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Question Number 64392 by mathmax by abdo last updated on 17/Jul/19
1)calculate A_n =∫_0 ^∞ ((sin(x^(2n) ))/((x^2 +1)^2 ))dx with n integr natural 2) study the convergene of Σ A_n
$$\left.\mathrm{1}\right){calculate}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({x}^{\mathrm{2}{n}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergene}\:{of}\:\Sigma\:{A}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 18/Jul/19
1) we have 2A_n =∫_(−∞) ^(+∞) ((sin(x^(2n) ))/((x^2 +1)^2 ))dx =Im(∫_(−∞) ^(+∞) (e^(ix^(2n) ) /((x^2 +1)^2 ))) let W(z) =(e^(iz^(2n) ) /((z^2 +1)^2 )) ⇒W(z)=(e^(iz^(2n) ) /((z−i)^2 (z+i)^2 )) so the poles of W are +^− i (doubles) 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^(iz^(2n) ) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2niz^(2n−1) e^(iz^(2n) ) (z+i)^2 −2(z+i)e^(iz^(2n) ) )/((z+i)^4 )) =lim_(z→i) ((2niz^(2n−1) e^(iz^(2n) ) (z+i)−2e^(iz^(2n) ) )/((z+i)^3 )) lim_(z→i) (({2niz^(2n−1) (z+i)−2}e^(iz^(2n) ) )/((z+i)^3 )) =(({(2ni)(2i)(−1)^n i^(−1) −2})/((2i)^3 ))e^(i(−1)^n ) =(({4ni(−1)^n −2)e^(i(−1)^n ) )/(−8i)) =(((1−2ni(−1)^n )e^(i(−1)^n ) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =((2iπ)/(4i)){1−2ni(−1)^n )e^(i(−1)^n ) =(π/2)(1−2ni(−1)^n )e^(i(−1)^n ) ) =(π/2)(1−2n(−1)^n i)(cos(−1)^n +isin(−1)^n ) =(π/2)(cos(−1)^n +isin(−1)^n −2n(−1)^n cos(−1)^n i +2n(−1)^n sin(−1)^n } =(π/2){ cos(−1)^n +2n(−1)^n sin(−1)^n +i(sin(−1)^n −2ncos(−1)^n )} 2A =Im(∫_(−∞) ^(+∞) W(z)dz) =(π/2){sin(−1)^n −2n cos(−1)^n } ⇒ A =(π/4){ sin(−1)^n −2ncos(−1)^n } .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\frac{{sin}\left({x}^{\mathrm{2}{n}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}^{\mathrm{2}{n}} } }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$${let}\:{W}\left({z}\right)\:=\frac{{e}^{{iz}^{\mathrm{2}{n}} } }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{W}\left({z}\right)=\frac{{e}^{{iz}^{\mathrm{2}{n}} } }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:{W}\:{are} \\ $$$$\overset{−} {+}{i}\:\:\left({doubles}\right)\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\frac{{e}^{{iz}^{\mathrm{2}{n}} } }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{2}{niz}^{\mathrm{2}{n}−\mathrm{1}} \:{e}^{{iz}^{\mathrm{2}{n}} } \left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{{iz}^{\mathrm{2}{n}} } }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{2}{niz}^{\mathrm{2}{n}−\mathrm{1}} {e}^{{iz}^{\mathrm{2}{n}} } \left({z}+{i}\right)−\mathrm{2}{e}^{{iz}^{\mathrm{2}{n}} } }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$${lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left\{\mathrm{2}{niz}^{\mathrm{2}{n}−\mathrm{1}} \left({z}+{i}\right)−\mathrm{2}\right\}{e}^{{iz}^{\mathrm{2}{n}} } }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left\{\left(\mathrm{2}{ni}\right)\left(\mathrm{2}{i}\right)\left(−\mathrm{1}\right)^{{n}} {i}^{−\mathrm{1}} −\mathrm{2}\right\}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }{e}^{{i}\left(−\mathrm{1}\right)^{{n}} } \\ $$$$=\frac{\left\{\mathrm{4}{ni}\left(−\mathrm{1}\right)^{{n}} −\mathrm{2}\right){e}^{{i}\left(−\mathrm{1}\right)^{{n}} } }{−\mathrm{8}{i}}\:=\frac{\left(\mathrm{1}−\mathrm{2}{ni}\left(−\mathrm{1}\right)^{{n}} \right){e}^{{i}\left(−\mathrm{1}\right)^{{n}} } }{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\left.\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{4}{i}}\left\{\mathrm{1}−\mathrm{2}{ni}\left(−\mathrm{1}\right)^{{n}} \right){e}^{{i}\left(−\mathrm{1}\right)^{{n}} } \:=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}{ni}\left(−\mathrm{1}\right)^{{n}} \right){e}^{{i}\left(−\mathrm{1}\right)^{{n}} } \right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} {i}\right)\left({cos}\left(−\mathrm{1}\right)^{{n}} \:+{isin}\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left({cos}\left(−\mathrm{1}\right)^{{n}} \:+{isin}\left(−\mathrm{1}\right)^{{n}} −\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} {cos}\left(−\mathrm{1}\right)^{{n}} {i}\:\:+\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} {sin}\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:{cos}\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} {sin}\left(−\mathrm{1}\right)^{{n}} \:+{i}\left({sin}\left(−\mathrm{1}\right)^{{n}} −\mathrm{2}{ncos}\left(−\mathrm{1}\right)^{{n}} \right)\right\} \\ $$$$\mathrm{2}{A}\:={Im}\left(\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\right)\:=\frac{\pi}{\mathrm{2}}\left\{{sin}\left(−\mathrm{1}\right)^{{n}} −\mathrm{2}{n}\:{cos}\left(−\mathrm{1}\right)^{{n}} \right\}\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{4}}\left\{\:{sin}\left(−\mathrm{1}\right)^{{n}} −\mathrm{2}{ncos}\left(−\mathrm{1}\right)^{{n}} \right\}\:. \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 18/Jul/19
2) we have A_n =(π/4)sin(−1)^n −((nπ)/2) cos(−1)^n ⇒ ΣA_n =(π/4)Σ sin(−1)^n −(π/2)Σn cos(−1)^n but lim_(n→+∞) sin(−1)^n ≠0 also lim_(n→+∞) ncos(−1)^n ≠0 ⇒ Σ A_n diverges.
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{A}_{{n}} =\frac{\pi}{\mathrm{4}}{sin}\left(−\mathrm{1}\right)^{{n}} \:−\frac{{n}\pi}{\mathrm{2}}\:{cos}\left(−\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$$\Sigma{A}_{{n}} =\frac{\pi}{\mathrm{4}}\Sigma\:{sin}\left(−\mathrm{1}\right)^{{n}} \:\:−\frac{\pi}{\mathrm{2}}\Sigma{n}\:{cos}\left(−\mathrm{1}\right)^{{n}} \:\:{but} \\ $$$${lim}_{{n}\rightarrow+\infty} {sin}\left(−\mathrm{1}\right)^{{n}} \:\neq\mathrm{0}\:\:{also}\:{lim}_{{n}\rightarrow+\infty} {ncos}\left(−\mathrm{1}\right)^{{n}} \:\neq\mathrm{0}\:\:\Rightarrow \\ $$$$\Sigma\:{A}_{{n}} \:\:{diverges}. \\ $$

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