$$\int_{0}^{\infty} \ln{(1+a x)}{x^{-b-1}} dx$$
I defined two branch cuts along the real axis: $[-\infty ,-\frac{1}{a}]$ & $[0,\infty]$ with the following contour: 
I defined the $arg{(z)} =0$ above the positive branch cut and $arg(z)=2\pi$ below the positive branch cut. Similarly $arg(z)=\pi$ above the negative branch cut and $arg(z)=-\pi$ below the negative branch cut.
Using the triangle inequality for integrals it can easily (but with care) be shown that the integrals along all parts of the circles tend to $0$ as $R \Rightarrow \infty$ (radius of the outer circle) and $\epsilon \Rightarrow 0$ (radius of the smaller circles around the singularities).
By doing this we get the necessary bounds for a and b: $a>0$ and $0<b<1$
The integrals along the positive branch cut work out nicely and result in: $$(1-e^{-2 b \pi i}) \int_{0}^{\infty} \ln{(1+a x)}{x^{-b-1}} dx$$
When I work out the integrals along the I end up with the following for the integral along the top side of the negative branch cut:
$$e^{-b \pi i} \int_{\infty}^{\frac{1}{a}} \ln{(1-a x)}{x^{-b-1}} dx$$
After factoring out $-1$ from the inside of the ln you are left with:
$$e^{-b \pi i} \int_{\infty}^{\frac{1}{a}} (\ln{(a x-1)}-i \pi){x^{-b-1}} dx$$
Applying the same procedure to the integral along the bottom side of the negative branch cut yields: $$e^{b \pi i} \int_{\frac{1}{a}}^{\infty} (\ln{(a x-1)}+i \pi){x^{-b-1}} dx$$
This is troublesome because the $e^{b \pi i}$ stops me from combining the two integrals along the negative branch cut in order to cancel out the integrals involving the $\ln$
In the following article the author magically has $e^{-b \pi i}$ in front of the integral which allows him to cancel out the parts including the $\ln$ which simplifies it enormously.
Can someone please explain to me what I am doing wrong with the integrals along the negative branch cut?
Any help is greatly appreciated!
Best Answer
On the negative branch cut, $\arg z$ stays continuous (and is equal to $\pi$; it is $\arg(1+az)$ that gets discontinuous). So your last two integrals should get $e^{-b\pi i}$ both.