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Question Number 33128 by prof Abdo imad last updated on 10/Apr/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2)(1+x^4)}.$$

Commented by prof Abdo imad last updated on 13/Apr/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2)(1+x^4)}wehave$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(1+x^2)(1+x^4)}.letintroducethe$$$$complexfunction\phi \left(z\right)=\frac{1}{(1+z^2)(1+z^4)}$$$$\phi \left(z\right)=\frac{1}{(z-i)(z+i)(z^2-i)(z^2-(-i))}$$$$=\frac{1}{(z-i)(z+i)(z-\sqrt{i})(z+\sqrt{i})(z-\sqrt{-i})(z+\sqrt{-i})}$$$$\sqrt{i}=e^{i\frac{\pi }{4}}\sqrt{-i}=e^{-i\frac{\pi }{4}}sothepolesof?\phi are$$$$i,-i,e^{i\frac{\pi }{4}},-e^{i\frac{\pi }{4}},e^{-i\frac{\pi }{4}},-e^{-i\frac{\pi }{4}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,i)+Res(\phi ,e^{i\frac{\pi }{4}})+Res(\phi ,-e^{-i\frac{\pi }{4}})$$$$letp\left(z\right)=(1+z^2)(1+z^4)$$$$p\left(z\right)=1+z^4+z^2+z^6\Rightarrow p^{{}^{\prime }}\left(z\right)=2z+4z^3+6z^5$$$$Res(\phi ,i)=\frac{1}{p^{{}^{\prime }}\left(i\right)}=\frac{1}{2i+4i^3+6i^5}=\frac{1}{2i-4i+6i}$$$$=\frac{1}{4i}$$$$Res(\phi ,e^{i\frac{\pi }{4}})=\frac{1}{p^{{}^{\prime }}\left(e^{i\frac{\pi }{4}}\right)}=\frac{1}{2e^{i\frac{\pi }{4}}+4e^{i\frac{3\pi }{4}}+6e^{i\frac{5\pi }{4}}}$$$$but2e^{i\frac{\pi }{4}}+4e^{i\frac{3\pi }{4}}+6e^{i\frac{5\pi }{4}}$$$$=2e^{i\frac{\pi }{4}}-4e^{-i\frac{\pi }{4}}-6e^{i\frac{\pi }{4}}=-4(e^{i\frac{\pi }{4}}+e^{-i\frac{\pi }{4}})$$$$=-4.2.\frac{\sqrt{2}}{2}=-4\sqrt{2}$$$$Res(\phi ,-e^{-i\frac{\pi }{4}})=\frac{1}{p^{{}^{\prime }}(-e^{-i\frac{\pi }{4}})}$$$$=\frac{1}{-2e^{-i\frac{\pi }{4}}-4e^{-i\frac{3\pi }{4}}-6e^{-i\frac{5\pi }{4}}}$$$$=\frac{-1}{2e^{-i\frac{\pi }{4}}+4e^{-i\frac{3\pi }{4}}+6e^{-i\frac{5\pi }{4}}}=\frac{-1}{2e^{-i\frac{\pi }{4}}-4e^{i\frac{\pi }{4}}-6e^{-i\frac{\pi }{4}}}$$$$=\frac{-1}{-4(e^{i\frac{\pi }{4}}+e^{-i\frac{\pi }{4}})}=\frac{1}{4.2.\frac{\sqrt{2}}{2}}=\frac{1}{4\sqrt{2}}\Rightarrow$$$$2I=2i\pi (\frac{1}{4i}-\frac{1}{4\sqrt{2}}+\frac{1}{4\sqrt{2}})=\frac{\pi }{2}\Rightarrow$$$$I=\frac{\pi }{4}.$$$$$$

Commented by prof Abdo imad last updated on 13/Apr/18

$$anothermethod$$$$thech.x=\frac{1}{t}give$$$$I=-{\int }_0^{\mathrm{\infty }}\frac{1}{(1+\frac{1}{t^2})(1+\frac{1}{t^4})}\frac{-dt}{t^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{t^{4}dt}{(1+t^2)(1+t^4)}={\int }_0^{\mathrm{\infty }}\frac{1+t^4-1}{(1+t^2)(1+t^4)}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}-{\int }_0^{\mathrm{\infty }}\frac{dt}{(1+t^2)(1+t^4)}=\frac{\pi }{2}-I\Rightarrow$$$$2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$$$$★I=\frac{\pi }{4}★$$

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