2 fascinating integral results you should know for an integration bee

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  • Опубліковано 25 лис 2024

КОМЕНТАРІ • 25

  • @txikitofandango
    @txikitofandango 5 місяців тому +6

    "integration with respect to u" With respect to you, sir

  • @saraandsammyb.9599
    @saraandsammyb.9599 9 місяців тому +16

    Could you possibly think about doing a sort of “study guide” or course for integration bees?

    • @Mario_Altare
      @Mario_Altare 9 місяців тому +1

      I agree, it would be cool

  • @axel_arno
    @axel_arno 9 місяців тому +4

    Aaaaah I was waiting for that ! Thank you for it.

    • @FroejMK
      @FroejMK 9 місяців тому +1

      ptdrrr incroyable de te trouver ici

    • @vit1leman14
      @vit1leman14 9 місяців тому

      @@FroejMKmais vraiment hahaha

  • @jmcsquared18
    @jmcsquared18 3 місяці тому

    I did this quite differently.
    I split each integral at x=1 and used power series near zero for one piece or near infinity for the other piece, approaching x=1 in a limit. This expressed the integrals as series of the sequence s that look like (p+n)⁻¹ where p=s/k.
    At that point, it's just a matter of recognizing that what you get are the Eisenstein series for cotangent or cosecant, depending on if the terms alternate. You get a (-1)⁻ⁿ in the sequence for the first integral, which gives cosecant.

  • @trelosyiaellinika
    @trelosyiaellinika 9 місяців тому

    This is truly gorgeous! I am falling in love with the plethora that the Gamma, Beta and Digamma functions have to offer...

  • @OOOOOIIOOOOOOIIO
    @OOOOOIIOOOOOOIIO 9 місяців тому +4

    Axel Arno, oui monsieur ! Excellent pronunciation by the way and really nice solving ;)

  • @mikeoffthebox
    @mikeoffthebox 9 місяців тому +4

    Nice to have these Mellin transforms handy!

  • @very_incredible
    @very_incredible 9 місяців тому +1

    Thanks for the video, very cool integrals
    These integrals converge for 0

  • @funcionzeta6039
    @funcionzeta6039 9 місяців тому +1

    A challenge for you 😊:
    Compute limit n->infinity of
    (Integral from 0 to 1 of 1/(1+x^n) dx)^n

  • @jejnsndn
    @jejnsndn 9 місяців тому +1

    - I think that i wrote it wrong- B(v,y)=integral(2xcosx^v-1 sinx^y-1) from 0 to pie by tow, if the limits of integration were something else, what do we do?

  • @MrWael1970
    @MrWael1970 9 місяців тому

    It is very interesting solution. Thank you

  • @sumanbag729
    @sumanbag729 9 місяців тому +2

    Amazing Video 😊❤
    I love mathematics like you so can I talk to you please ?

  • @questionmark8326
    @questionmark8326 9 місяців тому

    Hey could you do a vedio or even a series on integrals of JEE mains or Aadvanced - i would ve intesting and informative for students

  • @vladimir10
    @vladimir10 9 місяців тому +1

    What is known about s and k in both these above?
    Also, PV is due to the singularity at 1, isn't it?

  • @Ki0212
    @Ki0212 21 годину тому

    Couldn’t you use ramarjun’s master theorem?

  • @tomasstride9590
    @tomasstride9590 9 місяців тому

    I have only looked at the first integral so far. May be I have missed something. But immediately my first thought was that the integral could only exist for k sufficiently larger than s-1. Yet this does not figure anywhere in the solution. What am I missing?

  • @ayush10tharollno16
    @ayush10tharollno16 9 місяців тому

    If i share this to my friends they call me math nerd

  • @Ayush-yj5qv
    @Ayush-yj5qv 9 місяців тому

    Would be good for iit adv also

  • @giuseppemalaguti435
    @giuseppemalaguti435 9 місяців тому

    Il primo è ovvio,basta porre x^k=t...si ottiene β,e poi con le proprietà di Γ,si arriva al risultato...il 2 è meno ovvio,almeno per me

  • @djridoo
    @djridoo 9 місяців тому

    Axel Arno le boss !!!

  • @abdulllllahhh
    @abdulllllahhh 9 місяців тому +1

    I lose a small part of my soul every time a /gamma is cancelled out from the final solution

    • @maths_505
      @maths_505  9 місяців тому +2

      Well technically we didn't need the digamma functions so γ was doomed from the start.