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Question Number 191631 by Spillover last updated on 27/Apr/23

$${\int }_0^{\mathrm{\infty }}{x}^{\frac{1}{2}}{e}^{-x^2}dx$$

Answered by malwan last updated on 28/Apr/23

$$put{x}^2=y\Rightarrow x=y^{\frac{1}{2}}$$$$\therefore {}_0{\int }^{\mathrm{\infty }}{y}^{\frac{1}{4}}{e}^{-y}\left(\frac{1}{2}{y}^{-\frac{1}{2}}dy\right)$$$$=\frac{1}{2}{}_0{\int }^{\mathrm{\infty }}{y}^{-\frac{1}{4}}{e}^{-y}dy$$$$=\frac{1}{2}\times \frac{\mathrm{\Gamma }(-\frac{1}{4}+1)}{1^{-\frac{1}{4}+1}}=\frac{\mathrm{\Gamma }\left(\frac{3}{4}\right)}{2}$$$$MohammedAlwan$$$$28/4/2023$$

Commented by Spillover last updated on 28/Apr/23

$$thanks$$

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