分割し重ねることで無限区間積分を有限区間積分にする

${\displaystyle (\text{右辺})=\sum _{n=0}^{N-1}{\int }_a^{a+b}f(x+nb)dx=\sum _{n=0}^{N-1}{\int }_{a+nb}^{a+(n+1)b}f(x)dx=(\text{左辺})}$

$\begin{array}{rl}(\text{左辺})& ={\int }_0^1\sum _{n=0}^{\mathrm{\infty }}\frac{\prod _{k=1}^r\mathrm{\Gamma }(a_k+t+n)}{\prod _{k=1}^s\mathrm{\Gamma }(b_k+t+n)}{x}^{t+n}dt\\ & ={\int }_0^1\frac{\prod _{k=1}^r\mathrm{\Gamma }(a_k+t)}{\prod _{k=1}^s\mathrm{\Gamma }(b_k+t)}{x}^t\sum _{n=0}^{\mathrm{\infty }}\frac{(a_1+t,\cdots ,a_r+t{)}_n}{(b_1+t,\cdots ,b_s+t{)}_n}{x}^{n}dt\\ & =(\text{右辺})\end{array}$

$\begin{array}{rl}(\text{左辺})& ={\int }_0^1\frac{\mathrm{\Gamma }(a+t)}{\mathrm{\Gamma }(b+t)}{}_2{F}_1\left[\begin{array}{c}a+t,1\\ b+t\end{array};1\right]dt\\ & ={\int }_0^1\frac{\mathrm{\Gamma }(a+t)}{\mathrm{\Gamma }(b+t)}\frac{\mathrm{\Gamma }(b+t)\mathrm{\Gamma }(b-a-1)}{\mathrm{\Gamma }(b-a)\mathrm{\Gamma }(b+t-1)}dt\because \text{超幾何定理}\\ & =(\text{右辺})\end{array}$

$\begin{array}{rl}(\text{左辺})& ={\int }_0^{\mathrm{\infty }}\frac{dt}{(a+t{)}_n}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+t)}{\mathrm{\Gamma }(a+n+t)}dt\\ & =\frac{1}{n-1}{\int }_0^1\frac{\mathrm{\Gamma }(a+t)}{\mathrm{\Gamma }(a+n+t-1)}dt=\frac{1}{n-1}{\int }_0^1\frac{dt}{(a+t{)}_{n-1}}\\ & =(\text{右辺})\end{array}$

$\begin{array}{rl}(\text{左辺})& ={\int }_a^{\mathrm{\infty }}{t}^{-s}{x}^{t}dt\\ & =(-\ln x{)}^{s-1}{\int }_{-a\ln x}^{\mathrm{\infty }}{u}^{-s}{e}^{-u}du\left(t=-\frac{u}{\ln x}\right)\\ & =(\text{右辺})\end{array}$