تكامل مملوح

$\int_{0}^{\infty }\frac{sinx}{\sqrt[3]{x}}dx$

$\because \frac{3}{\Gamma (\frac{1}{3})}\int_{0}^{\infty }e^{-xy^{3}}sinxdy=\frac{sinx}{\sqrt[3]{x}}$

$\frac{3}{\Gamma (\frac{1}{3})}\int_{0}^{\infty }(\int_{0}^{\infty }e^{-xy^{3}}sinxdy)dx$

$=\frac{3}{\Gamma (\frac{1}{3})}\int_{0}^{\infty }(\int_{0}^{\infty }e^{-xy^{3}}sinxdx)dy$

$=\frac{3}{\Gamma (\frac{1}{3})}\int_{0}^{\infty }\frac{e^{-xy^{3}}(-y^{3}sinx-cosx)}{y^{6}+1}]_{0}^{\infty }dy$
$=\frac{3}{\Gamma (\frac{1}{3})}\int_{0}^{\infty }\frac{dy}{1+y^{6}}dy$

$=\frac{3}{\Gamma (\frac{1}{3})}.\frac{\frac{\pi }{6}}{sin(\frac{\pi }{6})}=\frac{\pi }{\Gamma (\frac{1}{3})}$