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Question Number 140768 by EDWIN88 last updated on 12/May/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+4}{{\mathrm{x}}^4+16}\mathrm{dx}=?$$

Answered by Ar Brandon last updated on 12/May/21

$$\mathcal{I}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+4}{{\mathrm{x}}^4+16}\mathrm{dx}=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+4}{{\mathrm{x}}^4+16}\mathrm{dx}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{1+\frac{4}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{16}{{\mathrm{x}}^2}}\mathrm{dx}=2{\int }_0^{\mathrm{\infty }}\frac{1+\frac{4}{{\mathrm{x}}^2}}{{(\mathrm{x}-\frac{4}{\mathrm{x}})}^2+8}\mathrm{dx}$$$$=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+8}=\frac{2}{2\sqrt{2}}{\left[{\tan }^{-1}\left(\frac{\mathrm{u}}{2\sqrt{2}}\right)\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}=\frac{1}{\sqrt{2}}(\frac{\pi }{2}+\frac{\pi }{2})=\frac{\pi }{\sqrt{2}}$$

Answered by liberty last updated on 12/May/21

$$\mathrm{consider}\mathrm{I}=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{x}}^2+{\mathrm{a}}^2}{{\mathrm{x}}^4+{\mathrm{a}}^4}\mathrm{dx};\mathrm{a}=2$$$$\mathrm{I}=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{x}}^2(1+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2})}{{\mathrm{a}}^2{\mathrm{x}}^2(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2})}\mathrm{dx}$$$$\mathrm{I}=\frac{2}{{\mathrm{a}}^2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{(1+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2})}{(\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2})}\mathrm{dx}$$$$\mathrm{I}=\frac{2}{{\mathrm{a}}^2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{(1+\frac{1}{{\mathrm{y}}^2})}{({\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2})}\left(\mathrm{a}\mathrm{dy}\right);\mathrm{x}=\mathrm{ay}$$$$\mathrm{I}=\frac{2}{\mathrm{a}}{\stackrel{\mathrm{\infty }}{\int }}_0\frac{(1+\frac{1}{{\mathrm{y}}^2})}{{(\mathrm{y}-\frac{1}{\mathrm{y}})}^2+{\left(\sqrt{2}\right)}^2}\mathrm{dy}$$$$\mathrm{I}=\frac{2}{\mathrm{a}}.\frac{1}{\sqrt{2}}{\left[\arctan \left(\frac{\mathrm{y}-\frac{1}{\mathrm{y}}}{\sqrt{2}}\right)\right]}_0^{\mathrm{\infty }}$$$$\mathrm{I}=\frac{2}{\mathrm{a}\sqrt{2}}.\pi ,\mathrm{put}\mathrm{a}=2$$$$\mathrm{I}=\frac{\pi }{\sqrt{2}}=\frac{\pi \sqrt{2}}{2}$$

Answered by Mathspace last updated on 12/May/21

$$let\mathrm{\Phi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+4}{x^4+16}dxand$$$$\mathrm{\Psi }\left(z\right)=\frac{z^2+4}{z^4+16}polesof\mathrm{\Psi }?$$$$\mathrm{\Psi })\left(z\right)=\frac{z^2+4}{(z^2-4i)(z^2+4i)}$$$$=$$$$\frac{z^2+4}{(z-2e^{\frac{i\pi }{4}})(z+2e^{\frac{i\pi }{4}})(z-2e^{-\frac{i\pi }{4}})(z+2e^{-\frac{i\pi }{4}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{\Psi }\left(z\right)dz=2i\pi \{Res(\mathrm{\Psi },2e^{\frac{i\pi }{4}})+Res(\mathrm{\Psi },-2e^{-\frac{i\pi }{4}})\}$$$$wehave$$$$Res(\mathrm{\Psi },2e^{\frac{i\pi }{4}})=\frac{4i+4}{4e^{\frac{i\pi }{4}}\left(8i\right)}=\frac{1+i}{8i}{e}^{-\frac{i\pi }{4}}$$$$Res(\mathrm{\Psi },-e^{-\frac{i\pi }{4}})=\frac{-4i+4}{-4e^{-\frac{i\pi }{4}}(-8i)}$$$$=\frac{1-i}{8i}{e}^{\frac{i\pi }{4}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{\Psi }\left(z\right)dz=2i\pi \{\frac{1+i}{8i}{e}^{-\frac{i\pi }{4}}+\frac{1-i}{8i}{e}^{\frac{i\pi }{4}}\}$$$$=\frac{\pi }{4}\{(1+i)e^{-\frac{i\pi }{4}}+(1-i)e^{\frac{i\pi }{4}}\}$$$$=\frac{\pi }{4}\left\{2Re\left((1+i)e^{-\frac{i\pi }{4}}\right)\right\}$$$$=\frac{\pi }{2}Re(1+i)e^{-\frac{i\pi }{4}}=\frac{\pi }{2}Re\left(\sqrt{2}\right)$$$$=\frac{\pi \sqrt{2}}{2}\Rightarrow \mathrm{\Phi }=\frac{\pi \sqrt{2}}{2}$$$$$$

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