Complex Analysis: Integral of log(sin(x))

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  • Опубліковано 23 гру 2024

КОМЕНТАРІ • 36

  • @txikitofandango
    @txikitofandango 2 роки тому +85

    The Chaotic Good to flammy's Chaotic Evil

  • @aritelramos4191
    @aritelramos4191 2 роки тому +8

    ¡Gracias!

  • @renesperb
    @renesperb 2 роки тому +41

    There is a very short and elementary way to calculate this integral I. First write sin[x] as 2*sin[x / 2]*sin[x / 2+pi/2] Then log[sin[x]= log[2]+log[sin[x] /2]
    +log[sin[x/2+pi/2] . The substitutions t= x/2 and t=x/2+pi/2] immediately lead to the equation I+ log[2]*pi = 2*I ,i.e. I = - pi*log[2] .

  • @pixel0818
    @pixel0818 4 місяці тому

    discovered your channel today and I got to say that I've fallen in love with the complex analysis series, keep it up!

  • @Alex-kp3hd
    @Alex-kp3hd 2 роки тому +14

    6:10 it depends on the branch cut of the logarithm, I'll pick it to be the negative real axis, then sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(x), the imaginary part has to be equal to 0, and the real part has to be less than 0, so sin(x)

  • @Alex-kp3hd
    @Alex-kp3hd 2 роки тому +4

    30:24 use the taylor expansion of e^x, then you have x ln( x(2ie^itheta+(2ie^itheta)^2 x/2! +...) = xlnx+xln(2ie^itheta+....), the limit as x goes to 0 is then 0.

  • @digxx
    @digxx 2 роки тому +4

    If you decompose log(sin(x))=iz-log(2)-i*Pi/2+log(1-exp(-2i*x)) and integrate from 0 to Pi, then Int_0^Pi log(sin(x)) dx = -Pi*log(2) + Int_0^Pi log(1-exp(-2i*x)) dx. You can then use complex analysis to show that the last integral is 0. Substitute z=exp(-2i*x), then this integral becomes Int_{|z|=1} log(1-z) dz/(-2i*z) where the contour is going clockwise. log(1-z)/z has a branch-cut on (1,infinity) and is holomorphic inside the disk of radius 1 about z=0. Hence this integral is zero by Cauchy. You may want to check the integral over the semi-circle of radius epsilon about z=1, but it is not difficult to see that it is zero as epsilon -> 0.
    Btw: What do you mean by standard solution? I once saw a solution using integration by parts twice (using the symmetry about Pi/2) and you end up with Int_0^{Pi/2} (x/sin(x))^2 dx and then there was a trick, but I don't remember.

    • @Alex_Deam
      @Alex_Deam 2 роки тому

      Not exactly the same integral, but maybe related to this? ua-cam.com/video/7wiybMkEfbc/v-deo.html

    • @digxx
      @digxx 2 роки тому +1

      @@Alex_Deam That's a neat method too, thanks.

    • @violintegral
      @violintegral 2 роки тому

      I know the solution you're thinking of. It uses integration by parts once and then Feynman's trick after substituting y = tan(x). So you were right about there being a trick! It's a nice alternative to the normal solution that exploits symmetry within the integrand.

  • @emanuellandeholm5657
    @emanuellandeholm5657 2 роки тому +1

    This is amazing! Made me sub as primitive functions of logarithms of trigonometric functions are currently in my scope.
    As much as I enjoy a purely real proof... :)

  • @rishabhjain2722
    @rishabhjain2722 2 роки тому +2

    Can you make videos on some of jee advanced tough complex number questions?

  • @freddyfozzyfilms2688
    @freddyfozzyfilms2688 2 роки тому

    i dont eevn know complex analysis. I just love your handwriting

    • @qncubed3
      @qncubed3  2 роки тому

      That is very kind of you

  • @anham6543
    @anham6543 2 роки тому +2

    Thank you for the integral and your way to explain it
    I need a help if you can sur
    I want integrat
    [0,inf] x^2*exp(-x)/(x^2+pi^2)

  • @Oofitu
    @Oofitu 2 дні тому

    solved this using derivative of the beta function.
    what an elegant problem!

  • @afzalsoomro7950
    @afzalsoomro7950 2 роки тому +1

    Can you suggest book for Complex analysis? Thanks

  • @jordanweir7187
    @jordanweir7187 2 роки тому +1

    rly nice explanation bruh subbed

  • @George-ij2gm
    @George-ij2gm 2 роки тому +1

    Do we really need to let n-> infinite? As long as n> ε, keep n as a constant should work well.

    • @qncubed3
      @qncubed3  2 роки тому

      The e^(-2n) in the logarithm won't vanish if n is finite

    • @George-ij2gm
      @George-ij2gm 2 роки тому

      @@qncubed3 Thank you!

  • @張鼎翊-t4y
    @張鼎翊-t4y 2 роки тому +5

    the opening with the chair is interesting though

  • @SylComplexDimensional
    @SylComplexDimensional Рік тому +1

    12:21 🚁👍

  • @thewatchman_returns
    @thewatchman_returns 2 роки тому +2

    Thanks professor

    • @qncubed3
      @qncubed3  2 роки тому +2

      I am merely a baby undergrad

    • @thewatchman_returns
      @thewatchman_returns 2 роки тому +1

      @@qncubed3 I dream of being like you someday

  • @MyOneFiftiethOfADollar
    @MyOneFiftiethOfADollar 2 роки тому +3

    So change the x to z. Then just show everyone that you know how to do it.

    • @qncubed3
      @qncubed3  2 роки тому +3

      Proof: Let x = z. This reduces to a trivial problem.

  • @pBERA0_0
    @pBERA0_0 2 роки тому +2

    Grade 12
    NCERT Problem :)

    • @kunalbose493
      @kunalbose493 2 роки тому

      Example 36
      very fav question of cbse

  • @samsan1914
    @samsan1914 2 роки тому +2

    he is really cute.

    • @qncubed3
      @qncubed3  2 роки тому +1

      That proposition is false

  • @marshalls36
    @marshalls36 2 роки тому

    大学题