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The question is:

Evaluate $\displaystyle \ \int_{- \infty}^{\infty} \frac{x e^{2ix}}{x^2 - 1}\,dx \ $ using the contour below.

(Explain what happens on each part of the contour.)

enter image description here

First of all, isn't this a bad choice of contour? (Since it branch cuts between the 2 singularities)

If we have to do it this way. Do we have to do it in 6 parts:

The upper curve CR

The lower left line segment

The lower right line segment

The center line segment

The 2 arcs around the singularities: Cr1 and Cr2

This seems to be a very complicated situation. How do I do each of these steps?

Is it residue theorem I have to use? How do I apply it in this situation?

I have spent an hour reading about this but didn't get it. I have very limited time to burn on this specific kind of problem. So helps are appreciated, either an answer with steps or intuitive hints are appreciated.

Thanks.

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The integral along the countour you have drawn is $0$, since there are no singularities inside. However, this seems like a strangely chosen example, because the real integral is not convergent in the usual sense. Close to $x=1$, $|f| \approx 1/|x-1|$, with a similar estimate near $x=-1$. (It may still be the case that the integral exists as a generalized principal value, but I haven't worked out the details.) –  mrf Dec 5 '12 at 21:59
    
Yup, I am pretty confused about why this question is asked this way too. I think the point is to ask what happens in each part of the contour and have everything cancelled out in the process of working it out. This is where my problem is too.. Just wondering if my guess of doing it in 6 parts is correct and how I work it our part by part.... –  Angus Leo Dec 5 '12 at 22:08
    
Are you sure you have $\,e^{2ix}\,$ in the numerator? This integral, as mrf wrote, is not convergent...or perhaps there's some other typo? –  DonAntonio Dec 5 '12 at 22:23
    
@DonAntonio: Yup, I checked for like 10 times because it apparently does not converge... But it is what it is in the question. –  Angus Leo Dec 5 '12 at 22:38

2 Answers 2

up vote 2 down vote accepted

As written $$ \int_{-\infty}^{\infty}\frac{xe^{2ix}}{x^2-1}\,\mathrm{d}x\tag{1} $$ diverges. That is, on each of the four intervals, $(-\infty,-1)$, $(-1,0)$,$(0,+1)$ and $(+1,\infty)$, the integral diverges. However, if the gaps on each side of $-1$ are equal and the gaps on each side of $+1$ are equal, then we get the Cauchy Principal Value.

We need to take something like the Cauchy Principal Value for $(1)$ to have a value. Otherwise, we can adjust $(1)$ to tend to any value in $\mathbb{C}$ by adjusting how we approach $-1$ and $+1$. The contour shown gives the Cauchy Principal Value because $C_{r1}$ and $C_{r2}$ are centered on $-1$ and $1$.

There are no singularities inside the contour, so $$ \int_C\frac{ze^{2iz}}{z^2-1}\,\mathrm{d}z=0\tag{2} $$ Along $C_R$, the absolute value of the integrand is less than $\frac1{R-1}e^{-2\mathrm{Im}(z)}$ and the integral vanishes as it is less than $$ \frac{R}{R-1}\int_0^\pi e^{-2R\sin(\theta)}\,\mathrm{d}\theta\tag{3} $$ $(3)$ vanishes as $R\to\infty$ by Monotone Convergence.

Along $C_{r1}$, the integral tends to $-\pi i$ times the residue of the integrand at $-1$. Along $C_{r2}$, the integral tends to $-\pi i$ times the residue of the integrand at $1$.

Thus, the Cauchy Principal Value of $(1)$ equals $\pi i$ times the sum of the residues of the integrand at $-1$ and $1$.

At $z=-1$ the residue of the integrand is $\dfrac{(-1)e^{-2i}}{-2}=\frac12(\cos(2)-i\sin(2))$

At $z=+1$ the residue of the integrand is $\dfrac{(+1)e^{+2i}}{+2}=\frac12(\cos(2)+i\sin(2))$

Therefore, $$ \mathrm{PV}\int_{-\infty}^{\infty}\frac{xe^{2ix}}{x^2-1}\,\mathrm{d}x=\pi i\cos(2)\tag{4} $$

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Awesome, this makes perfect sense to me! –  Angus Leo Dec 13 '12 at 16:23

You don't have to do six parts. What you need to do is the following: a) show that the integral along the big half-circle vanishes as you expand it to the infinity, b) compute the integral along the small half-circles, c) observe that you have obtained the correct result in the easiest possible way.

For a), note that since we are working in the upper half plane, ${\rm Im} x > 0$ and therefore the integrand contains $e^{-t^2}$ which decays very quickly as the half-circle gets bigger and bigger (also, near the real axis, the integrand behaves like $1 \over x$ and so also vanishes as $x \to \infty$).

For b), note that you can approximate the integrand using it's Laurent series around the center of the small contours. Only the $c_{-1}$ terms (the residues) will turn out to be important. You will get a $-i\pi {\rm Res_a} f$ contribution, where $f$ is the integrand, $a$ is the pole, $i\pi$ comes from the half circle (the full circle has length $2\pi$ and $-$ from the clockwise integration.

For c) just note that the integral along the full curve is zero because it contains no poles (Cauchy theorem) and therefore the original integral (obtained as a limit of the integrals of the lines segments) is precisely equal minus integral along the small half-circles. Or in short, sum over $i \pi$ times residue.

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Yes, but you're ignoring the fact that the integral in question doesn't exist (in the usual sense), and you have to argue how the final answer makes sense. Also, there is an error in your estimate over the large semi-circle. (Some version of Jordan's lemma is needed.) –  mrf Dec 5 '12 at 22:20
    
@mrf It'll be great if you can walk us through this. This problem just does not make any sense to me. –  Angus Leo Dec 5 '12 at 23:29
    
The residues at the 2 poles are 1/(2e^2) at x=-1 and e^2/2 at x=1. Clearly the part at the top part doesn't equal to 0. I guess. –  Angus Leo Dec 5 '12 at 23:37
    
Also, is it true that we only have to do it in 3 parts? Why are we ignoring the line segments? –  Angus Leo Dec 5 '12 at 23:43
    
@mrf: thank you for pointing this out. This answer was supposed to be an informative overview of the process (which is what I supposed OP was after) rather than a complete technical proof. There are certainly details to be flashed out. Cheers. –  Marek Dec 6 '12 at 8:43

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