Using the Residue Theorem for improper integrals involving multiple-valued functions

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  • Опубліковано 23 гру 2024

КОМЕНТАРІ • 18

  • @jbergamp
    @jbergamp 11 місяців тому

    Can someone tell me where the log(x) of the furthest on the right integral went to?

  • @rutapravad6364
    @rutapravad6364 3 роки тому

    Thank you prof Barrus...

  • @DavidJones-so1vk
    @DavidJones-so1vk 6 років тому

    I understand the methodology that you present in this video. However, when I apply the same techniques to a similar problem, namely the integrand ln(x+2) / ((x+5) (x+7)) evaluated from 0 to infinity, the methodology fails after a certain point. I have tried several different substitutions to shift the axis as one possible means of dealing with this integral, but so far I have not met with great success. Can you possibly include a video tutorial on how to deal with problems of this nature? Thank you.

    • @HotPepperLala
      @HotPepperLala 5 років тому +2

      I think your singularies lies on the contour, so that won't work

  • @princesscat2076
    @princesscat2076 4 роки тому

    How does the integral along Cr go to 0?

    • @catprincess9
      @catprincess9 4 роки тому

      @Medo Z I meant to ask for C_r and not C_R. I understand why C_R goes to 0 when R goes to infinity. But I am not sure why C_r goes to 0 when r goes to 0.

  • @JohnSmith-iu3fc
    @JohnSmith-iu3fc 5 років тому

    Pls, teach me why ln(i) shoudl be ipi/2.

    • @okan3028
      @okan3028 5 років тому +2

      i= e^ipi/2 so if you take natural logarithm of i you'll obtain ipi/2

    • @JohnSmith-iu3fc
      @JohnSmith-iu3fc 5 років тому +2

      @@okan3028
      Now, I understand it. Thank you.

    • @sayanjitb
      @sayanjitb 3 роки тому +2

      @@okan3028 complex argument for natural logarithm is not defined. Rather one can say that Log(i) = ln(mod(i)) + iπ/2 , hence Log i = iπ/2

    • @sayanjitb
      @sayanjitb 3 роки тому

      ..

  • @Nobody-Nowhere-Nothing
    @Nobody-Nowhere-Nothing 5 років тому +1

    You are straight God.

  • @Krantiveer-y7u
    @Krantiveer-y7u 2 роки тому

    thank you so much sir

  • @aronhegedus
    @aronhegedus 8 років тому +1

    Thank you:)

  • @ransarawijitharathna7566
    @ransarawijitharathna7566 3 роки тому +2

    are you sure you are not the "Casually explained" guy?

  • @francotomatillo
    @francotomatillo 4 роки тому

    Thanks!

  • @yegobernard206
    @yegobernard206 7 років тому

    Well put

  • @HarinduJayarathne
    @HarinduJayarathne 5 років тому

    Helpful!