you solve the integral from : 0 to infinity of (dx/((x²+1)(pi²+ln²(x))) , btw , i solved the integral the is result is "2/pi -1/2"

  • @charliecox13
    @charliecox13 Місяць тому

    I know this will get drowned out. But on mathstackexchange I say an integral I haven’t been able to solve easily. The integral from 0 to 1 of f(x)=(x^3)tan(x^2). I tried Feynmans, exponentiating it, by parts, and writing it using summation. Can you give me a solution, or better yet a step by step?

    • @maths_505
      @maths_505  Місяць тому +1

      @@charliecox13 I'll take a shot at it

  • @شعرکوتاه-ع7ظ
    @شعرکوتاه-ع7ظ 17 днів тому

    Was interesting

  • @Ownageffects
    @Ownageffects Місяць тому

    please make a video of how to do the leibniz rule by details please and when can we switch the operators

  • @user-x4x-g1e
    @user-x4x-g1e 29 днів тому

    Hello , can we solve \int_{-\infty}^{+\infty}\frac{dx}{\left( e^x-x
    ight)^{n}+\pi^{n}} for any natural number n greater than or equal to 2 ? Can u make a video about that ?
    However, I solved this problem for n=2 by using residue theorem and rectangle contour with height of i2pi and lengh of 2R : beautiful result with 1/(W(1)+1) where W is the principal branch of W lambert function but i don't know how to approch this problem for any natural number n greater than or equal to 2...
    Can i have a help ?
    Thx

  • @Lepl_Acie
    @Lepl_Acie Місяць тому

    Hello , sorry for being annoying , may you complete your complex analysis explanation , it's very nice but I want more 🙂 .
    because I hope I can use counter intergals for solving so many things ...

  • @Haxislive766
    @Haxislive766 28 днів тому

    I(ß) = 0??

  • @AyushRajput-xw2ru
    @AyushRajput-xw2ru Місяць тому +1

    Day by day kamaal questions are getting juicy . ❤

  • @tpgmflint
    @tpgmflint Місяць тому

    when can theta be transformed to pi/2 - theta? (6:23)

    • @lotaniq4449
      @lotaniq4449 Місяць тому

      If the limits are from (0 to a), you can always do a substitution theta’ = a - theta. Its just a substitution, easy to verify.

    • @absol4844
      @absol4844 Місяць тому

      ​@@lotaniq4449 But doesn't leave that us with extra negative sign? Since a - θ evaluated at a is 0 and at 0 it's a, so the bounds are upside down now...

    • @lotaniq4449
      @lotaniq4449 Місяць тому

      @@absol4844 yes but dtheta’=-dtheta, so that - cancels the other -

    • @atulyaroy8962
      @atulyaroy8962 Місяць тому

      ​​@@absol4844 the lower and upper limits will be interchanged if you do the substitution so to make the limits same you use the negative sign. This called the King property check it out :}

    • @atulyaroy8962
      @atulyaroy8962 Місяць тому

      It's called the King property, it can be easily proved by substitution (lower limit + upper limit -t)

  • @tanusreekar2839
    @tanusreekar2839 Місяць тому +3

    okayyy cool

  • @davidblauyoutube
    @davidblauyoutube Місяць тому

    Noice

  • @alikazemi1329
    @alikazemi1329 Місяць тому

    👏👏👏❤♥🙏👍

  • @Ownageffects
    @Ownageffects Місяць тому +1

    kamaal can i be the zeta to your (s) ? (no diddy)

  • @apnakaamkrelala
    @apnakaamkrelala Місяць тому +3

    Greetings and welcome back our kamaal😌✨🤌 my ears were eagerly waiting for your ohk cool and terribly sorry about that😭🤌✨💓