## integral and sum

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