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

$$\begin{array}{rcl}{\int }_0^1\frac{u^{z-1}}{1-u}\mathrm{d}u& =& {\int }_0^1{u}^{z-1}\frac{1}{1-u}\mathrm{d}u\\ & =& {\int }_0^1{u}^{z-1}(1+u+u^2+\cdots )\mathrm{d}u\\ & & \dots \frac{1}{1-u}=1+u+u^2+\cdots (|u|<1)\\ & =& {\int }_0^1(u^{z-1}+u^{z-1}u+u^{z-1}{u}^2+\cdots )\mathrm{d}u\\ & =& {\int }_0^1(u^{z-1}+u^{z-1+1}+u^{z-1+2}+\cdots )\mathrm{d}u\\ & =& {\int }_0^1(u^{z-1}+u^z+u^{z+1}+\cdots )\mathrm{d}u\\ & =& {[\frac{1}{z-1+1}{u}^{z-1+1}+\frac{1}{z+1}{u}^{z+1}+\frac{1}{z+1+1}{u}^{z+1+1}+\cdots ]}_0^1\\ & =& {[\frac{1}{z}{u}^z+\frac{1}{z+1}{u}^{z+1}+\frac{1}{z+2}{u}^{z+2}+\cdots ]}_0^1\\ & =& [\frac{1}{z}{1}^z-\frac{1}{z}{0}^z+\frac{1}{z+1}{1}^{z+1}-\frac{1}{z+1}{0}^{z+1}+\frac{1}{z+2}{1}^{z+2}-\frac{1}{z+2}{0}^{z+2}+\cdots ]\\ & =& \frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots \\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{z+k}\end{array}$$