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 (f)=(1/(Γ(n−a)))∫_b ^t (t−x)^(n−a−1) f^((n)) (x)dx n=⌈a⌉⇒D_0 ^(1/2) (t)=(1/(Γ((1/2))))∫_0 ^t (t−x)^(−(1/2)) 1dx =(1/(Γ((1/2))))[−2(√(t−x))]_0 ^t ==((2(√t))/(Γ((1/2)))) ((√π)/2)∫_0 ^∞ 2(t^(1/2) /(Γ((1/2))))e^(−t^2 ) =∫_0 ^∞ t^(1/2) e^(−t^2 ) dt;t^2 =x⇒t=(1/2)x^(−(1/2)) dx =(1/2)∫_0 ^∞ x^(−(1/4)) e^(−x) dx=(1/2)Γ((3/4))](https://www.tinkutara.com/question/Q207795.png)
$${D}_{{b}} ^{{a}} \left({f}\right)=\frac{\mathrm{1}}{\Gamma\left({n}−{a}\right)}\int_{{b}} ^{{t}} \left({t}−{x}\right)^{{n}−{a}−\mathrm{1}} {f}^{\left({n}\right)} \left({x}\right){dx} \\ $$$${n}=\lceil{a}\rceil\Rightarrow{D}_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left({t}\right)=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{{t}} \left({t}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{1}{dx} \\ $$$$=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\left[−\mathrm{2}\sqrt{{t}−{x}}\right]_{\mathrm{0}} ^{{t}} ==\frac{\mathrm{2}\sqrt{{t}}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \mathrm{2}\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{e}^{−{t}^{\mathrm{2}} } =\int_{\mathrm{0}} ^{\infty} {t}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{t}^{\mathrm{2}} } {dt};{t}^{\mathrm{2}} ={x}\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−{x}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$ \\ $$