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Question Number 134070 by mohammad17 last updated on 27/Feb/21

$${\int }_0^x{(x-t)}^{-\frac{1}{2}}{\left(t\right)}^{\frac{5}{2}}dt$$

Commented by mohammad17 last updated on 27/Feb/21

$$helpmesir$$

Answered by Dwaipayan Shikari last updated on 27/Feb/21

$${\int }_0^x{(x-t)}^{-\frac{1}{2}}{t}^{\frac{5}{2}}dtt=xu$$$$=x{\int }_0^1\frac{1}{\sqrt{x}}{(1-u)}^{-\frac{1}{2}}{x}^{\frac{5}{2}}{u}^{\frac{5}{2}}du$$$$=x^3{\int }_0^1{(1-u)}^{-\frac{1}{2}}{u}^{\frac{5}{2}}du=x^3\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{7}{2}\right)}{\mathrm{\Gamma }\left(3\right)}=\frac{5\pi }{16}{x}^3$$

Answered by mathmax by abdo last updated on 27/Feb/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{x}}{(\mathrm{x}-\mathrm{t})}^{-\frac{1}{2}}{\mathrm{t}}^{\frac{5}{2}}\mathrm{dt}\Rightarrow \mathrm{\Phi }{=}_{\mathrm{t}=\mathrm{xu}}{\int }_0^1{(\mathrm{x}-\mathrm{xu})}^{-\frac{1}{2}}{\left(\mathrm{xu}\right)}^{\frac{5}{2}}\mathrm{xdu}$$$$={\mathrm{x}}^{\frac{1}{2}}.{\mathrm{x}}^{\frac{5}{2}}{\int }_0^1{\mathrm{u}}^{\frac{5}{2}+1-1}{(1-\mathrm{u})}^{1-\frac{1}{2}-1}\mathrm{du}={\mathrm{x}}^3{\int }_0^1{\mathrm{u}}^{\frac{7}{2}-1}{(1-\mathrm{u})}^{\frac{1}{2}}\mathrm{du}$$$$={\mathrm{x}}^3\mathrm{B}(\frac{7}{2},\frac{1}{2})={\mathrm{x}}^3\frac{\mathrm{\Gamma }\left(\frac{7}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{7}{2}+\frac{1}{2})}={\mathrm{x}}^3\sqrt{\pi }\frac{\mathrm{\Gamma }\left(\frac{7}{2}\right)}{\mathrm{\Gamma }\left(4\right)}=\frac{\sqrt{\pi }}{3!}{\mathrm{x}}^3\mathrm{\Gamma }\left(\frac{7}{2}\right)\mathrm{but}$$$$\mathrm{\Gamma }\left(\frac{7}{2}\right)=\mathrm{\Gamma }(\frac{5}{2}+1)=\frac{5}{2}\mathrm{\Gamma }\left(\frac{5}{2}\right)=\frac{5}{2}\mathrm{\Gamma }(\frac{3}{2}+1)=\frac{5}{2}.\frac{3}{2}\mathrm{\Gamma }\left(\frac{3}{2}\right)$$$$=\frac{15}{4}\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{15}{4}\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{15}{8}\sqrt{\pi }\Rightarrow$$$$\mathrm{\Phi }=\frac{\sqrt{\pi }}{6}{\mathrm{x}}^3\frac{15}{8}\sqrt{\pi }=\frac{5\pi }{16}{\mathrm{x}}^3$$

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