MONSTER INTEGRAL

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  • Опубликовано: 15 дек 2024

Комментарии • 37

  • @farfa2937
    @farfa2937 9 месяцев назад +74

    Today's integral is sponsored by an existencial crisis

  • @Akhulud
    @Akhulud 9 месяцев назад +39

    the integrals are your friends

  • @thomasblackwell9507
    @thomasblackwell9507 9 месяцев назад +37

    Yes, you have friends!

    • @maths_505
      @maths_505  9 месяцев назад +13

      Yes my friend that intro was just for comedic relief 😂 lovely hearing from you once again. We haven't spoken on Instagram in quite a while, is everything okay?

  • @kavimahajan8412
    @kavimahajan8412 9 месяцев назад +6

    Being early on a maths 505 video, there’s nothing greater

  • @MrWael1970
    @MrWael1970 7 месяцев назад

    Thanks for your featured effort.

  • @beautyofmath6821
    @beautyofmath6821 4 месяца назад

    Wonderful integral

  • @rv178
    @rv178 8 месяцев назад +3

    Dont worry bro, every single positive integer is your personal friend

  • @aravindakannank.s.
    @aravindakannank.s. 9 месяцев назад +17

    why u need friends when u got bros (us daily viewers)😊

    • @maths_505
      @maths_505  9 месяцев назад +6

      SUIIIIIIIIIIIIIIIIII

  • @abdulllllahhh
    @abdulllllahhh 9 месяцев назад +6

    I had to put aside the time that I’d spend with my (not |R) friends to watch this video

  • @Patrickoliveirajf
    @Patrickoliveirajf 8 месяцев назад +2

    You are my friend ❤

  • @polpotify
    @polpotify 8 месяцев назад +1

    In my language, Kamaal means Maximum, so you are the roof function of all Maths RUclipsrs 😅

    • @maths_505
      @maths_505  8 месяцев назад

      It has a similar meaning in Arabic 😂 thanks bro

    • @trelosyiaellinika
      @trelosyiaellinika 7 месяцев назад

      Kamaal كمال in Arabic means perfection (it may also mean completion and by extension, be interpreted as maximum in some languages that have borrowed from Arabic). I have no idea what nationality you belong to, but, in any case, the name Perfection suits you well! The evidence is in your videos! And, never let it cross your mind that you lack friends. You have a huge community of math lovers who appreciate you and wish you well.

  • @gamermessi1375
    @gamermessi1375 8 месяцев назад +1

    Can you evaluate this sum:
    Sum of cos(bln(x))/x^a with x going from 1 to infinity .
    a and b are like parametres

  • @mikefischbein3230
    @mikefischbein3230 9 месяцев назад +3

    You differentiate with respect and integrate with zeal.

  • @CloneAtWork
    @CloneAtWork 8 месяцев назад

    Here's a kinda cool Integral of a double sum, maybe you like it :3 (Also F for friends)
    Integral from 0 to 1 of (Sum from n=1 to inf of x^n * (Sum from k=1 to k=2n-1 of (-1)^(n+k) / binomial(2n,k))) dx
    Answer should be 1-(pi^2 / 12)

  • @MatthisDayer
    @MatthisDayer 8 месяцев назад +2

    11:37 i suspect a phi will show up

  • @kavimahajan8412
    @kavimahajan8412 9 месяцев назад +2

    How do you figure out how to solve stuff

    • @maths_505
      @maths_505  9 месяцев назад +7

      Well....it helps when I invent the problem in the first place 😂😂😂

  • @charlesbrook746
    @charlesbrook746 8 месяцев назад

    Before using the digamma you could just factorise the numerator and have a simple polynomial integral

  • @Ghaith7702
    @Ghaith7702 9 месяцев назад +3

    awesome, also F

  • @zatsun2733
    @zatsun2733 7 месяцев назад

    I just have a question about the derivative regarding s of the integral;
    you seem to do it freely but shouldn't we first check for continuity definition and domination of the integrated function ?

  • @shivanshnigam4015
    @shivanshnigam4015 9 месяцев назад +4

    Heres a fun series problem
    Summation cos(2nx)/(4n²-1) from n= 1 to n--> infinity
    Ans is (1/2)-(π/4)sinx

  • @shivanshnigam4015
    @shivanshnigam4015 9 месяцев назад +3

    The actual fact is that you have friends but none of them are math nerds like you

  • @yoav613
    @yoav613 8 месяцев назад

    Noice,now give us nice integral with the trigamma function.😊💯

  • @rudransh118
    @rudransh118 8 месяцев назад

    respect brother F

  • @giuseppemalaguti435
    @giuseppemalaguti435 8 месяцев назад

    I'(s)=I(0-->1)=(x^s-1)(1-x)/(1-x^k)...semplifico I'(s)=I(0->1) (x^s-1)/(1+x+x^2+...x^(k-1)..

  • @fredfred9847
    @fredfred9847 9 месяцев назад

    F

  • @MohamedachrafKadim-jm5yr
    @MohamedachrafKadim-jm5yr 9 месяцев назад

    I solve this hhh😂

  • @edmundwoolliams1240
    @edmundwoolliams1240 9 месяцев назад

    Surely you're joking?!

  • @xdShaty
    @xdShaty 9 месяцев назад

    F

  • @Ijkbeauty
    @Ijkbeauty 8 месяцев назад

    F