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Complexifying an integral

  1. Mar 13, 2014 #1
    Watching a video on differential equations by Arthur Mattuck of MIT, I came across a method which was new to me for solving certain integrals, such as [tex]\int e^{-x}cos(x)dx[/tex] . part of the video is here:

    Given that [tex]e^{ix}=cosx+isinx[/tex], this integral can be re-written as [tex]Re\int e^{ix}\cdot e^{-x}dx[/tex] = [tex]Re\int e^{x(-1+i)}dx[/tex] and integrated this way, avoiding the need for reduction formulas...

    I have looked up a few examples of this and think I can understand it when cos(x) is part of the integral, but am unsure when it comes to sin(x).

    There is a (now closed) thread on this forum https://www.physicsforums.com/showthread.php?t=511534&highlight=complexifying+integral which I have been attempting to follow but I can't see quite how it works. I was wondering if anyone could help me to understand this.

    The following is effectively the working from that post:

    [tex]\int e^x sin2x dx[/tex]

    =[tex]Re\int e^x \cdot e^{i(\frac{\pi}{2}-2x)}dx[/tex]

    I can see that since the real part of polar co-ordinates is contained in the "cos(2x)" part, they have put the Sin(2x) in terms of Cos(2x) by subtracting it from [tex]\frac{\pi}{2}[/tex].

    But I don't see how the next step is reached:

    [tex]=Re\frac{i}{1-2i}e^{(1-2i)x}[/tex]


    I'm not sure how they got to this, nor what the intermediate steps might be. If I was doing this, I would have gone along the lines of [tex] Re\int e^{x+i(\frac{\pi}{2}-2x)}dx[/tex]. While I can see that that expression doesn't look very attractive to try to integrate, I can not see how the author got to [tex]e^{(1-2)x}[/tex]
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    pasmith

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    If you want sin(x), take the imaginary part of [itex]e^{ix}[/itex] instead of the real part.
     
    Last edited by a moderator: Sep 25, 2014
  4. Mar 13, 2014 #3
    OK, thanks Pasmith...
     
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