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Question Number 140768 by EDWIN88 last updated on 12/May/21
∫_(−∞) ^∞ ((x^2 +4)/(x^4 +16)) dx =?
$$\:\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\:\mathrm{dx}\:=? \\ $$
Answered by Ar Brandon last updated on 12/May/21
I=∫_(−∞) ^∞ ((x^2 +4)/(x^4 +16))dx=2∫_0 ^∞ ((x^2 +4)/(x^4 +16))dx =2∫_0 ^∞ ((1+(4/x^2 ))/(x^2 +((16)/x^2 )))dx=2∫_0 ^∞ ((1+(4/x^2 ))/((x−(4/x))^2 +8))dx =2∫_(−∞) ^∞ (du/(u^2 +8))=(2/(2(√2)))[tan^(−1) ((u/(2(√2))))]_(−∞) ^∞ =(1/( (√2)))((π/2)+(π/2))=(π/( (√2)))
$$\mathcal{I}=\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}−\frac{\mathrm{4}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{8}}\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\int_{−\infty} ^{\infty} \frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{8}}=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{u}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right]_{−\infty} ^{\infty} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$
Answered by liberty last updated on 12/May/21
consider I=2∫_0 ^∞ ((x^2 +a^2 )/(x^4 +a^4 )) dx ; a = 2 I=2∫_0 ^∞ ((x^2 (1+(a^2 /x^2 )))/(a^2 x^2 ((x^2 /a^2 )+(a^2 /x^2 )))) dx I=(2/a^2 ) ∫_0 ^∞ (((1+(a^2 /x^2 )))/(((x^2 /a^2 )+(a^2 /x^2 )))) dx I= (2/a^2 ) ∫_0 ^∞ (((1+(1/y^2 )))/((y^2 +(1/y^2 )))) (a dy ) ; x=ay I= (2/a) ∫_0 ^( ∞) (((1+(1/y^2 )))/((y−(1/y))^2 +((√2))^2 )) dy I= (2/a) .(1/( (√2))) [arctan (((y−(1/y))/( (√2)))) ]_0 ^∞ I= (2/(a(√2))).π , put a = 2 I= (π/( (√2))) = ((π(√2))/2)
$$\mathrm{consider}\:\mathrm{I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} +\mathrm{a}^{\mathrm{4}} }\:\mathrm{dx}\:;\:\mathrm{a}\:=\:\mathrm{2} \\ $$$$\mathrm{I}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}\:\mathrm{dx} \\ $$$$\mathrm{I}=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)}\:\mathrm{dx}\: \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)}{\left(\mathrm{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)}\:\left(\mathrm{a}\:\mathrm{dy}\:\right)\:;\:\mathrm{x}=\mathrm{ay} \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}}\:\overset{\:\infty} {\int}_{\mathrm{0}} \:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\right)}{\left(\mathrm{y}−\frac{\mathrm{1}}{\mathrm{y}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:\mathrm{dy} \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\mathrm{arctan}\:\left(\frac{\mathrm{y}−\frac{\mathrm{1}}{\mathrm{y}}}{\:\sqrt{\mathrm{2}}}\right)\:\right]_{\mathrm{0}} ^{\infty} \\ $$$$\mathrm{I}=\:\frac{\mathrm{2}}{\mathrm{a}\sqrt{\mathrm{2}}}.\pi\:,\:\mathrm{put}\:\mathrm{a}\:=\:\mathrm{2} \\ $$$$\mathrm{I}=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Answered by Mathspace last updated on 12/May/21
let Φ=∫_(−∞) ^(+∞) ((x^2 +4)/(x^(4 ) +16))dx and Ψ(z)=((z^2 +4)/(z^4 +16)) poles of Ψ? Ψ)(z)=((z^2 +4)/((z^2 −4i)(z^2 +4i))) = ((z^2 +4)/((z−2e^((iπ)/4) )(z+2e^((iπ)/4) )(z−2e^(−((iπ)/4)) )(z+2e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) Ψ(z)dz=2iπ{ Res(Ψ,2e^((iπ)/4) )+Res(Ψ,−2e^(−((iπ)/4)) )} we have Res(Ψ,2e^((iπ)/4) ) =((4i+4)/(4e^((iπ)/4) (8i)))=((1+i)/(8i))e^(−((iπ)/4)) Res(Ψ,−e^(−((iπ)/4)) )=((−4i+4)/(−4e^(−((iπ)/4)) (−8i))) =((1−i)/(8i))e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) Ψ(z)dz =2iπ{((1+i)/(8i))e^(−((iπ)/4)) +((1−i)/(8i))e^((iπ)/4) } =(π/4){(1+i)e^(−((iπ)/4)) +(1−i)e^((iπ)/4) } =(π/4){2Re((1+i)e^(−((iπ)/4)) )} =(π/2) Re(1+i)e^(−((iπ)/4)) =(π/2)Re((√2)) =((π(√2))/2) ⇒Φ=((π(√2))/2)
$${let}\:\Phi=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} +\mathrm{4}}{{x}^{\mathrm{4}\:} +\mathrm{16}}{dx}\:{and} \\ $$$$\Psi\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{4}}{{z}^{\mathrm{4}} \:+\mathrm{16}}\:\:{poles}\:{of}\:\Psi? \\ $$$$\left.\Psi\right)\left({z}\right)=\frac{{z}^{\mathrm{2}} +\mathrm{4}}{\left({z}^{\mathrm{2}} −\mathrm{4}{i}\right)\left({z}^{\mathrm{2}} +\mathrm{4}{i}\right)} \\ $$$$= \\ $$$$\frac{{z}^{\mathrm{2}} +\mathrm{4}}{\left({z}−\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\Psi\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{\:{Res}\left(\Psi,\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\Psi,−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${we}\:{have} \\ $$$${Res}\left(\Psi,\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{4}{i}+\mathrm{4}}{\mathrm{4}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{8}{i}\right)}=\frac{\mathrm{1}+{i}}{\mathrm{8}{i}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\Psi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)=\frac{−\mathrm{4}{i}+\mathrm{4}}{−\mathrm{4}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(−\mathrm{8}{i}\right)} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{8}{i}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\Psi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}+{i}}{\mathrm{8}{i}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} +\frac{\mathrm{1}−{i}}{\mathrm{8}{i}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\left\{\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\left(\mathrm{1}−{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\left\{\mathrm{2}{Re}\left(\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{Re}\left(\mathrm{1}+{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\frac{\pi}{\mathrm{2}}{Re}\left(\sqrt{\mathrm{2}}\right) \\ $$$$=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\Phi=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$ \\ $$

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