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Question Number 122791 by mohammad17 last updated on 19/Nov/20

Answered by mathmax by abdo last updated on 19/Nov/20

I =∫_0 ^∞ ((cos(mx))/((x^2 +1)^2 ))dx ⇒ 2I =∫_(−∞) ^(+∞) ((cos(mx))/((x^2 +1)^2 ))dx =Res(∫_(−∞) ^(+∞) (e^(imx) /((x^2 +1)^2 ))dx) let ϕ(z)=(e^(imz) /((z^2 +1)^2 )) we have lim_(z→∞) ∣zϕ(z)∣=0 and ϕ(z) =(e^(imz) /((z−i)^2 (z+i)^2 )) [rssidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) and Res(ϕ,i)=lim_(z→i) (1/((2−1)!)) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(e^(imz) /((z+i)^2 ))}^((1)) =lim_(z→i) ((ime^(imz) (z+i)^2 −2(z+i)e^(imz) )/((z+i)^4 )) =lim_(z→i) ((ime^(imz) (z+i)−2e^(imz) )/((z+i)^3 )) =lim_(z→i) (((im(z+i)−2)e^(imz) )/((z+i)^3 )) =((2i(im)−2)e^(−m) )/((2i)^3 )) =(((−2m−2)e^(−m) )/(−8i)) =(((m+1)e^(−m) )/(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((m+1)e^(−m) )/(4i)) =(π/2)(m+1)e^(−m) ⇒ 2I =(π/2)(m+1)e^(−m) ⇒★I =(π/4)(m+1)e^(−m) ★

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{mx}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\:\mathrm{2I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{mx}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:=\mathrm{Res}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{imx}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{imz}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{lim}_{\mathrm{z}\rightarrow\infty} \mid\mathrm{z}\varphi\left(\mathrm{z}\right)\mid=\mathrm{0}\:\mathrm{and} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{imz}} }{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\:\left[\mathrm{rssidus}\:\mathrm{theorem}\:\mathrm{give}\right. \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:\:\mathrm{and}\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\:\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\mathrm{e}^{\mathrm{imz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\:\frac{\mathrm{ime}^{\mathrm{imz}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)\mathrm{e}^{\mathrm{imz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\:\frac{\mathrm{ime}^{\mathrm{imz}} \left(\mathrm{z}+\mathrm{i}\right)−\mathrm{2e}^{\mathrm{imz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\left(\mathrm{im}\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{2}\right)\mathrm{e}^{\mathrm{imz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left.\mathrm{2i}\left(\mathrm{im}\right)−\mathrm{2}\right)\mathrm{e}^{−\mathrm{m}} }{\left(\mathrm{2i}\right)^{\mathrm{3}} }\:=\frac{\left(−\mathrm{2m}−\mathrm{2}\right)\mathrm{e}^{−\mathrm{m}} }{−\mathrm{8i}}\:=\frac{\left(\mathrm{m}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{m}} }{\mathrm{4i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi×\frac{\left(\mathrm{m}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{m}} }{\mathrm{4i}}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{m}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{m}} \:\Rightarrow \\ $$$$\mathrm{2I}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{m}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{m}} \:\Rightarrow\bigstar\mathrm{I}\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{m}+\mathrm{1}\right)\mathrm{e}^{−\mathrm{m}} \bigstar \\ $$

Answered by TANMAY PANACEA last updated on 19/Nov/20

(1/3)∫_0 ^∞ (((4x^2 +4)−(x^2 +4))/((x^2 +1)(x^2 +4)))dx (4/3)∫_0 ^∞ (dx/(x^2 +4))−(1/3)∫_0 ^∞ (dx/(x^2 +1)) ∣(4/3)×(1/2)tan^(−1) ((x/2))−(1/3)tan^(−1) x∣_ ^∞ (2/3)×(π/2)−(1/3)×(π/2)=(π/6)

$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}\right)−\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mid\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{3}}{tan}^{−\mathrm{1}} {x}\mid_{} ^{\infty} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}×\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{6}} \\ $$

Commented by mohammad17 last updated on 19/Nov/20

thank you mis

$${thank}\:{you}\:{mis} \\ $$

Answered by mathmax by abdo last updated on 19/Nov/20

A =∫_0 ^∞ ((sin^2 x)/x^2 ) dx by parts A =[−((sin^2 x)/x)]_0 ^∞ +∫_0 ^∞ ((2sinx cosx)/x)dx =∫_0 ^∞ ((sin(2x))/x)dx =_(2x=t) ∫_0 ^∞ ((sin(t))/(t/2))×(dt/2) =∫_0 ^∞ ((sint)/t)dt =(π/2)

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{A}\:=\left[−\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2sinx}\:\mathrm{cosx}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{x}}\mathrm{dx}\:=_{\mathrm{2x}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{t}\right)}{\frac{\mathrm{t}}{\mathrm{2}}}×\frac{\mathrm{dt}}{\mathrm{2}}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 19/Nov/20

4) I =∫_0 ^∞ (x^2 /((x^2 +1)(x^2 +4)))dx ⇒I =(1/2)∫_(−∞) ^(+∞) (x^2 /((x^2 +1)(x^2 +4)))dx let ϕ(z)=(z^2 /((z^2 +1)(z^2 +4))) ⇒ϕ(z) =(z^2 /((z−i)(z+i)(z−2i)(z+2i))) residus theorem →∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i)+Res(ϕ,2i)} Res(ϕ,i)=lim_(z→i) (z−i)ϕ(z) =((−1)/(2i(3))) =−(1/(6i)) Res(ϕ,2i) =lim_(z→2i) (z−2i)ϕ(z) =((−4)/(4i(−3))) =(1/(3i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{−(1/(6i))+(1/(3i))} =(π/3) +((2π)/3) =π ⇒ I =(π/2) error at the question ...!

$$\left.\mathrm{4}\right)\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}\right)}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}\right)}\mathrm{dx} \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)}\:\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}−\mathrm{2i}\right)\left(\mathrm{z}+\mathrm{2i}\right)} \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)+\mathrm{Res}\left(\varphi,\mathrm{2i}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \left(\mathrm{z}−\mathrm{i}\right)\varphi\left(\mathrm{z}\right)\:=\frac{−\mathrm{1}}{\mathrm{2i}\left(\mathrm{3}\right)}\:=−\frac{\mathrm{1}}{\mathrm{6i}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{2i}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2i}} \:\:\left(\mathrm{z}−\mathrm{2i}\right)\varphi\left(\mathrm{z}\right)\:=\frac{−\mathrm{4}}{\mathrm{4i}\left(−\mathrm{3}\right)}\:=\frac{\mathrm{1}}{\mathrm{3i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{−\frac{\mathrm{1}}{\mathrm{6i}}+\frac{\mathrm{1}}{\mathrm{3i}}\right\}\:=\frac{\pi}{\mathrm{3}}\:+\frac{\mathrm{2}\pi}{\mathrm{3}}\:=\pi\:\Rightarrow\:\mathrm{I}\:=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{error}\:\mathrm{at}\:\mathrm{the}\:\mathrm{question}\:...! \\ $$

Answered by mathmax by abdo last updated on 19/Nov/20

A =∫_0 ^∞ ((xsin(ax))/(x^2 +4))dx ⇒ A =(1/2)∫_(−∞) ^(+∞) ((xsin(ax))/(x^2 +4))dx let =(1/2) Im(∫_(−∞) ^(+∞) (x/(x^2 +4))e^(iax) dx) let w(z)=((ze^(iaz) )/(z^2 +4)) lim_(z→∞) ∣zw(z)∣=0 and w(z)=((ze^(iaz) )/((z−2i)(z+2i))) ∫_(−∞) ^(+∞) W(z)dz =2iπRes(w,2i) =2iπ.((2i e^(ia(2i)) )/(4i)) =iπ e^(−2a) ⇒ A =(π/2)e^(−2a) error at the question ′.!

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dx}\:\mathrm{let} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{e}^{\mathrm{iax}} \:\mathrm{dx}\right)\:\:\mathrm{let}\:\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{iaz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$\mathrm{lim}_{\mathrm{z}\rightarrow\infty} \mid\mathrm{zw}\left(\mathrm{z}\right)\mid=\mathrm{0}\:\:\mathrm{and}\:\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{iaz}} }{\left(\mathrm{z}−\mathrm{2i}\right)\left(\mathrm{z}+\mathrm{2i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{W}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\mathrm{Res}\left(\mathrm{w},\mathrm{2i}\right)\:=\mathrm{2i}\pi.\frac{\mathrm{2i}\:\mathrm{e}^{\mathrm{ia}\left(\mathrm{2i}\right)} }{\mathrm{4i}}\:=\mathrm{i}\pi\:\mathrm{e}^{−\mathrm{2a}} \:\Rightarrow \\ $$$$\mathrm{A}\:=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{2a}} \:\:\:\:\:\:\mathrm{error}\:\mathrm{at}\:\mathrm{the}\:\mathrm{question}\:'.! \\ $$

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