Question Number 39712 by math khazana by abdo last updated on 10/Jul/18
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({x}^{{n}} \right)\:+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)^{{n}} }\:{dx} \\ $$
Commented by math khazana by abdo last updated on 10/Jul/18
$${n}\:{integr}\:{natural}. \\ $$
Commented by abdo mathsup 649 cc last updated on 10/Jul/18
$${let}\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({x}^{{n}} \right)\:+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)^{{n}} }{dx}\:{changement} \\ $$$${x}=\mathrm{3}{t}\:{give} \\ $$$${A}_{{n}} =\:\mathrm{3}\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{3}^{{n}} {t}^{{n}} \right)\:+{sin}\left(\mathrm{3}^{{n}} {t}^{{n}} \right)}{\mathrm{9}^{{n}} \left({t}^{\mathrm{2}\:} \:+\mathrm{1}\right)^{{n}} }\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\:\:\left\{\:\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\mathrm{3}^{{n}} {t}^{{n}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\right)\:+{Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\:\mathrm{3}^{{n}} {t}^{{n}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\right)\right\}{dt} \\ $$$${let}\:{calculate}\:{I}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\:\mathrm{3}^{{n}} \:{t}^{{n}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }{dt}\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } }{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$${with}\:{multiplicity}\:{n} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$
Commented by abdo mathsup 649 cc last updated on 10/Jul/18
$${Res}\left(\varphi,{i}\right)\:={lim}_{{n}\rightarrow+\infty} \frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{{n}} \varphi\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \:{leibniz} \\ $$$${formula}\:{give}\: \\ $$$$\left\{\:\left({z}+{i}\right)^{−{n}} \:\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:{C}_{{n}−\mathrm{1}} ^{{k}} \:\left(\left({z}+{i}\right)^{−{n}} \right)^{\left({k}\right)} \:\left(\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \right)^{\left({n}−\mathrm{1}−{k}\right)} \\ $$$${we}\:{have} \\ $$$$\left.\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} \:=−{n}\left({z}+{i}\right)^{−{n}−\mathrm{1}} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\:\left({z}+{i}\right)^{−{n}−\mathrm{2}} \:{by}\:{recurence} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({k}\right)} =\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{k}} \\ $$$${also}\:{we}\:{have} \\ $$$$\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \right\}^{\left(\mathrm{1}\right)} \:={ni}\:\mathrm{3}^{{n}} \:{z}^{{n}−\mathrm{1}} \:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \:\:={p}_{\mathrm{1}} \left({z}\right){e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \\ $$$$\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \right\}^{\left(\mathrm{2}\right)} ={p}_{\mathrm{2}} \left({z}\right)\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } … \\ $$$$\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \right\}^{\left({n}−\mathrm{1}−{k}\right)} ={p}_{{n}−\mathrm{1}−{k}} \left({z}\right)\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \:{and}\:{we}\:{can} \\ $$$${find}\:{a}\:{recurrence}\:{relation}\:{between}\:{the}\:{p}_{{n}} \\ $$$${Res}\left(\varphi,{i}\right)=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}−\mathrm{1}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} \:{p}_{{n}−\mathrm{1}−{k}} \left({i}\right)\:{e}^{\mathrm{3}^{{n}} {i}^{{n}+\mathrm{1}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\left({n}−\mathrm{1}\right)!}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right){p}_{{n}−\mathrm{1}−{k}} \left({i}\right)\:{e}^{\mathrm{3}^{{n}} {i}^{{n}+\mathrm{1}} } \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
$${y}={cos}\left({x}^{{n}} \right)+{sin}\left({x}^{{n}} \right) \\ $$$$=\sqrt{\mathrm{2}}\:\left\{{sin}\left(\frac{\Pi}{\mathrm{4}}+{x}^{{n}} \right)\right\} \\ $$$$\sqrt{\mathrm{2}}\:\:\geqslant{y}\geqslant−\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\int_{−−\infty} ^{+\infty} \frac{{cos}\left({x}^{{n}} \right)+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\geqslant−\sqrt{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} } \\ $$$${let}\:{I}_{{k}} =\sqrt{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\:\:{x}=\mathrm{3}{tana} \\ $$$$=\sqrt{\mathrm{2}}\:\int_{−\frac{\Pi}{\mathrm{2}}\:\:} ^{\frac{\Pi}{\mathrm{2}}} \frac{\mathrm{3}{sec}^{\mathrm{2}} {ada}}{\left\{\mathrm{9}\left({sec}^{\mathrm{2}} {a}\right)\right\}^{{n}} } \\ $$$$=\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} {ada} \\ $$$$=\frac{\sqrt{\mathrm{2}}\:×\mathrm{2}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} {ada} \\ $$$${now}\:{using}\:{gamma}\:−{beta}\:{function} \\ $$$$\frac{\lceil{p}.\lceil{q}}{\lceil\left({p}+{q}\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \theta{cos}^{\mathrm{2}{q}−\mathrm{1}} \theta{d}\theta \\ $$$$ \\ $$$$\mathrm{2}{p}−\mathrm{1}=\mathrm{0}\:\:\:\:{p}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\: \\ $$$$\mathrm{2}{q}−\mathrm{1}=\mathrm{2}{n}−\mathrm{2}\:\: \\ $$$${q}=\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{2}}\:×\mathrm{2}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {a}\:\:{cos}^{\mathrm{2}×\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {a}\:{da} \\ $$$$=\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil{n}} \\ $$$${so}\: \\ $$$$\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil{n}}\geqslant\int_{−\infty} ^{+\infty} \frac{{cos}\left({x}^{{n}} \right)+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\geqslant \\ $$$$\:\:\frac{−\sqrt{\mathrm{2}}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil{n}} \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
$${pls}\:{check} \\ $$