Proving $-\text{Li}_2(x^{-1}-1)+\text{Li}_2(1-x^{-1})-\text{Li}_2(2 x)+\text{Li}_2(2-2 x)+\text{Li}_2(2) = i \pi \log(x)$ for $x>1/2$.

Question

While working on a physics problem, I have stumbled upon the following identity. For $x>\frac{1}{2}$ note $$ -\text{Li}_2\left(\frac{1}{x}-1\right)+\text{Li}_2\left(1-\frac{1}{x}\right)-\text{Li}_2(2 x)+\text{Li}_2(2-2 x)+\text{Li}_2(2) \ = \ i \pi \log(x) $$ where $\mathrm{Li}_2$ is the polylogarithm of order 2.

My physics problem suggests LHS = RHS in the above (and I have confirmed numerically this is true).

How does one prove this? I have not been able to figure out how to use standard polylog identities to show this.

Answer 1

Let

$$f(x)=-\text{Li}_2\left(\frac{1}{x}-1\right)+\text{Li}_2\left(1-\frac{1}{x}\right)-\text{Li}_2(2 x)+\text{Li}_2(2-2 x)+\text{Li}_2(2),$$

$$f(x)=\int \mathrm{d} f(x)=\int\left(\frac{\ln\left(\frac{2x-1}{x}\right)}{x(x-1)}+\frac{\ln(x)}{x(x-1)}+\frac{\ln(1-2x)}{x}+\frac{\ln(2x-1)}{1-x}\right)\mathrm{d}x$$

$$=\int \left(\frac{\ln(2x-1)}{x(x-1)}+\frac{\ln(1-2x)}{x}+\frac{\ln(2x-1)}{1-x}\right)\mathrm{d}x$$

$$=\int\left(-\frac{\ln(2x-1)}{x}-\frac{\ln(2x-1)}{1-x}+\frac{\ln(1-2x)}{x}+\frac{\ln(2x-1)}{1-x}\right)\mathrm{d}x$$

$$=\int\frac{\ln\left(\frac{1-2x}{2x-1}\right)}{x}\mathrm{d}x=\ln(-1)\ln(x)=i\pi\ln(x)+c.$$

Setting $x=1$, we have $c=0$ and the proof is finalized.