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
$${D}_{{b}} ^{{a}} \left({f}\right)=\frac{\mathrm{1}}{\Gamma\left({n}−{a}\right)}\int_{{b}} ^{{t}} \left({t}−{x}\right)^{{n}−{a}−\mathrm{1}} {f}^{\left({n}\right)} \left({x}\right){dx} \\ $$$${n}=\lceil{a}\rceil\Rightarrow{D}_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left({t}\right)=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{{t}} \left({t}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{1}{dx} \\ $$$$=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\left[−\mathrm{2}\sqrt{{t}−{x}}\right]_{\mathrm{0}} ^{{t}} ==\frac{\mathrm{2}\sqrt{{t}}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \mathrm{2}\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{e}^{−{t}^{\mathrm{2}} } =\int_{\mathrm{0}} ^{\infty} {t}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{t}^{\mathrm{2}} } {dt};{t}^{\mathrm{2}} ={x}\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−{x}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$ \\ $$