I'm trying to use a contour integral to integrate $$K = \int_0^{\infty}\frac{z^{1/2}\log(z)}{(1+z)^2}dz$$
I used the "squaring to log" trick with a keyhole contour on the postive real axis. I must be bungling something though.
When I compute the residue of $\frac{z^{1/2}\log(z)^2}{(1+z)^2}$ at $-1$, I obtain $2\pi i(2\pi + \frac{i\pi^2}{2})$ (obtained by taking the derivative of the numerator and evaluating at $-1$). On the otherhand, when I simplify $$\int_0^{\infty}\frac{z^{1/2}\log(z)^2}{(1+z)^2}-\frac{z^{1/2}(\log(z)+2\pi i)^2}{(1+z)^2}dz = \\-4\pi i K+2\pi^3$$
I suspect something is wrong because the real parts of this don't match my residue computation. What am I missing?
Best Answer
We provide support for the main steps. Introducing $\mathrm{Log}(z)$, the branch with argument in $[0,2\pi)$ we integrate
$$f(z) = \exp((1/2)\mathrm{Log}(z)) \frac{\mathrm{Log}(z)}{(1+z)^2}$$
along a keyhole contour with the slit on the positive real axis. We get in the limit above the slit
$$K = \int_0^\infty \sqrt{x}\frac{\log(x)}{(1+x)^2} \; dx.$$
Below the slit we find
$$\int_\infty^0 \exp(\pi i) \sqrt{x} \frac{\log(x)+2\pi i}{(1+x)^2} \; dx = \int_0^\infty \sqrt{x} \frac{\log(x)+2\pi i}{(1+x)^2} \; dx \\ = K + 2\pi i \int_0^\infty \sqrt{x} \frac{1}{(1+x)^2} \; dx = K + 2\pi i J.$$
We have the two contributions separated into real and imaginary parts, so we just need to compute the residue at $z=-1$ of $f(z).$ Differentiating we find
$$\frac{1}{2z} \exp((1/2)\mathrm{Log}(z)) \mathrm{Log}(z) + \frac{1}{z} \exp((1/2)\mathrm{Log}(z)).$$
Evaluate at $z=-1$ to get
$$-\frac{1}{2} \exp((1/2)\pi i) (\pi i) - \exp((1/2)\pi i) = \frac{\pi}{2} - i.$$
With $K$ and $J$ real we collect the contributions to obtain
$$2K + 2\pi i J = 2\pi i \times \left(\frac{\pi}{2} - i\right) = \pi^2 i + 2\pi.$$
This means that
$$\bbox[5px,border:2px solid #00A000]{ K = \pi \quad\text{and}\quad J = \frac{\pi}{2}.}$$
The required ML estimates for the circular components go through with $\lim_{R\to\infty} 2\pi R \times \sqrt{R} \log(R)/(1+R)^2 = 0$ and $\lim_{\epsilon\to 0} 2\pi \epsilon \times \sqrt{\epsilon} \log(\epsilon) /(1+\epsilon)^2 = 0.$