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Question Number 25682 by abdo imad last updated on 13/Dec/17

$$wegive{\int }_0^{\mathrm{\infty }}{t}^{a-1}{(1+t)}^{-1}dt=\pi {\left(sin\left(\pi a\right)\right)}^{-1}with0<a<1findthevalueof{\int }_0^{\mathrm{\infty }}{(1+x^{16})}^{-1}dx$$

Answered by ajfour last updated on 13/Dec/17

$$If{x}^{16}=\mathit{t}andlet\int \frac{dx}{1+x^{16}}=I$$ $$ThenI=\int \frac{dt}{16x^{15}(1+t)}$$ $$=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{t^{-15/16}dt}{1+t}$$ $$=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{16}-1}dt}{1+t}=\frac{1}{16}\times \frac{\pi }{\sin \left(\frac{\pi }{16}\right)}$$ $$\sin \frac{\pi }{16}=\sqrt{\frac{1-\cos \left(\pi /8\right)}{2}}$$ $$=\sqrt{\frac{1-\sqrt{\frac{1+1/\sqrt{2}}{2}}}{2}}.$$

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