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

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