To say that $\displaystyle \int_{0}^{1} \frac{\ln x}{x-a} \ dx = -\frac{1}{a} \int_{0}^{1} \ln x \sum_{n=0}^{\infty} \Big( \frac{x}{a} \Big)^{n} \ dx $ musn't $\displaystyle \Big| \frac{x}{a} \Big| < 1 $ for $\displaystyle 0 < x < 1 $ ?