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Using complex numbers for evaluating integrals

  1. Oct 16, 2012 #1
    How can I use complex numbers to evaluate an integral? For instance I'm reading a book on complex numbers and it says that to evaluate the integral from 0 to pi { e^2x cos 4x dx }, I must take the real part of the integral from 0 to pi { e^((2 + 4i)x) dx}.

    It totally skips how you do that. I don't see how taking the real part of the second integral relates to the first one. If I try to integrate e^((2 + 4i)x) and then take the real part, I eventually get:

    Re(z) = 1/5 e^2x (1/2 cos 4x + sin 4x)

    But I don't see any relation whatsoever, or how they did this.
     
  2. jcsd
  3. Oct 16, 2012 #2

    tiny-tim

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    welcome to pf!

    hi genjix! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    show us how you did this :smile:
     
  4. Oct 16, 2012 #3
    Well I went onto #math on Freenode and someone gave me the missing puzzle piece. Basically that Re(integral f(x)) = integral Re(f(x)). With that in hand, everything is obvious:

    Take the real part of e^2x e^4ix
    Re(e^2x e^4ix) = e^2x (cos 4x + i sin 4x) ... from de moivre and because the Re(integral f(x)) = integral Re(f(x)), we can say that Re(integral e^2x e^4ix) = integral Re(e^2x e^4ix) = integral Re(e^2x (cos 4x + i sin 4x)) = integral e^2x cos 4x

    now the integral e^2x e^4ix = integral e^(2 + 4i)x and to integrate exponentials we have the usual integral e^cx = 1/c e^cx

    so we get:
    1/(2 + 4i) e^{(2 + 4i)x}
    http://mathbin.net/110072 [Broken] (multiply top and bottom of fraction by (2 - 4i) on first line)
    now we have the bottom line to be evaluated from 0 to +pi (as per the original integral)
    cos 4pi = cos 0 = 1
    sin 4pi = sin 0 = 0

    so we get the final line as being:
    1/5 e^2pi (1/2 cos 4pi + sin 4 pi) - 1/5 e^0 (1/2 cos 0 + sin 0) = 1/10 (e^2 - 1)
     
    Last edited by a moderator: May 6, 2017
  5. Oct 16, 2012 #4
    I just did, but my post isn't showing up? I try to click back in my browser and submit it, but it is saying I've made a duplicate post strangely.

    I found someone on #math in Freenode who told me that the Re(integral f(x)) = integral Re(f(x)) which is the missing puzzle piece I needed to work everything out. The post I've tried to make here shows all my working out for others to see how it is done.
     
  6. Oct 17, 2012 #5

    tiny-tim

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    hi genjix! :smile:

    (just got up :zzz:)
    yes, that's right :smile:

    (if you're still unsure, try submitting your proof again)
     
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