積分解説6

${\displaystyle \frac{x}{x+1}=t}$の置換により、
${\displaystyle \begin{array}{rcl}{\int }_0^{\infty }\frac{\sqrt[3]{x+1}-\sqrt[3]{x}}{\sqrt{x}}dx& =& {\int }_0^1\frac{\sqrt[3]{\frac{1}{1-t}}-\sqrt[3]{\frac{t}{1-t}}}{\sqrt{\frac{t}{1-t}}}\frac{dt}{(1-t{)}^2}\\ & =& {\int }_0^1{t}^{-\frac{1}{2}}(1-t{)}^{-\frac{11}{6}}{\int }_t^1\frac{\partial }{\partial s}\sqrt[3]{s}dsdt\\ & =& {\int }_0^1{t}^{-\frac{1}{2}}(1-t{)}^{-\frac{11}{6}}{\int }_t^1\frac{1}{3}{s}^{-\frac{2}{3}}dsdt\\ & =& {\int }_0^1\frac{1}{3}{s}^{-\frac{2}{3}}{\int }_0^s{t}^{-\frac{1}{2}}(1-t{)}^{-\frac{11}{6}}dtds\\ & =& {\int }_0^1\frac{1}{3}{s}^{-\frac{2}{3}}{\int }_0^1(st{)}^{-\frac{1}{2}}(1-st{)}^{-\frac{11}{6}}sdtds\\ & =& {\int }_0^1\frac{1}{3}{s}^{-\frac{1}{6}}\frac{\mathrm{\Gamma }(\frac{1}{2})\mathrm{\Gamma }(1)}{\mathrm{\Gamma }(\frac{3}{2})}F(\frac{1}{2},\frac{11}{6};\frac{3}{2};s)ds\\ & =& \frac{1}{3}\frac{6}{5}\frac{1}{\frac{1}{2}}F(\frac{1}{2},\frac{5}{6};\frac{3}{2};1)\\ & =& \frac{4}{5}\frac{\mathrm{\Gamma }(\frac{3}{2})\mathrm{\Gamma }(\frac{1}{6})}{\mathrm{\Gamma }(1)\mathrm{\Gamma }(\frac{2}{3})}\\ & =& \frac{2}{5}\frac{\sqrt{\pi }\mathrm{\Gamma }(\frac{1}{6})}{\mathrm{\Gamma }(\frac{2}{3})}\end{array}}$