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Question Number 62856 by mathmax by abdo last updated on 26/Jun/19

$$letf\left(\lambda \right)={\int }_0^{+\mathrm{\infty }}\frac{x^4}{x^6+{\lambda }^6}dxwith\lambda >0$$$$1)calculatef\left(\lambda \right)$$$$2)calculatealsog\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{x^4}{{(x^6+{\lambda }^6)}^2}dx$$$$3)findthevaluesof{\int }_0^{\mathrm{\infty }}\frac{x^4}{x^6+1}dx,{\int }_0^{\mathrm{\infty }}\frac{x^4}{x^6+8}dxand{\int }_0^{\mathrm{\infty }}\frac{x^4}{{(x^6+8)}^2}dx.$$

Commented by mathmax by abdo last updated on 26/Jun/19

$$1)weusetheresult{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}if0<a<1\left(resultproved\right)$$$$changementx=\lambda tgivef\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{{\left(\lambda t\right)}^4}{{\left(\lambda t\right)}^6+{\lambda }^6}\lambda dt=\frac{{\lambda }^5}{{\lambda }^6}{\int }_0^{\mathrm{\infty }}\frac{t^4}{1+t^6}dt$$$$=\frac{1}{\lambda }{\int }_0^{\mathrm{\infty }}\frac{t^{4}dt}{1+t^6}{=}_{t=u^{\frac{1}{6}}}\frac{1}{\lambda }{\int }_0^{\mathrm{\infty }}\frac{{\left(u^{\frac{1}{6}}\right)}^4}{1+u}\frac{1}{6}{u}^{\frac{1}{6}-1}du$$$$=\frac{1}{6\lambda }{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{2}{3}+\frac{1}{6}-1}}{1+u}du=\frac{1}{6\lambda }{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{5}{6}-1}}{1+u}du=\frac{1}{6\lambda }\frac{\pi }{sin\left(\frac{5\pi }{6}\right)}=\frac{\pi }{6\lambda sin(\pi -\frac{\pi }{6})}$$$$=\frac{\pi }{6\lambda sin\left(\frac{\pi }{6}\right)}=\frac{\pi }{6\lambda \frac{1}{2}}=\frac{\pi }{3\lambda }\Rightarrow ★f\left(\lambda \right)=\frac{\pi }{3\lambda }★$$$$2)letderivatef\left(\lambda \right)wehave{f}^{{}^{\prime }}\left(\lambda \right)=-{\int }_0^{\mathrm{\infty }}\frac{6{\lambda }^5{x}^4}{{(x^6+{\lambda }^6)}^2}dx$$$$=-6{\lambda }^5{\int }_0^{\mathrm{\infty }}\frac{x^4}{{(x^6+{\lambda }^6)}^2}dx=-6{\lambda }^{5}g\left(\lambda \right)\Rightarrow g\left(\lambda \right)=-\frac{1}{6{\lambda }^5}{f}^{{}^{\prime }}\left(\lambda \right)$$$$wehave{f}^{{}^{\prime }}\left(\lambda \right)=-\frac{\pi }{3}\frac{1}{{\lambda }^2}\Rightarrow g\left(\lambda \right)=-\frac{1}{6{\lambda }^5}(-\frac{\pi }{3{\lambda }^2})=\frac{\pi }{18{\lambda }^7}\Rightarrow$$$$★g\left(\lambda \right)=\frac{\pi }{18{\lambda }^7}★$$$$$$$$$$

Commented by mathmax by abdo last updated on 26/Jun/19

$$3){\int }_0^{\mathrm{\infty }}\frac{x^4}{1+x^6}dx=f\left(1\right)=\frac{\pi }{3}letcalculate$$$${\int }_0^{\mathrm{\infty }}\frac{x^4}{x^6+8}dxhere{\lambda }^6=8\Rightarrow \lambda =8^{\frac{1}{6}}={\left(2^3\right)}^{\frac{1}{6}}=2^{\frac{1}{2}}=2\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{x^4}{x^6+8}dx=f\left(\sqrt{2}\right)=\frac{\pi }{3\sqrt{2}}$$$${\int }_0^{\mathrm{\infty }}\frac{x^4}{{(x^6+8)}^2}dx=g\left(\sqrt{2}\right)=\frac{\pi }{18{\left(\sqrt{2}\right)}^7}=\frac{\pi }{18.2^{\frac{7}{2}}}=\frac{\pi }{18.3.\sqrt{2}}=\frac{\pi }{54\sqrt{2}}.$$$$$$

Commented by mathmax by abdo last updated on 26/Jun/19

$$erroroftypo{2}^{\frac{1}{2}}=\sqrt{2}$$

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