This is a contour integration problem, so I preemptively say that the integral along the semi-circle of radius goes to as by Jordan's lemma or something like that since that's not the main concern of the problem.

So where is the singularity inside our contour.

, where and .

So let's solve .

Note .

Consider

Now

Therefore we want our term to get a " " and change sign to cancel out the 's. Doing this though luckily can cancel out and too!

Thus our solution to solve this equation is .

Observe

We now get by L'Hopital's rule.

Therefore .