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Integral(s) involving differential equations or contour integration

  1. Jul 8, 2011 #1
    Hello, all:
    I have a few questions concerning a definite integral where I am meeting with limited success. The most important problem concerns:
    [tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]
    Contour integration always has been my last resort, but differentiating under the integral sign, I soon found that some of the conditions were violated and simply could not find the right parameterization so that they held.
    Relevant equations:
    Naturally the residue theorem in complex analysis. I don't really have anything else...

    The attempt at a (wrong) solution:
    I attack the doubly improper integral over the entire real line (this will give me twice the integral because the integrand is an even function).
    I note that there are no branch cuts and that poles exist at i and -i, so I construct the usual semicircular contour and evaluate the integral:
    [tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-r}^{r}\frac{\cos{x}}{x^2+1}dx+\int_{0}^{\pi}\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}dx[/tex]
    In both cases I take the limit as r goes to infinity. In big-O notation:
    [tex]\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}\in \frac{O(r)}{O(r^2)}[/tex]
    Which goes to zero as r goes to infinity, so the second integral goes to zero.
    Therefore:
    [tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-\infty}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]
    [tex]\frac{1}{2}\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]
    Then I calculate the complex residue of i (the only pole enclosed in the contour). Since this is a simple pole of a ratio of two functions f(z)=cos(z) and g(z)=(z^2+1), the residue at i is:
    [tex]\frac{f(i)}{g'(i)}=\frac{\cos{i}}{2i}=\frac{\cosh{1}}{2i}=\frac{e+e^{-1}}{4i}[/tex]
    The residue theorem tells me that [itex]2i\pi[/itex] times the sum of the residues gives me the value of the closed contour integral, so I have:
    [tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=2i\pi\frac{e+e^{-1}}{4i}=\frac{\pi}{2}(e+e^{-1})[/tex]
    By the transitive property, this shows that:
    [tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx=\frac{\pi}{4}(e+e^{-1})[/tex]

    When I did the contour integration I must have done something wrong because http://www.wolframalpha.com/input/?i=integrate+cos[x]/[x^2+1]+from+0+to+infinity" gave me this:
    [tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}=\frac{\pi}{2e}[/tex]
    My question is this: what did I do wrong? Also, is there a clever t-parameterization I can use which doesn't violate the conditions of differentiation under the integral sign? If there is a method to subdue this integral outside of contour integration, I would be pleased to know: a series expansion, differentiation under the integral sign (I presume this would require a differential equation solution, as e appears in the result)...

    All help is appreciated, as always.
    QM
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 8, 2011 #2

    micromass

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    Fixed LaTeX. You have to use / in the [ /tex] brackets, not \.
     
    Last edited by a moderator: Apr 26, 2017
  4. Jul 8, 2011 #3
    Thanks, micromass. I'm completely new to this LaTeX equation thing...I've been on Math Equation Object for so long now...
     
  5. Jul 8, 2011 #4

    micromass

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    This is where your mistake lies. You have [itex]\cos(re^{ix})[/itex], but this is the complex cosine and it can grow exponentially large!! I.e. you don't have [itex]|\cos(re^{ix})|\leq 1[/itex] anymore.

    In general, this method won't work because the second integral does not go to 0.

    What to do then? Well, it turns out that we can consider (with C being the semicircular contour):

    [tex]\int_{C}{\frac{e^{iz}dz}{z^2+1}}[/tex]

    This can be split up into two integrals, and in this case, the second integral will go to 0. So we have

    [tex]\int_{C}{\frac{e^{iz}dz}{z^2+1}}\rightarrow \int_{-\infty}^{+\infty}{\frac{e^{ix}dx}{x^2+1}}[/tex]

    And your integral can be found as the real part of this.
     
  6. Jul 8, 2011 #5
    Ahh! Thanks, that does make sense now. In my haste such a simple observation eluded me! You can see clearly why I am not cut out for contour integration. I will attack the problem with this new integrand now.

    Thanks again,
    QM
     
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