Here is the problem:
With help from Jack D'Aurizio
We were able to prove that the contour integral of the big semi circle $=0$ as $R \to \infty$.
Now the problem is the small semi-circle.
How do I prove as $\delta \to 0$ that the contour integral of the small semi-circle $=0$?
I can try to eliminate the $\log$
We start with
$$\left | \int f \right| < \int |f| $$
$$|f| < \frac{\sqrt{\log^2(\delta) - \theta^2}}{\delta^2 - 1}$$
We take the limit as $\delta \to 0$, the RHS $\to -\infty$ which is no good.
But we see that, $x^2 > \log^2(x)$
So:
$$|f| < \frac{\sqrt{\log^2(\delta) - \theta^2}}{\delta^2 - 1} < \frac{\sqrt{\delta^2 - \pi^2}}{\delta^2 - 1} $$
But that is still no good.
What we can however do is prove that:
$$x^2 > \log^2(x) - \pi^2$$
It is already true that: for x > 0
$$x^2 > \log^2(x)$$ $$0 > -\pi^2$$
$$\therefore x^2 > \log^2(x) - \pi^2$$
The limit if $x^2$ is there will be 0.
But will the estimation lemma work?
