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Question Number 35983 by abdo mathsup 649 cc last updated on 26/May/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{2\sqrt{t}+1}{t^5+3}dt.$$

Commented by prof Abdo imad last updated on 27/May/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{2\sqrt{t}+1}{t^5+3}dtchangement\sqrt{t}=x$$$$giveI={\int }_0^{\mathrm{\infty }}\frac{2x+1}{x^{10}+3}2xdx=4{\int }_0^{\mathrm{\infty }}\frac{x^2}{x^{10}+3}dx$$$$+2{\int }_0^{\mathrm{\infty }}\frac{x}{x^{10}+3}dxbutchangementx=3^{\frac{1}{10}}ugive$$$${\int }_0^{\mathrm{\infty }}\frac{x^2}{x^{10}+3}dx={\int }_0^{\mathrm{\infty }}\frac{3^{\frac{1}{5}}{u}^2}{3u^{10}+3}{3}^{\frac{1}{10}}du$$$$=\frac{3^{\frac{3}{10}}}{3}{\int }_0^{\mathrm{\infty }}\frac{u^2}{1+u^{10}}duthenweusethechang.$$$$u^{10}=t\Rightarrow u=t^{\frac{1}{10}}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{u^2}{1+u^{10}}du$$$$={\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{5}}}{1+t}\frac{1}{10}{t}^{\frac{1}{10}-1}dt=\frac{1}{10}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{3}{10}-1}}{1+t}dt$$$$=\frac{1}{10}\frac{\pi }{sin\left(\frac{3\pi }{10}\right)}bythesamechang.weget$$$${\int }_0^{\mathrm{\infty }}\frac{x}{x^{10}+3}dx={\int }_0^{\mathrm{\infty }}\frac{3^{\frac{1}{10}}u}{3u^{10}+3}{3}^{\frac{1}{10}}du$$$$=\frac{3^{\frac{1}{5}}}{3}{\int }_0^{\mathrm{\infty }}\frac{udu}{1+u^{10}}=3^{\frac{-4}{5}}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{10}}}{1+t}\frac{1}{10}{t}^{\frac{1}{10}-1}dt$$$$=\frac{3^{-\frac{4}{5}}}{10}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{2}{5}-1}}{1+t}dt=\frac{3^{-\frac{4}{5}}}{10}\frac{\pi }{sin\left(\frac{2\pi }{5}\right)}\Rightarrow$$$$I=\frac{4}{10}\frac{\pi }{sin\left(\frac{3\pi }{10}\right)}.\frac{3^{\frac{3}{10}}}{3}+2\frac{3^{-\frac{4}{5}}}{10}\frac{\pi }{sin\left(\frac{2\pi }{5}\right)}$$$$I=\frac{2\pi }{5}{3}^{-\frac{7}{10}}\frac{1}{sin\left(\frac{3\pi }{10}\right)}+\frac{\pi }{5}{3}^{-\frac{4}{5}}.\frac{1}{sin\left(\frac{2\pi }{5}\right)}$$$$Ihaveusedtheformula$$$${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}with0<a<1.$$

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