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Integration using residue theorem (part 2)

  1. Jul 24, 2014 #1
    Hello.
    I need some explanation here. I got the solution but I don't understand something.

    Question:
    Find the integral using Residue Theorem.

    $$\int_{-\infty}^{\infty}\frac{dx}{(x^2+4)^2}$$

    Here is the first part of the solution that I don't understand:

    To evaluate ##\int_{-\infty}^{\infty}\frac{dx}{(x^2+4)^2}##, consider ##\oint_c\frac{dz}{(z^2 + 4)^2}##,
    where C consists of the real axis [-R, R] with R > 2, and the upper half of Γ: |z| = R (all with counterclockwise orientation).

    My question: Why is R>2?
     
  2. jcsd
  3. Jul 24, 2014 #2

    vanhees71

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    Where are the singularities of your integrand? Where are the residues relevant for the integral? Why are you allowed to close the contour with the semi circle (and in which limit)?
     
  4. Jul 24, 2014 #3
    The singularites are at z=2i and 2i. I think R>2 so that the singularity lies in the semicircle so that residue theorem can be applied. But I don't know why it is allowed to to close the contour with the semi circle. I don't understand why this only involves the upper part of the circle.
     
  5. Jul 24, 2014 #4

    vanhees71

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    Just draw the integration path in the complex [itex]z[/itex] plane. The original integral was along the [itex]x[/itex] axis. Now argue, why you are allowed to add the semi-circle in the upper half-plane when taking [itex]R \rightarrow \infty[/itex], i.e., why do you get the correct value for the original integral!
     
  6. Jul 24, 2014 #5
    I am not sure but at some point in the calculation, we're going to let ##R \to \infty## to find the value of ##\int_{-\infty}^\infty \frac{dx}{(x^2 + 4)^2}##. As ##R## increases without bound, ##C## will contain the singularity of ##f## at ##z = 2i##. So that's why we need to choose a R large enough. Why the semicircle in the upper half-plane? I guess because R is going to positive infinity and f(x) is a real number so only the real axis is actually needed. I'm not sure, though. Why do we get the same value as the original integral is because of the residue theorem itself, I guess, but I can't explain why.
     
  7. Jul 24, 2014 #6

    HallsofIvy

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    Integration by summing the residues inside the path applies only to closed paths. That's the reason you need the semicircle from (R, 0) back to (-R, 0). You could have used other path- say, the line from (R, 0) to (0, R), then to (-R, 0) or the lines from (R, 0) to (R, R) then to (-R, R) then to (-R, 0) but they would be much harder to integrate. If you had used the semicircle in the lower half plane, you would get exactly the same answer.
     
  8. Jul 24, 2014 #7

    vanhees71

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    Ok, write down the integral for the semi circle explicitly and then let [itex]R \rightarrow \infty[/itex]!
     
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