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Question Number 207769 by 073 last updated on 25/May/24

Commented by 073 last updated on 26/May/24

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Answered by Berbere last updated on 26/May/24

$$D_b^a\left(f\right)=\frac{1}{\mathrm{\Gamma }(n-a)}{\int }_b^t{(t-x)}^{n-a-1}{f}^{\left(n\right)}\left(x\right)dx$$$$n=\lceil a\rceil \Rightarrow D_0^{\frac{1}{2}}\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\int }_0^t{(t-x)}^{-\frac{1}{2}}1dx$$$$=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{[-2\sqrt{t-x}]}_0^t==\frac{2\sqrt{t}}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}$$$$\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}2\frac{t^{\frac{1}{2}}}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{e}^{-t^2}={\int }_0^{\mathrm{\infty }}{t}^{\frac{1}{2}}{e}^{-t^2}dt;t^2=x\Rightarrow t=\frac{1}{2}{x}^{-\frac{1}{2}}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{x}^{-\frac{1}{4}}{e}^{-x}dx=\frac{1}{2}\mathrm{\Gamma }\left(\frac{3}{4}\right)$$$$$$

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