Using a contour integral about a branch cut to compute $\int \limits ^\infty _0 \frac {\ln x} {x^a (x+1)} dx$

Question

Find the value of $I = \int \limits ^\infty _0 \frac {\ln x} {x^a (x+1)} dx$ for $a \in (0,1)$, placing the branch cut of the logarithm on the positive real axis.

You can use the result that $\int \limits ^\infty _0 \frac 1 {x^a (x+1)} dx = \frac \pi {\sin(\pi a)}$.

Using a keyhole contour to skirt around the branch cut we can determine that the value of the closed integral $\oint\frac {\ln z} {z^a (z+1)} dz$ excluding the real axis is $ 2\pi i \cdot \text{Res(f,-1)} = -2\pi^2 e^{-\pi i a}$ due to the pole at $ z = -1$.

Using this information, I need to find the value of the integral along the real line.

The next step in the question should be to split the contour integral into separate parts and evaluate them individually to deduce the value of the part on the real axis but I haven't have any success in doing so.

Answer 1

HINT:

Let $f(a)$ be the integral defined by

$$f(a)=\int_0^\infty \frac{\log(x)}{x^a(1+x)}\,dx$$

We evaluate $f(a)$ by analyzing the contour integral

$$I=\oint_C\frac{\log(z)}{z^a(1+z)}\,dz$$

where $C$ is the classical keyhole contour integral. We can write

$$\int_0^\infty\frac{\log(x)}{x^a(1+x)}\,dx+\int_\infty^0\frac{\log(x)+i2\pi}{e^{i2\pi a}\,x^a(1+x)}=2\pi i\text{Res}\left(\frac{\log(z)}{z^a(1+z)},z=-1\right) \tag 1$$

---

SPOILER ALERT

From $(1)$, we can write $$\begin{align}(1-e^{-i2\pi a})\int_0^\infty \frac{\log(x)}{x^a(1+x)}&=2\pi i\left(e^{-i2\pi a}\int_0^\infty \frac{1}{x^a(1+x)}\,dx+\text{Res}\left(\frac{\log(z)}{z^a(1+z)},z=-1\right)\right)\\\\\int_0^\infty \frac{\log(x)}{x^a(1+x)}&=\frac{\pi e^{i\pi a}}{\sin(\pi a)}\left(\frac{\pi e^{-i2\pi a}}{\sin(\pi a)}+\frac{i\pi}{e^{i\pi a}}\right)\\\\&=\pi^2\cot(\pi a)\csc(\pi a)\end{align}$$