I'm supposed to solve the real integral using a contour integral (The Cauchy Principal Value). Can someone give me a hand? I cannot seem to be able to do it…
This is what I've tried so far:
-
I tried computing the contour integral first using the Residue theorem. I simplified the total residue to :$$-\cos(\pi a) – i\sin(\pi a)$$
-
I also tried showing that the part of the integral that doesn't lie on the real line goes to zero as $R$ goes to infinity. (where $R$ is the bound on the integral). I tried using the ML equality in combination with the triangle equality to no avail.
Thank you.
Best Answer
After we have agreed that this is just an ordinary integral, the rest is not tough:
Let's denote the four paths by $I_1,I_2,I_3,I_4$ where $I_1$ corresponds to the integral in question, $I_2, I_4$ are the parts parallel to the imaginary axis and $I_3$ is the horizontal one which passes $z=2\pi i$. Using the contour above, we easily see that the horizontal parts only differ by phase so $$ I_3=-e^{2\pi a i}I_1 $$
Furthermore it's easy to show that $I_2,I_4$ are vanishing in the limit $|r|\rightarrow\infty$. We can also easily check that there is only one residue at $z=\pi i$ inside the contour. We can conclude that
$$ I_1-e^{2\pi a i}I_1=2\pi i Res[z=i\pi] $$ Using $Res[z=i\pi]=-i e^{i \pi a}$ we get $$ I_1=\frac{\pi}{\sin(\pi a)} $$