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Easy complex analysis question

  1. Dec 4, 2013 #1
    [solved]Easy complex analysis question

    Hi. In the complex plain, since y = 0 (in z=x+iy) at the x axis, shouldn't the following be true? :

    ##y=0##
    [tex] \int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+2x+5} dx = \int_{-\infty}^{\infty} \frac{e^{iaz}}{z^2+2z+5} dz = \int_{-\infty}^{\infty} f(z) dz[/tex]
    Or does it only go like this:
    ##y=0##
    [tex] \int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+2x+5} dx = Re(\int_{-\infty}^{\infty} \frac{e^{iaz}}{z^2+2z+5} dz)= Re( \int_{-\infty}^{\infty} f(z) dz)[/tex]

    I am asking because I need to find the value of the integral ##\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+2x+5} dx## using the residue theorem, but I don't know whether I need to use the residue of ##Re(\int_{-\infty}^{\infty} f(z) dz)## or ##\int_{-\infty}^{\infty} f(z) dz##. The problem is that if I use the latter, my logic seems coherent but I get a residue which has both a real and imaginary part, which hints that I should find the residue of the former instead.
     
    Last edited: Dec 4, 2013
  2. jcsd
  3. Dec 4, 2013 #2

    ShayanJ

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    In fact,both are wrong!
    This is how it should be:

    [itex]
    \int_{-\infty}^{\infty} \frac{\cos{ax}}{x^2+2x+5}dx=Re (\int_{-\infty}^{\infty} \frac{e^{iax}}{x^2+2x+5}dx)
    [/itex]
     
  4. Dec 4, 2013 #3
    Could you please explain:

    1) why my equations are wrong
    2) why yours is right

    thanks :)

    EDIT: oops I forgot to specify that y=0 for BOTH of those two integrals. They were both integrated along the x-axis.
     
  5. Dec 4, 2013 #4

    vela

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    No, how could this be true since ##\cos ax \ne e^{iax}##?
     
  6. Dec 4, 2013 #5
    but it's true if y =0, isn't it? sorry I forgot to specify this condition in the OP. All the integrals are evaluated on the real x-axis.
     
  7. Dec 4, 2013 #6

    vela

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    No. Euler's formula tells you that ##e^{iax} = \cos ax + i\sin ax##. For complex z, you have ##e^{ia(x+iy)} = e^{-y}e^{iax}##, which is equal to ##e^{iax}## when y=0, so you still have a problem.
     
  8. Dec 4, 2013 #7
    oh crap I forgot about the ##i## infront of the ##az## (i've been cramming 70-80 hours a week now in preparation for the exams, duno if this is a good excuse). OK, I see the problem now,t hx
     
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