積分解説6

$\displaystyle\frac{x}{x+1}=t$の置換により、
$\displaystyle\begin{eqnarray} \int_0^∞\frac{\sqrt[3]{x+1}-\sqrt[3]{x}}{\sqrt{x}}dx &=& \int_0^1\frac{\sqrt[3]{\frac{1}{1-t}}-\sqrt[3]{\frac{t}{1-t}}}{\sqrt{\frac{t}{1-t}}}\frac{dt}{(1-t)^2} \\ &=& \int_0^1 t^{-\frac{1}{2}}(1-t)^{-\frac{11}{6}}\int_t^1\frac{\partial}{\partial s}\sqrt[3]{s}dsdt \\ &=& \int_0^1 t^{-\frac{1}{2}}(1-t)^{-\frac{11}{6}}\int_t^1\frac{1}{3}s^{-\frac{2}{3}}dsdt \\ &=& \int_0^1 \frac{1}{3}s^{-\frac{2}{3}}\int_0^s t^{-\frac{1}{2}}(1-t)^{-\frac{11}{6}}dtds \\ &=& \int_0^1 \frac{1}{3}s^{-\frac{2}{3}}\int_0^1(st)^{-\frac{1}{2}}(1-st)^{-\frac{11}{6}}sdtds \\ &=& \int_0^1\frac{1}{3}s^{-\frac{1}{6}}\frac{\Gamma(\frac{1}{2})\Gamma(1)}{\Gamma(\frac{3}{2})}F\big(\frac{1}{2},\frac{11}{6};\frac{3}{2};s\big)ds \\ &=& \frac{1}{3}\frac{6}{5}\frac{1}{\frac{1}{2}}F\big(\frac{1}{2},\frac{5}{6};\frac{3}{2};1\big) \\ &=& \frac{4}{5}\frac{\Gamma(\frac{3}{2})\Gamma(\frac{1}{6})}{\Gamma(1)\Gamma(\frac{2}{3})} \\ &=& \frac{2}{5}\frac{\sqrt{\pi}\Gamma(\frac{1}{6})}{\Gamma(\frac{2}{3})} \end{eqnarray}$