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Contour integrals

  1. Mar 26, 2013 #1
    In trying to solve [itex]\int^{\infty}_{-\infty} x + \frac{1}{x} dx[/itex] could it be split up and solved using the Cauchy Principle Value theorem and a contour integral along a semi-circle. Thus;
    [itex]PV\int^{\infty}_{-\infty}x dx =0 [/itex] [itex]+\int \frac{1}{x} dx = \int^{\pi}_{0} i d\theta [/itex]

    Is this valid reasoning?
  2. jcsd
  3. Mar 27, 2013 #2


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    No, it isn't. For one thing, you have a pole at x= 0 which is on your contour. You would need another semicircle around x= 0 to avoid that.
  4. Mar 28, 2013 #3
    The initial intergal was [itex] f(x)=\int^{\infty}_{-\infty} \sqrt{x^{2}+y^{2}}dx[/itex] so I taylor expanded it to get [itex]f(x) \approx \int^{\infty}_{-\infty} x + \frac{y^{2}}{2x} dx [/itex]

    I thought one could then justify that the cauchy principle value of [itex]\int^{\infty}_{-\infty} x dx =0 [/itex] and then what I have done with the [itex]\frac{1}{x} [/itex] integral. I am doubting my approach because the Taylor series was about x=0 which seems odd, is there a better way? I read that one take x to be complex then contour interagtes it, I am just not sure how?
    Last edited: Mar 28, 2013
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