An incredible integral solved using Feynman's trick

Поділитися
Вставка
  • Опубліковано 25 лис 2024

КОМЕНТАРІ •

  • @manstuckinabox3679
    @manstuckinabox3679 11 місяців тому +17

    we can also evaluate these integrals (after applying Feynman technique)using contour integration, which is a problem from Gamelin's Complex Analysis.
    Exotic integral indeed.

    • @maths_505
      @maths_505  11 місяців тому +7

      Bro I don't think even gamelin knew his text this good 😂

  • @pnintetr
    @pnintetr 11 місяців тому +5

    Absolute beauty.
    I thought 1/log(x) in the integrand could be dreadful, but it was nothing in front of Feynman.

  • @vit1leman14
    @vit1leman14 11 місяців тому +1

    It’s so nice to catch up all your latest video! Quite awesome !

  • @krisbrandenberger544
    @krisbrandenberger544 11 місяців тому +3

    @ 7:36 The first term of I'(alpha) should have a minus sign.

    • @TMH2007
      @TMH2007 11 місяців тому +1

      yeah

  • @jieyuenlee1758
    @jieyuenlee1758 7 місяців тому +1

    8:04 first term should have a negative sign in front

  • @ericthegreat7805
    @ericthegreat7805 11 місяців тому

    What is great is that this means there is a connection and equivalence between this number and the binary integral
    Int(oo,0) (exp(-x)/(1 + exp(-x)) dx = ln2
    Which I mentioned was used in logistic regression for binary outcomes. Perhaps this can be related to the Lhopitals rule and considered an integral equivalent (due to the lnx in the denominator) i.e. a "derivative binary integral" with information/entropy equal to ln2, equivalent to an ordinary binary integral representing the rate of change of information or score function.

    • @ericthegreat7805
      @ericthegreat7805 11 місяців тому

      Also, in the limit the sequence at the bottom becomes a geometric series. So the relation is dependent on the fraction of the geometric series to the geometric series represented by (x^alpha - 1).

  • @nathanmenezes7914
    @nathanmenezes7914 11 місяців тому

    That PFD blew my mind. That was so fast

  • @MrWael1970
    @MrWael1970 11 місяців тому

    Very smart way. Thank you

  • @jannesfilgerdamm1419
    @jannesfilgerdamm1419 11 місяців тому +1

    Are there some table sharts, that shows all, of the many possible transformation, into f.e. the gamma gunction etc ?

  • @jhacklack
    @jhacklack 11 місяців тому +2

    wonderful

  • @antoinehedin6608
    @antoinehedin6608 8 місяців тому

    Hi!
    Since Euler's reflection formula can only be applied with 0

  • @TMH2007
    @TMH2007 11 місяців тому +1

    Very cool!

  • @wagsman9999
    @wagsman9999 11 місяців тому

    Whoa. Monster. Cool result tho.

  • @giuseppemalaguti435
    @giuseppemalaguti435 11 місяців тому

    Applico feyman, semplicemente...I(a)=[...x^a-1.…],con I(0)=0,I(1)=I...I'(a)=Σ(-1)^k*Β(a+2k+1,-a-2k)...ma poi non riesco ad integrare la beta...come si fa?

  • @arkadelik
    @arkadelik 11 місяців тому

    KNG 👑

  • @kristim1
    @kristim1 11 місяців тому

    🫡

  • @Noam_.Menashe
    @Noam_.Menashe 11 місяців тому +2

    I think I once saw an easier way to solve integrals of this form, but I don't remember it.

    • @daddy_myers
      @daddy_myers 11 місяців тому +1

      Are you Indian by any chance?

    • @Noam_.Menashe
      @Noam_.Menashe 11 місяців тому

      no@@daddy_myers