I am trying to compute the following integral, for complex $a$ and $b$ $$ \int^1_0 dx \frac{\ln(x-a)}{x-b} $$ by turning it into something in terms of dilogarithms. But for certain values of $a$ and $b$ the logarithmic cut gets in the way, and is impeding my progress. I need help with keeping track of the cut in my computation.
I've tried to take this step by step, checking myself against a numerical integration carried out using Mathematica.
My steps are as follows:
- Shift integration variables $x\rightarrow x+a$: $$I = \int_{-a}^{1-a}dx\frac{\ln(x)}{x+a-b}$$
- Split integration (according to Mathematica, this is NOT correct for all $a$ and $b$): $$I = \int_{0}^{1-a}dx\frac{\ln(x)}{x+a-b}-\int_{-0}^{-a}dx\frac{\ln(x)}{x+a-b}$$
- In each term rescale the integral: $x\rightarrow (1-a)x$ in the first integral, $x\rightarrow-a x$ in the second. $$I=\int_0^1 dx \frac{\ln((1-a)x)}{x+\frac{a-b}{1-a}}-\int_0^1 dx \frac{\ln(-a x)}{x-\frac{a-b}{a}}$$
- Next, write the denominators as derivatives of logarithms $1/(x+\lambda)=\frac{d}{dx}\ln(1+x/\lambda)$ $$\int_0^1 dx \frac{d}{dx} \Big[\ln\Big(\frac{1-a}{a-b}x+1\Big)\Big]\ln((1-a)x)+\int_0^1 dx \frac{d}{dx} \Big[\ln\Big(\frac{-a}{a-b}x+1\Big)\Big]\ln{(-ax)}$$
- Perform a partial integration $$I=-\int_0^1\frac{dx}{x}\ln\Big(1+\frac{1-a}{a-b}x\Big)+\Big[\ln\Big(1+\frac{1-a}{a-b}x\Big)\ln((1-a)x)\Big]^1_0\\ +\int_0^1\frac{dx}{x}\ln\Big(1+\frac{-a}{a-b}x\Big)+\Big[\ln\Big(1+\frac{-a}{a-b}x\Big)\ln(-ax)\Big]^1_0$$
- Convert to dilogarithms: $$I=\text{Li}_2\Big(\frac{-1+a}{a-b}\Big)-\text{Li}_2\Big(\frac{a}{a-b}\Big)+\ln\Big(1+\frac{1-a}{a-b}\Big)\ln(1-a)-\ln\Big(1-\frac{a}{a-b}\Big)\ln(-a)$$
But this expression is not true for any complex values of $a$ and $b$. How do I modify my analysis to take into account of the logarithmic cut?