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Question Number 72016 by mathmax by abdo last updated on 23/Oct/19
find U_n =∫_0 ^∞ ((cos(nx))/((3+nx^2 )^2 ))dx with n integr
$${find}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({nx}\right)}{\left(\mathrm{3}+{nx}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:{with}\:{n}\:{integr} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/19
we have U_n =∫_0 ^∞ ((cos(nx))/((nx^2 +3)^2 ))dx =_((√n)x=(√3)u) ∫_0 ^∞ ((cos(n(((√3)u)/( (√n)))))/((3u^2 +3)^2 )) (((√3)du)/( (√n))) =((√3)/(9(√n))) ∫_0 ^∞ ((cos((√(3n))u))/((u^2 +1)^2 ))du =((√3)/(18(√n))) ∫_(−∞) ^(+∞) ((cos((√(3n))u))/((u^2 +1)^2 ))du =((√3)/(18(√n))) Re( ∫_(−∞) ^(+∞) (e^(i(√(3n))u) /((u^2 +1)^2 )))du let W(z)=(e^(i(√(3n))z) /((z^2 +1)^2 )) ⇒ W(z)=(e^(i(√(3n))z) /((z−i)^2 (z+i)^2 )) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπRes(W,i) Res(W,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) { (e^(i(√(3n))z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((i(√(3n))e^(i(√(3n))z) (z+i)^2 −2(z+i)e^(i(√(3n))z) )/((z+i)^4 )) =lim_(z→i) (((i(√(3n))(z+i)−2)e^(i(√(3n))z) )/((z+i)^3 )) =(((−2 (√(3n))−2)e^(−(√(3n))) )/((2i)^3 )) =(((−2(√(3n))−2)e^(−(√(3n))) )/(−8i)) =(((1+(√(3n)))e^(−(√(3n))) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(((1+(√(3n)))e^(−(√(3n))) )/(4i)) =(π/2)(1+(√(3n)))e^(−(√(3n))) ⇒U_n =((√3)/(18(√n)))×(π/2)(1+(√(3n)))e^(−(√(3n))) =((π(√3))/(36(√n)))(1+(√(3n)))e^(−(√(3n)))
$${we}\:{have}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{\left({nx}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:=_{\sqrt{{n}}{x}=\sqrt{\mathrm{3}}{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({n}\frac{\sqrt{\mathrm{3}}{u}}{\:\sqrt{{n}}}\right)}{\left(\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}{du}}{\:\sqrt{{n}}} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}\sqrt{{n}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\sqrt{\mathrm{3}{n}}{u}\right)}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{du}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}\sqrt{{n}}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\sqrt{\mathrm{3}{n}}{u}\right)}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}\sqrt{{n}}}\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\sqrt{\mathrm{3}{n}}{u}} }{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\right){du}\:{let}\:{W}\left({z}\right)=\frac{{e}^{{i}\sqrt{\mathrm{3}{n}}{z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${W}\left({z}\right)=\frac{{e}^{{i}\sqrt{\mathrm{3}{n}}{z}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left\{\:\frac{{e}^{{i}\sqrt{\mathrm{3}{n}}{z}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:\:\frac{{i}\sqrt{\mathrm{3}{n}}{e}^{{i}\sqrt{\mathrm{3}{n}}{z}} \left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{{i}\sqrt{\mathrm{3}{n}}{z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\:\:\frac{\left({i}\sqrt{\mathrm{3}{n}}\left({z}+{i}\right)−\mathrm{2}\right){e}^{{i}\sqrt{\mathrm{3}{n}}{z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\:=\frac{\left(−\mathrm{2}\:\sqrt{\mathrm{3}{n}}−\mathrm{2}\right){e}^{−\sqrt{\mathrm{3}{n}}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(−\mathrm{2}\sqrt{\mathrm{3}{n}}−\mathrm{2}\right){e}^{−\sqrt{\mathrm{3}{n}}} }{−\mathrm{8}{i}}\:=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}{n}}\right){e}^{−\sqrt{\mathrm{3}{n}}} }{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}{n}}\right){e}^{−\sqrt{\mathrm{3}{n}}} }{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}{n}}\right){e}^{−\sqrt{\mathrm{3}{n}}} \\ $$$$\Rightarrow{U}_{{n}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{18}\sqrt{{n}}}×\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}{n}}\right){e}^{−\sqrt{\mathrm{3}{n}}} \\ $$$$=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{36}\sqrt{{n}}}\left(\mathrm{1}+\sqrt{\mathrm{3}{n}}\right){e}^{−\sqrt{\mathrm{3}{n}}} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/19
n≥1
$${n}\geqslant\mathrm{1} \\ $$
Answered by mind is power last updated on 23/Oct/19
u=nx U_n =n∫_0 ^(+∞) ((cos(u))/((3n+u^2 )^2 ))du withe residu bee continued
$$\mathrm{u}=\mathrm{nx} \\ $$$$\mathrm{U}_{\mathrm{n}} =\mathrm{n}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{cos}\left(\mathrm{u}\right)}{\left(\mathrm{3n}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{du} \\ $$$$\mathrm{withe}\:\mathrm{residu}\:\mathrm{bee}\:\mathrm{continued} \\ $$$$ \\ $$

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