Contour Integration: $\int_0^\infty \frac{\ln{x}^2}{1+x^2}dx = \frac{\pi^3}{8}$

Question

This is question VI.2.14a in Lang, 3rd edition. The contour it's being taken on is a closed half annulus counterclockwise, from a value R on the real axis along a semicircle to -R along a straight line on the real axis to -$\delta$ along a semicircle to $\delta$ then along a straight line back to R.

Shakarchi defines a function $f(z)=\frac{(\ln{z}-i\pi/2)^2}{1+z^2}$.

I'm having trouble understanding why the integral of $f(z)$ along the semicircles is 0 - he explains that it's something about the length of the semicircle behaving like log functions, but that doesn't make sense to me.

Any help would be really appreciated!

Answer 1

It's more obvious why the larger semicircle goes to $0$ than the smaller one. For the smaller one, note that

$$\Bigr|\int_\delta f(z)dz\Bigr| \leq \pi \delta \cdot \left(\frac{|\log \delta^2| + r \pi}{1+0^2}\right) \to 0$$

where $r\in\mathbb{Q}$ is dependent on your choice of branch cut. This limit holds because $x\ln x \to 0$ as $x\to 0$

Answer 2

As $|z|$ increases you have $(\ln z-i\pi/2)^2 =o(|z|)=o(R)$. Then the absolute value of the semicircle integral must satisfy

$I_{semicirc}=o((\pi R)(R/(1+R^2)))<=o(2\pi)=o(1)$

which forces $I_{semicirc}$ to zero as claimed.