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Question Number 53114 by maxmathsup by imad last updated on 17/Jan/19
let A_n =∫_0 ^∞ ((x sin(nx))/((x^2 +n^2 )^2 ))dx with n integr natural not 0 1) find the value of A_n 2) study the convergence of Σ A_n
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}\:{sin}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:{with}\:{n}\:{integr}\:{natural}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{A}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 18/Jan/19
1) changement x=nt give A_n =∫_0 ^∞ ((nt sin(n^2 t))/(n^4 (t^2 +1)^2 )) ndt =(1/n^2 )∫_0 ^∞ ((tsin(n^2 t))/((t^2 +1)^2 ))dt ⇒ 2n^2 A_n =∫_(−∞) ^(+∞) ((tsin(n^2 t))/((t^2 +1)^2 ))dt =Im( ∫_(−∞) ^(+∞) ((t e^(in^2 t) )/((t^2 +1)^2 ))dt) let consider the complex function ϕ(z) =((z e^(in^2 z) )/((z^2 +1)^2 )) ⇒ϕ(z)=((z e^(in^2 z) )/((z−i)^2 (z+i)^2 )) so the poles of ϕ are i and −i (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(in^2 z) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(in^2 z) +in^2 z e^(in^2 z) )(z+i)^2 −2(z+i)z e^(in^2 z) )/((z+i)^4 )) =lim_(z→i) ((((1+in^2 z)(z+i)−2z)e^(in^2 z) )/((z+i)^3 )) =((((1−n^2 )(2i)−2i)e^(−n^2 ) )/((2i)^3 )) =((−2in^2 e^(−n^2 ) )/(−8i)) =(n^2 /4) e^(−n^2 ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (n^2 /4) e^(−n^2 ) =((iπn^2 )/2) e^(−n^2 ) 2n^2 A_n =((πn^2 )/2) e^(−n^2 ) ⇒ A_n =(π/4) e^(−n^2 ) . we have Σ_(n=0) ^∞ A_n =(π/4) Σ_(n=0) ^∞ e^(−n^2 ) but n^2 ≥n ⇒−n^2 ≤−n ⇒e^(−n^2 ) ≤e^(−n) ⇒ Σ_(n=0) ^∞ A_n ≤(π/4) Σ_(n=0) ^∞ ((1/e))^n =(π/4)(1/(1−(1/e))) =(π/4) (e/(e−1)) ⇒ 0< Σ_(n=0) ^∞ A_n ≤((πe)/(4(e−1))) .
$$\left.\mathrm{1}\right)\:{changement}\:{x}={nt}\:{give}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{nt}\:{sin}\left({n}^{\mathrm{2}} {t}\right)}{{n}^{\mathrm{4}} \left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{ndt}\:=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tsin}\left({n}^{\mathrm{2}} {t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\mathrm{2}{n}^{\mathrm{2}} \:{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\:\frac{{tsin}\left({n}^{\mathrm{2}} {t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}\:{e}^{{in}^{\mathrm{2}} {t}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right)\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\frac{{z}\:{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{z}\:{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\left({doubles}\right)\:{residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \left\{\:\frac{{z}\:{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \:\:\frac{\left({e}^{{in}^{\mathrm{2}} {z}} \:+{in}^{\mathrm{2}} {z}\:{e}^{{in}^{\mathrm{2}} {z}} \right)\left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){z}\:{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left(\left(\mathrm{1}+{in}^{\mathrm{2}} {z}\right)\left({z}+{i}\right)−\mathrm{2}{z}\right){e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\:=\frac{\left(\left(\mathrm{1}−{n}^{\mathrm{2}} \right)\left(\mathrm{2}{i}\right)−\mathrm{2}{i}\right){e}^{−{n}^{\mathrm{2}} } }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{2}{in}^{\mathrm{2}} \:{e}^{−{n}^{\mathrm{2}} } }{−\mathrm{8}{i}}\:=\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\:{e}^{−{n}^{\mathrm{2}} \:} \:\Rightarrow\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\:{e}^{−{n}^{\mathrm{2}} } =\frac{{i}\pi{n}^{\mathrm{2}} }{\mathrm{2}}\:{e}^{−{n}^{\mathrm{2}} } \\ $$$$\mathrm{2}{n}^{\mathrm{2}} \:{A}_{{n}} =\frac{\pi{n}^{\mathrm{2}} }{\mathrm{2}}\:{e}^{−{n}^{\mathrm{2}} } \:\Rightarrow\:{A}_{{n}} =\frac{\pi}{\mathrm{4}}\:{e}^{−{n}^{\mathrm{2}} } \:. \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} =\frac{\pi}{\mathrm{4}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}^{\mathrm{2}} } \:\:\:{but}\:\:{n}^{\mathrm{2}} \geqslant{n}\:\Rightarrow−{n}^{\mathrm{2}} \leqslant−{n}\:\Rightarrow{e}^{−{n}^{\mathrm{2}} } \leqslant{e}^{−{n}} \:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \leqslant\frac{\pi}{\mathrm{4}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}}{{e}}\right)^{{n}} \:=\frac{\pi}{\mathrm{4}}\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{e}}}\:=\frac{\pi}{\mathrm{4}}\:\frac{{e}}{{e}−\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{0}<\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:\leqslant\frac{\pi{e}}{\mathrm{4}\left({e}−\mathrm{1}\right)}\:. \\ $$

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