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Question Number 33027 by prof Abdo imad last updated on 09/Apr/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{x^3}{1+x^5}dx.$$

Answered by Joel578 last updated on 14/Apr/18

$$I={\int }_0^{\mathrm{\infty }}\frac{x^3}{1+x^5}dx$$$$u=x^5\to du=5x^{4}dx$$$$$$$$I=\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{x^{-1}}{1+u}du=\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{u^{-\frac{1}{5}}}{1+u}du$$$$=\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{4}{5}-1}}{1+u}du({\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{\sin \left(\pi a\right)})$$$$=\frac{\pi }{5\sin \left(\frac{4\pi }{5}\right)}$$$$=\frac{4\pi }{5\sqrt{10-2\sqrt{5}}}\approx 1.07$$

Commented by Joel578 last updated on 10/Apr/18

$$\sin \left(\frac{4\pi }{5}\right)=\sin \left(\frac{\pi }{5}\right)=\frac{1}{4}\sqrt{10-2\sqrt{5}}$$

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