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Question Number 201266 by Mathspace last updated on 02/Dec/23

$$calculate{\int }_1^{\mathrm{\infty }}\frac{dx}{\sqrt{1+x^3}}$$

Answered by witcher3 last updated on 03/Dec/23

$$\frac{1}{{\mathrm{x}}^3}=\mathrm{y}$$$$\iff \frac{1}{3}{\int }_0^1\frac{{\mathrm{y}}^{-\frac{5}{6}}}{\sqrt{1+\mathrm{y}}}\mathrm{dy}$$$${(1+\mathrm{y})}^{-\frac{1}{2}}=1+\sum _{\mathrm{n}⩾1}(-\frac{1}{2})....(-\frac{1}{2}-\mathrm{n}+1).\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}$$$$\frac{{\mathrm{y}}^{-\frac{1}{6}}}{\sqrt{1+\mathrm{y}}}=\sum _{\mathrm{n}⩾0}\frac{\mathrm{\Gamma }(\frac{1}{2}+\mathrm{n}){(-\mathrm{y})}^{\mathrm{n}}{\mathrm{y}}^{-\frac{5}{6}}}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{n}!}$$$$\iff \frac{1}{3}{\int }_0^1\sum _{\mathrm{n}⩾0}\frac{\mathrm{\Gamma }(\frac{1}{2}+\mathrm{n})}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{n}!}{(-1)}^{\mathrm{n}}.{\mathrm{y}}^{-\frac{5}{6}+\mathrm{n}}\mathrm{dy}=\frac{1}{3}\sum _{\mathrm{n}⩾0}{(-1)}^{\mathrm{n}}\frac{\mathrm{\Gamma }(\frac{1}{2}+\mathrm{n})}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{n}!}.\frac{1}{\mathrm{n}+\frac{1}{6}}$$$$=2\sum _{\mathrm{n}⩾0}\frac{\frac{\mathrm{\Gamma }(\mathrm{n}+\frac{1}{2})}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}.\frac{\mathrm{\Gamma }(\mathrm{n}+\frac{1}{6})}{\mathrm{\Gamma }\left(\frac{1}{6}\right)}}{\frac{\mathrm{\Gamma }(\frac{7}{6}+\mathrm{n})}{\mathrm{\Gamma }\left(\frac{7}{6}\right)}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}$$$$=2{}_2{\mathrm{F}}_1(\frac{1}{2},\frac{1}{6},\frac{7}{6},-1)\approx 1,89$$

Answered by Calculusboy last updated on 03/Dec/23

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:{\int }_1^{\mathrm{\infty }}{(1+{\mathit{x}}^3)}^{-\frac{1}{2}}\mathit{d}\mathit{x}$$$$\mathit{l}\mathit{e}\mathit{t}{\mathit{x}}^{-3}+1={\mathit{t}}^2\mathit{x}={({\mathit{t}}^2-1)}^{-\frac{1}{3}}\mathit{d}\mathit{x}=-\frac{1}{3}{({\mathit{t}}^2-1)}^{-\frac{4}{3}}\cdot 2\mathit{t}\mathit{d}\mathit{t}=-\frac{2}{3}{({\mathit{t}}^2-1)}^{-\frac{4}{3}}\mathit{t}\mathit{d}\mathit{t}$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=\mathrm{\infty }\mathit{t}=1\mathit{a}\mathit{n}\mathit{d}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=1\mathit{t}=\mathrm{\infty }$$$$\mathit{I}={\int }_{\mathrm{\infty }}^1\frac{-\frac{2}{3}{({\mathit{t}}^2+1)}^{-\frac{4}{3}}\mathit{t}\mathit{d}\mathit{t}}{{[1+{({\mathit{t}}^2-1)}^{-1}]}^{\frac{1}{2}}}\iff \mathit{I}=\frac{2}{3}{\int }_1^{\mathrm{\infty }}\frac{{({\mathit{t}}^2-1)}^{-\frac{4}{3}}\mathit{t}\mathit{d}\mathit{t}}{\frac{{\left({\mathit{t}}^2\right)}^{\frac{1}{2}}}{{({\mathit{t}}^2-1)}^{\frac{1}{2}}}}$$$$\mathit{I}=\frac{2}{3}{\int }_1^{\mathrm{\infty }}\frac{\mathit{d}\mathit{t}}{{({\mathit{t}}^2-1)}^{\frac{5}{6}}}=1.8947$$

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