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Question Number 28262 by abdo imad last updated on 22/Jan/18

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

Commented by abdo imad last updated on 24/Jan/18

$$letput{x}^5=t\iff x=t^{\frac{1}{5}}$$$$I={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^5}={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{5}{t}^{\frac{1}{5}-1}}{1+t}dt=\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{5}-1}}{1+t}dtbutwe$$$$knowthat{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt{}^{}=\frac{\pi }{sin\left(\pi a\right)}with0<a<1$$$$soI=\frac{\frac{\pi }{5}}{sin\left(\frac{\pi }{5}\right)}butcos\left(\frac{\pi }{5}\right)=\frac{1+\sqrt{5}}{4}and$$$$sin\left(\frac{\pi }{5}\right)=\sqrt{1-\frac{{(1+\sqrt{5})}^2}{16}}=\frac{\sqrt{16-(6+2\sqrt{5})}}{4}$$$$=\frac{\sqrt{10-2\sqrt{5}}}{4}sothevalueofIisknown.$$

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