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Show that $\int_{0}^{\infty}\,\sqrt{x}\,e^{-\sqrt[3]{x}}\,dx=\frac{315}{16}\sqrt{\pi}$ .
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written 6.1 years ago by | • modified 6.1 years ago |
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$ Put \thinspace x^{ \frac{1}{3} } = t \Longrightarrow x = t^3 \\ $
$ dx = 3t^2 dt \\ $
$ \int_0^{\infty} t^{\frac{3}{2}} e^{-t} . 3t^2 dt = 3 \int_0^{\infty}.e^{-t} . t^{\frac{7}{2}} dt \\ $
$ = 3 \Gamma{\frac{9}{2}} \\ $
$ = 3 * \frac{7}{2}* \frac{5}{2}* \frac{3}{2}* \frac{1}{2}* \Gamma{\frac{1}{2}} \\ $
$ = \frac{315}{16} \Gamma{\pi} \\ $