Integral representation of Dilogarithm

Question

I have a question regarding how to obtain a certain integral representation of the dilogarithm, namely:

$$\DeclareMathOperator{\li}{Li_2}\li(x) = -\int_{0}^{1}\frac{x\ln t}{1-tx}\mathrm dt\quad ,|x|<1\qquad (a)$$

I know that the dilogarithm is commonly defined as

$$\li(x) = -\int_{0}^{x}\frac{\ln(1-t)}{t}\mathrm dt \overset{t\to xt}{=} -\int_{0}^{1}\frac{\ln(1-xt)}{t}\mathrm dt\qquad(b)$$

But how do we go from $(b)$ to $(a)$? My main issue is reconciling the limits of integration after substitution. For example,

$$\li(x) = -\int_{0}^{x}\frac{\ln(1-t)}{t}\mathrm dt \stackrel{t\to 1-p}{=} -\int_{1}^{1-x}\frac{\ln p}{1-p}\mathrm dp\stackrel{p\to xy}{=}-\int_{\frac{1}{x}}^{\frac{1}{x}-1}\frac{x\ln y}{1-xy}\mathrm dy\qquad (c)$$

I checked $(c)$ and discovered that the integral is undefined...I can't figure out why that is the case. The integrands in $(a)$ and $(c)$ are the same, but the limits differ. Are the limits I calculated from my substitutions incorrect? Also, why are the limits in $(a)$ correct?

Answer 1

But how do we go from (b) to (a)? - just integrate by parts © $$(b)=-\ln t\ln(1-xt)\Big|_{t=0}^{t=1}-\int_0^1\frac{x\ln t}{1-xt}\,dt=(a).$$