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Question Number 154823 by talminator2856791 last updated on 21/Sep/21

$$$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-x}}{x^{\frac{3}{4}}}dx$$$$$$

Answered by Jonathanwaweh last updated on 21/Sep/21

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x}}{x^{\frac{3}{4}}}dx={\int }_0^{\mathrm{\infty }}{x}^{-\frac{3}{4}}{e}^{-x}dx={\int }_0^{\mathrm{\infty }}{x}^{1/4-1}{e}^{-\frac{3}{4}}dx=\mathrm{\Gamma }\left(1/4\right)$$$$$$

Commented by puissant last updated on 21/Sep/21

$$Mrjonathan,c^{\prime }est:$$$${\int }_0^{\mathrm{\infty }}{x}^{\frac{1}{4}-1}{e}^{-x}dx=\mathrm{\Gamma }\left(\frac{1}{4}\right)=\frac{\pi \sqrt{2}}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}..$$

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