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Question Number 134866 by mathmax by abdo last updated on 07/Mar/21

let U_n =∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^2 ))dx calculate lim_(n→+∞) e^n^2 U_n

$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{calculate}\:\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{e}^{\mathrm{n}^{\mathrm{2}} } \mathrm{U}_{\mathrm{n}} \\ $$

Answered by Ar Brandon last updated on 08/Mar/21

U_n =∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^2 ))dx=∫_0 ^∞ (1/((x^2 +n^2 )^2 ))Σ_(k=0) ^∞ (−1)^k (((nx)^(2k) )/((2k)!)) =Σ_(k=0) ^∞ (−1)^k (n^(2k) /((2k)!))∫_0 ^∞ (x^(2k) /((x^2 +n^2 )^2 ))dx =Σ_(k=0) ^∞ (−1)^k (n^(2k−4) /((2k)!))∫_0 ^∞ (x^(2k) /(((x^2 /n^2 )+1)^2 ))dx , u=(x^2 /n^2 ) du=((2x)/n^2 )dx =Σ_(k=0) ^∞ (−1)^k (n^(2k−4) /((2k)!))∙(n^(2k+1) /2)∫_0 ^∞ (u^(k−(1/2)) /((u+1)^2 ))du =Σ_(k=0) ^∞ (−1)^k (n^(4k−3) /((2k)!2))β(k+(1/2), (3/2)−k) =Σ_(k=0) ^∞ (−1)^k (n^(4k−3) /((2k)!2))∙((Γ(k+(1/2))Γ((3/2)−k))/(Γ(2))) Γ(k+(1/2))Γ((3/2)−k)=((√π)/2^(2k−1) )∙((Γ(2k))/(Γ(k)))∙((√π)/2^(1−2k) )∙((Γ(2−2k))/(Γ(1−k))) U_n =Σ_(k=0) ^∞ (−1)^k (n^(4k−3) /((2k)!2))∙((π(2k−1)!)/((k−1)!))∙(((1−2k)!)/((−k)!))

$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\left(\mathrm{nx}\right)^{\mathrm{2k}} }{\left(\mathrm{2k}\right)!} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{n}^{\mathrm{2k}} }{\left(\mathrm{2k}\right)!}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2k}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{n}^{\mathrm{2k}−\mathrm{4}} }{\left(\mathrm{2k}\right)!}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2k}} }{\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:,\:\mathrm{u}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\:\mathrm{du}=\frac{\mathrm{2x}}{\mathrm{n}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{n}^{\mathrm{2k}−\mathrm{4}} }{\left(\mathrm{2k}\right)!}\centerdot\frac{\mathrm{n}^{\mathrm{2k}+\mathrm{1}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\mathrm{k}−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{u}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{du} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{n}^{\mathrm{4k}−\mathrm{3}} }{\left(\mathrm{2k}\right)!\mathrm{2}}\beta\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{k}\right) \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{n}^{\mathrm{4k}−\mathrm{3}} }{\left(\mathrm{2k}\right)!\mathrm{2}}\centerdot\frac{\Gamma\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{k}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$\Gamma\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{k}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2k}−\mathrm{1}} }\centerdot\frac{\Gamma\left(\mathrm{2k}\right)}{\Gamma\left(\mathrm{k}\right)}\centerdot\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{1}−\mathrm{2k}} }\centerdot\frac{\Gamma\left(\mathrm{2}−\mathrm{2k}\right)}{\Gamma\left(\mathrm{1}−\mathrm{k}\right)} \\ $$$$\mathrm{U}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{n}^{\mathrm{4k}−\mathrm{3}} }{\left(\mathrm{2k}\right)!\mathrm{2}}\centerdot\frac{\pi\left(\mathrm{2k}−\mathrm{1}\right)!}{\left(\mathrm{k}−\mathrm{1}\right)!}\centerdot\frac{\left(\mathrm{1}−\mathrm{2k}\right)!}{\left(−\mathrm{k}\right)!} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 08/Mar/21

U_n =∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^2 ))dx changement x=nt give U_n =∫_0 ^∞ ((cos(n^2 t))/(n^4 (t^2 +1)^2 ))ndt =(1/n^3 )∫_0 ^∞ ((cos(n^2 t))/((t^2 +1)^2 ))dt ⇒ 2n^3 U_n =∫_(−∞) ^(+∞) ((cos(n^2 t))/((t^(2 ) +1)^2 ))dt =Re(∫_(−∞) ^(+∞) (e^(in^2 t) /((t^2 +1)^2 ))dt) let ϕ(z)=(e^(in^2 z) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(in^2 z) /((z−i)^2 (z+i)^2 )) residus give ∫_R 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^(in^2 z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((in^2 e^(in^2 z) (z+i)^2 −2(z+i)e^(in^2 z) )/((z+i)^4 )) =lim_(z→i) ((in^2 e^(in^2 z) (z+i)−2e^(in^2 z) )/((z+i)^3 )) =((in^2 e^(−n^2 ) (2i)−2e^(−n^2 ) )/((2i)^3 )) =(((−2n^2 −2)e^(−n^2 ) )/(−8i)) =(((n^2 +1)e^(−n^2 ) )/(4i)) ⇒∫_R ϕ(z)dz=2iπ.(((n^2 +1)e^(−n^2 ) )/(4i)) =(π/2)(n^2 +1)e^(−n^2 ) ⇒2n^3 U_n =(π/2)(n^2 +1)e^(−n^2 ) ⇒ U_n =(π/(4n^3 ))(n^2 +1)e^(−n^2 ) ⇒e^(n^2 ) U_n =((π(n^2 +1))/(4n^3 )) ⇒ lim_(n→+∞) e^n^2 U_n =lim_(n→+∞) (π/(4n))=0

$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{nt}\:\mathrm{give} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}^{\mathrm{2}} \mathrm{t}\right)}{\mathrm{n}^{\mathrm{4}} \left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{ndt}\:=\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}^{\mathrm{2}} \mathrm{t}\right)}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{2n}^{\mathrm{3}} \mathrm{U}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}^{\mathrm{2}} \mathrm{t}\right)}{\left(\mathrm{t}^{\mathrm{2}\:} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}\:=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{t}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} }{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\:\mathrm{residus}\:\mathrm{give} \\ $$$$\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:\mathrm{and}\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\mathrm{in}^{\mathrm{2}} \mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)\mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\:\frac{\mathrm{in}^{\mathrm{2}} \mathrm{e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} \left(\mathrm{z}+\mathrm{i}\right)−\mathrm{2e}^{\mathrm{in}^{\mathrm{2}} \mathrm{z}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\:=\frac{\mathrm{in}^{\mathrm{2}} \mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } \left(\mathrm{2i}\right)−\mathrm{2e}^{−\mathrm{n}^{\mathrm{2}} } }{\left(\mathrm{2i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(−\mathrm{2n}^{\mathrm{2}} −\mathrm{2}\right)\mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } }{−\mathrm{8i}}\:=\frac{\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } }{\mathrm{4i}}\:\Rightarrow\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi.\frac{\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } }{\mathrm{4i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } \:\Rightarrow\mathrm{2n}^{\mathrm{3}} \:\mathrm{U}_{\mathrm{n}} =\frac{\pi}{\mathrm{2}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } \:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\pi}{\mathrm{4n}^{\mathrm{3}} }\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{e}^{−\mathrm{n}^{\mathrm{2}} } \:\:\Rightarrow\mathrm{e}^{\mathrm{n}^{\mathrm{2}} \:} \mathrm{U}_{\mathrm{n}} =\frac{\pi\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4n}^{\mathrm{3}} }\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{e}^{\mathrm{n}^{\mathrm{2}} } \mathrm{U}_{\mathrm{n}} \:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\pi}{\mathrm{4n}}=\mathrm{0} \\ $$

Commented by Ar Brandon last updated on 08/Mar/21

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