The integral is similar to \(\displaystyle I(z) =\int_{0}^{1} \frac{x^z }{\ln x} dx=\frac{1}{z+1} \int_{-\infty}^{0} \frac{e^{t(z+1)}}{t(z+1)}dt=-\frac{1}{z+1 }\int_{0}^{-\infty} (tz+t)^{-1}\cdot \sum_{n=0}^{\infty}\frac{(tz+t)^n }{n!}dt=-\frac{1}{z+1 }\int_{0}^{-\infty} \sum_{n=0}^{\infty}\frac{(tz+t)^{n-1} }{n!}dt\). How can I continue from here ?