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Question Number 62220 by maxmathsup by imad last updated on 17/Jun/19

$$letf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{t^2}{x^6+t^6}dtwithx>0$$$$1)calculatef\left(x\right)$$$$2)calculateg\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{t^2}{{(x^6+t^6)}^2}dt$$$$3)findvaluesofintegrals{\int }_0^{\mathrm{\infty }}\frac{t^2}{t^6+8}dtand{\int }_0^{\mathrm{\infty }}\frac{t^2}{{(t^6+8)}^2}dt.$$

Commented by maxmathsup by imad last updated on 19/Jun/19

$$1)f\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{t^2}{x^6+t^6}dtchangementt=xugive$$$$f\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{x^2{u}^2}{x^6+x^6{u}^6}xdu=\frac{1}{x^3}{\int }_0^{\mathrm{\infty }}\frac{u^2}{1+u^6}duchangement{u}^6=\alpha giveu={\alpha }^{\frac{1}{6}}$$$$x^{3}f\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{{\left({\alpha }^{\frac{1}{6}}\right)}^2}{1+\alpha }\frac{1}{6}{\alpha }^{\frac{1}{6}-1}du={\int }_0^{\mathrm{\infty }}\frac{{\alpha }^{\frac{1}{3}+\frac{1}{6}-1}}{1+\alpha }du=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\alpha }^{\frac{1}{2}-1}}{1+\alpha }d\alpha$$$$=\frac{1}{6}\frac{\pi }{sin\left(\frac{\pi }{2}\right)}=\frac{\pi }{6}\Rightarrow f\left(x\right)=\frac{\pi }{6x^3}(x>0)$$$$2)wehave{f}^{{}^{\prime }}\left(x\right)=-{\int }_0^{\mathrm{\infty }}\frac{6x^5{t}^2}{{(x^6+t^6)}^2}dt=-6x^5{\int }_0^{\mathrm{\infty }}\frac{t^2}{{(x^6+t^6)}^2}dt=-6x^{5}g\left(x\right)\Rightarrow$$$$g\left(x\right)=-\frac{f^{{}^{\prime }}\left(x\right)}{6x^5}butf\left(x\right)=\frac{\pi }{6}{x}^{-3}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{\pi }{6}(-3)x^{-4}=-\frac{\pi }{2x^4}\Rightarrow$$$$g\left(x\right)=-\frac{1}{6x^5}(-\frac{\pi }{2x^4})=\frac{\pi }{12x^9}$$$$3){\int }_0^{\mathrm{\infty }}\frac{t^2}{t^6+8}dt={\int }_0^{\mathrm{\infty }}\frac{t^2}{{\left(\sqrt{2}\right)}^6+t^6}f\left(\sqrt{2}\right)=\frac{\pi }{6{\left(\sqrt{2}\right)}^3}=\frac{\pi }{12\sqrt{2}}=\frac{\pi \sqrt{2}}{24}$$$${\int }_0^{\mathrm{\infty }}\frac{t^2}{{(t^6+8)}^2}dt=g\left(\sqrt{2}\right)=\frac{\pi }{12{\left(\sqrt{2}\right)}^9}=\frac{\pi }{12.\sqrt{2}\left(2^4\right)}=\frac{\pi }{12\times 16\times \sqrt{2}}=...$$

Commented by maxmathsup by imad last updated on 19/Jun/19

$$ihaveusedtheresult{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}if0<a<1$$

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