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Improper Integral Using Complex Analysis

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Compute the Integral: ##\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx##

    2. Relevant equations
    ##\int_C \space f(z) = 2\pi i \sum \space res \space f(z)##

    3. The attempt at a solution
    At first I tried doing this using a bounded integral but couldn't seem to get the right answer from there. I could get it using Jordan's Lemma on a semi circular curve traced in the lower half plane. I'm just curious why the original approach didn't work.

    $$\int_{-R}^R \space \frac{e^{-2ix}}{x^2+4}dx + \int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz = 2\pi i (\lim_{z\to 2i} (z-2i)\frac{e^{-2iz}}{z^2+4}) = 2\pi i (\lim_{z\to 2i} \frac{e^{-2iz}}{z+2i})$$

    $$|\int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz| \leq \int_{C_R} \space |\frac{e^{-2iz}}{z^2+4}|dz \leq \int_{C_R} \space \frac{|e^{-2iz}|}{|z+2i||z-2i|}dz \leq \int_{C_R} \space \frac{1}{(R-2i)(R+2i)}dz$$
    $$|\int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz| \leq \frac{1}{(R-2i)(R+2i)} \int_{C_R} \space dz = \frac{1}{(R-2i)(R+2i)} \pi R = \frac{\pi R}{(R-2i)(R+2i)}$$
    which goes to 0 as R approaches infinity. The last inequality in the 2nd line comes from using the form of the triangle inequality saying ##|a+b| \geq |a|-|b|##.

    Then as ## R\to\infty##,
    $$\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx + \int_{C_\infty} \space \frac{e^{-2iz}}{z^2+4}dz = 2\pi i (\lim_{z\to 2i} \frac{e^{-2iz}}{z+2i})$$
    $$\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx + 0 = 2\pi i \frac{e^{-2i*2i}}{2i+2i} = 2\pi i \frac{e^4}{4i}$$
    $$\int_{-\infty}^\infty \space \frac{e^{-2ix}}{x^2+4}dx = \frac{\pi e^4}{2}$$
    where using Jordan's Lemma in the negatively traced curve gives ##\frac{\pi}{2e^4}##. Any ideas as to why the bounded integral method does not work, or where I made a mistake in the derivation at? Thanks ahead of time.
     
  2. jcsd
  3. Apr 26, 2017 #2

    Ray Vickson

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    What do you mean by the "bounded integral" method? I have never heard of it.
     
  4. Apr 26, 2017 #3
    Rather than just using Jordan's Lemma to say an integral goes to zero, we try to find an upper bound for the integral instead. For these problems the upper bound depended on the radius of the contour, and then as you took that radius to infinity, the upper bound went to 0, so the integral had to go to zero. The 2nd and 3rd lines of integrals are what I was referring to. I'm sure its not called "the bounded integral method", I just didn't know how else to refer to it.
     
  5. Apr 26, 2017 #4

    Ray Vickson

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    I am not sure the method has a name; it is just using the definition
    $$\int_{-\infty}^{\infty} f(x) \, dx = \lim_{R_1,R_2 \to \infty} \int_{-R_1}^{R_2} f(x) \, dx.$$
    Usually we can take ##R_1 = R_2 = R## in that definition, but we should not do that automatically because in some cases the double limit (with independent ##R_1,R_2##) may not exist, while the ##R_1 = R_2## limit does exist---it is called the Cauchy Principal Value. Anyway, we then try to complete the contour in such a way that the integration along the rest of the contour (joining ##R_2 + i 0## to ##-R_1 + i 0## out in the complex plane) goes to zero. It is more-or-less obvious that we can take ##R_1 = R_2 = R## in your case.

    What we should never, ever do is just automatically write the integral as ##\int_C f(z) \, dz## without first checking somehow that the integration over the remainder of the contour goes to zero.
     
  6. Apr 26, 2017 #5
    Is that not what I did when I came up with the bound ##|\int_{C_R} \space \frac{e^{-2iz}}{z^2+4}dz| \leq \frac{\pi R}{(R-2i)(R+2i)}## which goes to 0 as ##R\to\infty##?
     
  7. Apr 26, 2017 #6

    Ray Vickson

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    Yes, of course it is, but that did not deal with my original question, which is what you meant by the term "bounded integral method". As far as I can make out, you are just using the residue method correctly, so trying to rename it only leads to confusion.
     
  8. Apr 26, 2017 #7
    I said what I was talking about in my first reply. But the whole the answer I get using this method is incorrect, the whole residue method and showing the integral over the contour is bounded by 0. I get the wrong value for my residue doing it this way. I believe the value for the residue at the singularity in the semicircle is correct, but wrong for this given situation.
     
  9. Apr 26, 2017 #8

    Dick

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    If you are using a contour closed in the lower half plane, as you should - it's where the exponential has a negative real part, then why are you computing the residue from the pole ##2i## in the upper half plane?
     
  10. Apr 26, 2017 #9

    Ray Vickson

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    I don't know what you mean when you say that "the residue at the singularity ...is correct, but wrong for this given situation".

    When I do the method, I use a correct residue and get the correct answer (= ##\pi e^{-4}/2##), with no ambiguities or confusion whatsoever. Basically, I apply the Cauchy integral
    $$\oint_C \frac{f(z)}{z-z_0} \, dz = - 2 \pi i f(z_0),$$
    with ##f(z) = e^{-2 i z}/(z - 2 i)##, ##z_0 = - 2i## and with ##C## a clockwise semi-circle or clockwise-oriented rectangle in the lower half-plane. (Using a rectangle is a lot easier than using a semi-circle.)
     
    Last edited: Apr 26, 2017
  11. Apr 26, 2017 #10
    That's my question. Why does it HAVE to be in the lower half plane? I showed the upper half plane was bounded by zero, which obviously isn't true otherwise the upper half plane would work. I assume there is a flaw in showing that the upper half plane is bounded.
     
  12. Apr 26, 2017 #11

    Dick

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    The problem with closing it in the upper half plane is that the exponential isn't bounded by 1.
     
  13. Apr 26, 2017 #12
    How so? ##|re^{i\theta}|=|r|## no matter where it is placed, is it not? So in this case ##r=1##, ##\theta = -2iz## and ##|e^{-2iz}| = 1##?
     
  14. Apr 26, 2017 #13

    Ray Vickson

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    Yes, it has to be the lower half-plane: if you have ##e^{i a z}## with ##z = x + i y##, the exponential equals ##e^{i a x} e^{-a y}##. Thus when ##a > 0## you need to complete in the upper half plane (to get a damped exponential ##e^{-ay}##) and when ##a < 0##---as in your problem--- you need to go into the lower half-plane in order to get damping.
     
    Last edited: Apr 26, 2017
  15. Apr 26, 2017 #14

    Dick

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    ##\theta## supposed to be a real angle. Try finding ##e^{-2iz}## when ##z=8i##.
     
  16. Apr 26, 2017 #15
    Those last 2 replies helped a lot with that. Especially since I wasn't thinking and didn't think about ##\theta## having to be a real number. Thanks for the help.
     
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