Results 1 to 2 of 2

Thread: Real integral with complex analysis

  1. #1
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411

    Real integral with complex analysis

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

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

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

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

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

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

    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 $\displaystyle \int_{\gamma_2}g(z)dz$, 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Let suppose to 'attack' the slightly different integral...

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

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

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

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





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

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

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

    $\displaystyle \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 $\displaystyle r \rightarrow 0$ and $\displaystyle R \rightarrow \infty$ , tacking into account that ...

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

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

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

    From (6) is evident that is...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Complex Integral in Complex Analysis
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: Aug 21st 2011, 09:46 PM
  2. Real Analysis - Riemann Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 10th 2009, 07:06 PM
  3. improper integral using complex analysis
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Apr 14th 2008, 09:56 AM
  4. complex analysis integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 6th 2007, 12:44 PM
  5. Complex Analysis-- Integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Dec 21st 2006, 09:02 AM

Search Tags


/mathhelpforum @mathhelpforum