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Question Number 56345 by maxmathsup by imad last updated on 14/Mar/19

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^n+a^n}withnintegr⩾2anda>0$$$$1)calculatef\left(a\right)intemsofa$$$$2)letg\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^n+a^n)}^2}calculateg\left(a\right)intermsofa$$$$3)findthevaluesofintegrals{\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+16}and{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^8+16)}^2}$$

Commented by maxmathsup by imad last updated on 17/Mar/19

$$1)changementx=atgivef\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{adt}{a^n{t}^n+a^n}=\frac{1}{a^{n-1}}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^n+1}$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{t^n+1}{=}_{t=u^{\frac{1}{n}}}{\int }_0^{\mathrm{\infty }}\frac{1}{n(1+u)}{u}^{\frac{1}{n}-1}du=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{n}-1}}{1+u}du$$$$=\frac{1}{n}\frac{\pi }{sin\left(\frac{\pi }{n}\right)}=\frac{\pi }{nsin\left(\frac{\pi }{n}\right)}(wehaveprovedthat{\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x}dx=\frac{\pi }{sin\left(\pi a\right)}with0<a<1)\Rightarrow$$$$★f\left(a\right)=\frac{\pi }{n{a}^{n-1}sin\left(\frac{\pi }{n}\right)}★withn⩾2$$$$2)wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial a}\left(\frac{1}{x^n+a^n}\right)dx=-{\int }_0^{\mathrm{\infty }}\frac{{na}^{n-1}}{{(x^n+a^n)}^2}dx$$$$=-{na}^{n-1}g\left(a\right)\Rightarrow g\left(a\right)=-\frac{f\left(a\right)}{{na}^{n-1}}=-\frac{1}{{na}^{n-1}}\frac{\pi }{n{a}^{n-1}sin\left(\frac{\pi }{n}\right)}\Rightarrow$$$$★g\left(a\right)=-\frac{\pi }{n^2{a}^{2n-2}sin\left(\frac{\pi }{n}\right)}★$$$$3)wehave{\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+16}={\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+{\left(\sqrt{2}\right)}^8}\Rightarrow n=8anda=\sqrt{2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{x^8+16}=\frac{\pi }{8{\left(\sqrt{2}\right)}^{7}sin\left(\frac{\pi }{8}\right)}alsowehave$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^8+16)}^2}={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2+{\left(\sqrt{2}\right)}^8)}^2}\Rightarrow n=8anda=\sqrt{2}atg\left(a\right)\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^8+16)}^2}=\frac{\pi }{64{\left(\sqrt{2}\right)}^{14}sin\left(\frac{\pi }{8}\right)}=\frac{\pi }{64.2^{7}sin\left(\frac{\pi }{8}\right)}with$$$$sin\left(\frac{\pi }{8}\right)=\frac{\sqrt{2-\sqrt{2}}}{2}.$$

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