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Math Help - Real integral with complex analysis

  1. #1
    Senior Member Dinkydoe's Avatar
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    Real integral with complex analysis

    I have some trouble evaluating the following integral using complex analysis:

     \int_{0}^{\infty} \frac{\cos(\alpha x)}{x^2}dx

    I believe the trick is to complexify the integrand, let g(z)=e^{\alpha z}/z^2 and then make a contour:

    \gamma_1(t) = t, t\in [-R,-1/R]
    \gamma_2(t)=e^{it}/R,t\in[\pi,2\pi]
    \gamma_3(t) = t, t\in [1/R,R]
    \gamma_4(t)= e^{it}, t\in [0,\pi]

    And \gamma_R = \bigcup \gamma_i . Then we can evaluate \lim_{R\to\infty}\int_{\gamma_R}g(z)dz = 2\alpha\pi i with the Cauchy-integral formula.

    By Jordan's Lemma \int_{\gamma_4}g(z)dz=0. We like to evaluate Re(\lim_{R\to\infty}\int_{\gamma_1\cup\gamma_3}g(z  )dz) = \int_{-\infty}^{\infty} \frac{\cos(\alpha x)}{x^2}dx

    Can someone tell me how far off I am, and maybe the correct approach for this. My problem is, I have no idea how to evaluate \int_{\gamma_2}g(z)dz, the partial residu around the second order singularity z=0.

    So is this the right approach, if not, what is? And if yes, how to evaluate the afore mentioned partial residue.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let suppose to 'attack' the slightly different integral...

    \int_{0}^{\infty} \frac{1 - \cos (a x)}{x^{2}}\cdot dx (1)

    At this scope we can compute the complex integral...

    \int_{\Gamma} \frac{1 - e^{i a z}}{z^{2}}\cdot dz (2)

    ... along the red path \Gamma indicated in figure...





    ... when r \rightarrow 0 and R \rightarrow \infty. Since no singularities lie inside the path is...

    \int_{\Gamma} \frac{1 - e^{i a z}}{z^{2}}\cdot dz = 0 (3)

    ... and for the Jordan's lemma the integral along the 'big half circle' tends to 0 if R \rightarrow \infty. Tacking into account that the (3) becomes...

    \int_{-R}^{-r} \frac{1 - e^{i a x}}{x^{2}}\cdot dx + i\cdot \int_{\pi}^{0} \frac{1 - e^{i a r e^{i \theta}}}{r}\cdot e^{-i \theta}\cdot d\theta + \int_{r}^{R} \frac{1 - e^{i a x}}{x^{2}}\cdot dx = 0 (4)

    Now if r \rightarrow 0 and R \rightarrow \infty , tacking into account that ...

    \lim_{r \rightarrow 0} i\cdot \int_{\pi}^{0} \frac{1 - e^{i a r e^{i \theta}}}{r}\cdot e^{-i \theta}\cdot d\theta = - a\cdot \pi (5)

    ... from (4) we derive that...

    \int_{0}^{\infty} \frac{1-\cos (a x)}{x^{2}}\cdot dx = \frac{a}{2}\cdot \pi (6)

    From (6) is evident that is...

    \int_{0}^{\infty} \frac{1}{x^{2}}\cdot dx = \int_{0}^{\infty} \frac{\cos (a x)}{x^{2}}\cdot dx + \frac{a}{2}\cdot \pi (7)

    ... and, because the integral in first member diverges, the integral in second member diverges too...

    Kind regards

    \chi \sigma
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