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