Question Number 140768 by EDWIN88 last updated on 12/May/21
$$\:\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\:\mathrm{dx}\:=? \\ $$
Answered by Ar Brandon last updated on 12/May/21
$$\mathcal{I}=\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}−\frac{\mathrm{4}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{8}}\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\int_{−\infty} ^{\infty} \frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{8}}=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{u}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right]_{−\infty} ^{\infty} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$
Answered by liberty last updated on 12/May/21
$$\mathrm{consider}\:\mathrm{I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} +\mathrm{a}^{\mathrm{4}} }\:\mathrm{dx}\:;\:\mathrm{a}\:=\:\mathrm{2} \\ $$$$\mathrm{I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}\:\mathrm{dx} \\ $$$$\mathrm{I}=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}\:\mathrm{dx}\: \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)}{\left(\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)}\:\left(\mathrm{a}\:\mathrm{dy}\:\right)\:;\:\mathrm{x}=\mathrm{ay} \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}}\:\overset{\:\infty} {\int}_{\mathrm{0}} \:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)}{\left(\mathrm{y}−\frac{\mathrm{1}}{\mathrm{y}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:\mathrm{dy} \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\mathrm{arctan}\:\left(\frac{\mathrm{y}−\frac{\mathrm{1}}{\mathrm{y}}}{\:\sqrt{\mathrm{2}}}\right)\:\right]_{\mathrm{0}} ^{\infty} \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}\sqrt{\mathrm{2}}}.\pi\:,\:\mathrm{put}\:\mathrm{a}\:=\:\mathrm{2} \\ $$$$\mathrm{I}=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Answered by Mathspace last updated on 12/May/21
$${let}\:\Phi=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} +\mathrm{4}}{{x}^{\mathrm{4}\:} +\mathrm{16}}{dx}\:{and} \\ $$$$\Psi\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{4}}{{z}^{\mathrm{4}} \:+\mathrm{16}}\:\:{poles}\:{of}\:\Psi? \\ $$$$\left.\Psi\right)\left({z}\right)=\frac{{z}^{\mathrm{2}} +\mathrm{4}}{\left({z}^{\mathrm{2}} −\mathrm{4}{i}\right)\left({z}^{\mathrm{2}} +\mathrm{4}{i}\right)} \\ $$$$= \\ $$$$\frac{{z}^{\mathrm{2}} +\mathrm{4}}{\left({z}−\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\Psi\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{\:{Res}\left(\Psi,\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\Psi,−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${we}\:{have} \\ $$$${Res}\left(\Psi,\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{4}{i}+\mathrm{4}}{\mathrm{4}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{8}{i}\right)}=\frac{\mathrm{1}+{i}}{\mathrm{8}{i}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\Psi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)=\frac{−\mathrm{4}{i}+\mathrm{4}}{−\mathrm{4}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(−\mathrm{8}{i}\right)} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{8}{i}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\Psi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}+{i}}{\mathrm{8}{i}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} +\frac{\mathrm{1}−{i}}{\mathrm{8}{i}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\left\{\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\left(\mathrm{1}−{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\left\{\mathrm{2}{Re}\left(\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{Re}\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\frac{\pi}{\mathrm{2}}{Re}\left(\sqrt{\mathrm{2}}\right) \\ $$$$=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\Phi=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$ \\ $$