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How to use contour, complex analysis to solve integrals?

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    [itex]\int_{-\infty}^{\infty} \frac{\sin(x)}{x}[/itex] using Complex Analysis

    2. Relevant equations

    Contour analysis on [itex]\int_{-\infty}^{\infty} \frac{\sin(x)}{x}[/itex]

    3. The attempt at a solution

    Hello,

    I am completely new to contour integration. I would really appreciate it if someone can walk me through contour integration. And residues and how to perform contour integration.

    So first we choose a contour. What are the guidelines for choosing a contour? Does it matter?

    Thanks, for helping me!
     
  2. jcsd
  3. Nov 12, 2014 #2

    HallsofIvy

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    Well, first you need a contour! And it is simplest to use a closed contour. Since you can't, technically, have a contour "at infinity", choose some number "R" and consider the integral [tex]\int_{-R}^R \frac{sin(x)}{x}dx[/tex]. Circular contours are typically easiest to work with so we might close the contour by taking the semi-circle, from R to -R in the upper half plane. There will be a problem at x= 0 where the is a (removable) singularity. You might, then, change the line from -R to R by taking a small semi-circle, of radius "r", say, from -r to r, also in the upper half plane.

    Then there will be no singularities inside the contour so the integral around the entire contour is 0. You can then say that the integral around the contour is the integral, on the real line, from -R to r and from r to R, plus the integral around the semi-circle with radius R plus the integral around the semi-circle with radius r and that sum is 0. You should be able to show that the integral around the semi-circle with radius R has limit 0 as R goes to infinity so your original integral is the negative of the limit of the integral around the semi-circle of radius r, as r goes to 0.
     
  4. Nov 12, 2014 #3
    Hello @HallsofIvy, thanks for the pleasant reply! Will you advise me with something?

    How does the circular contour give area of the function from negative infinity to infinity? Why do you need a "circle," anyway?

    We go:

    [tex] -R \to 0[/tex] and then [tex] 0 \to R[/itex] then finally, [tex] R \to -R[/tex]

    Here is what I dont understand. why do we need to go from [tex]R \to -R[/tex] if we already have it going from [tex] -R \to R[/tex]

    Or is the integral in the end:

    I = Full length - length of curvature (not diameter) = diameter

    But then again, how does length give area? Thanks HallsofIvy!
     
  5. Nov 12, 2014 #4

    Ray Vickson

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    You are suffering from many mis-conceptions about contour integration, so much so that I think you cannot really hope to learn it in a few PF postings. You really need to read up on it in a lengthier treatment. See. eg.,

    http://en.wikipedia.org/wiki/Methods_of_contour_integration or
    http://web.williams.edu/Mathematics/sjmiller/public_html/372/coursenotes/Trapper_MethodsContourIntegrals.pdf [Broken] or
    https://www.math.ust.hk/~maykwok/courses/ma304/06_07/Complex_4.pdf

    The first link is to a summary; the second link has numerous examples similar to yours (but assumes some background theory already); the third link is to a complete treatment of the subject, giving all the basic results and developing the methods.
     
    Last edited by a moderator: May 7, 2017
  6. Nov 12, 2014 #5

    nrqed

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    The question is of course why take this small semi-circle in the upper half plane and not in the lower half plane :-)
     
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