u^(z-1)/(1-u)における定積分[0,1],もしくは1/(z+k)の無限級数

\(\frac{u^{z-1}}{1-u}\)における定積分[0,1],もしくは\(\frac{1}{z+k}\)の無限級数

$$\begin{eqnarray} \int_{0}^1 \frac{u^{z-1}}{1-u} \mathrm{d}u &=&\int_{0}^1 u^{z-1}\frac{1}{1-u} \mathrm{d}u \\&=&\int_{0}^1 u^{z-1}\left(1+u+u^2+\cdots\right) \mathrm{d}u \\&&\;\ldots\href{https://shikitenkai.blogspot.com/2021/07/11-x-1.html}{\frac{1}{1-u}=1+u+u^2+\cdots\;\;\;\left(|u|\lt1\right)} \\&=&\int_{0}^1 \left(u^{z-1}+u^{z-1}u+u^{z-1}u^2+\cdots\right) \mathrm{d}u \\&=&\int_{0}^1 \left(u^{z-1}+u^{z-1+1}+u^{z-1+2}+\cdots\right) \mathrm{d}u \\&=&\int_{0}^1 \left(u^{z-1}+u^{z}+u^{z+1}+\cdots\right) \mathrm{d}u \\&=&\left[\frac{1}{z-1+1}u^{z-1+1}+\frac{1}{z+1}u^{z+1}+\frac{1}{z+1+1}u^{z+1+1}+\cdots\right]_0^1 \\&=&\left[\frac{1}{z}u^z+\frac{1}{z+1}u^{z+1}+\frac{1}{z+2}u^{z+2}+\cdots\right]_0^1 \\&=&\left[\frac{1}{z}1^z-\frac{1}{z}0^z+\frac{1}{z+1}1^{z+1}-\frac{1}{z+1}0^{z+1}+\frac{1}{z+2}1^{z+2}-\frac{1}{z+2}0^{z+2}+\cdots\right] \\&=&\frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots \\&=&\sum_{k=0}^\infty \frac{1}{z+k} \end{eqnarray}$$