Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 100368 by bemath last updated on 26/Jun/20

lim_(n→∞) ∫_(−∞) ^∞ cos (x^n ) dx =?  where n=2k, k∈N, k≠0

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{−\infty} ^{\infty} \mathrm{cos}\:\left({x}^{{n}} \right)\:{dx}\:=? \\ $$$${where}\:{n}=\mathrm{2}{k},\:{k}\in\mathbb{N},\:{k}\neq\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 26/Jun/20

A_n =∫_(−∞) ^(+∞)  cos(x^(2n) ) dx ⇒A_n =2∫_0 ^∞  cos(x^(2n) )dx =2 Re(∫_0 ^∞ e^(−ix^(2n) ) dx) but  changement ix^(2n)  =t give x^(2n)  =(t/i) =−it  =e^(−((iπ)/2))  t ⇒x =(e^(−((iπ)/2)) t)^(1/(2n))   =e^(−((iπ)/(4n)))  t^(1/(2n))  ⇒∫_0 ^∞  e^(−ix^(2n) ) dx =e^(−((iπ)/(4n)))  ∫_0 ^∞   e^(−t)   (1/(2n)) t^((1/(2n))−1)  dt  =(e^(−((iπ)/(4n))) /(2n))∫_0 ^∞   e^(−t)  t^((1/(2n))−1)  dt =((Γ((1/(2n))))/(2n))×e^(−((iπ)/(4n)))  ⇒ A_n  =2×((Γ((1/(2n))))/(2n))×cos((π/(4n))) ⇒  A_n =(1/n)Γ((1/(2n)))cos((π/(4n)))  rest to find lim_(n→+∞) A_n ....be continued...

$$\mathrm{A}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2n}} \right)\:\mathrm{dx}\:\Rightarrow\mathrm{A}_{\mathrm{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2n}} \right)\mathrm{dx}\:=\mathrm{2}\:\mathrm{Re}\left(\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2n}} } \mathrm{dx}\right)\:\mathrm{but} \\ $$$$\mathrm{changement}\:\mathrm{ix}^{\mathrm{2n}} \:=\mathrm{t}\:\mathrm{give}\:\mathrm{x}^{\mathrm{2n}} \:=\frac{\mathrm{t}}{\mathrm{i}}\:=−\mathrm{it}\:\:=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \:\mathrm{t}\:\Rightarrow\mathrm{x}\:=\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{2n}}} \\ $$$$=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4n}}} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2n}}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{ix}^{\mathrm{2n}} } \mathrm{dx}\:=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4n}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{2n}}\:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2n}}−\mathrm{1}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4n}}} }{\mathrm{2n}}\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2n}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2n}}\right)}{\mathrm{2n}}×\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4n}}} \:\Rightarrow\:\mathrm{A}_{\mathrm{n}} \:=\mathrm{2}×\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2n}}\right)}{\mathrm{2n}}×\mathrm{cos}\left(\frac{\pi}{\mathrm{4n}}\right)\:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2n}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{4n}}\right)\:\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} ....\mathrm{be}\:\mathrm{continued}... \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 26/Jun/20

thnk you sir.

$$\mathrm{thnk}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Ar Brandon last updated on 26/Jun/20

wow, cool !

Commented by maths mind last updated on 26/Jun/20

nice  sir i completΓ((1/(2n)))=(π/(Γ(1−(1/(2n)))sin((π/(2n)))))  A_n =(1/n)((.π)/(Γ(1−(1/(2n)))sin((π/(2n)))))cos((π/(4n))),(1/n)=t  ⇒A_n =((πt)/(Γ(1−(t/2))sin(((πt)/2))))cos((π/4)t)  lim_(t→0) ((πt)/(sin(((πt)/2))))=(2/π),we get (2/(πΓ(1)))cos(0)=(2/π)

$${nice}\:\:{sir}\:{i}\:{complet}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)=\frac{\pi}{\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right){sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{{n}}\frac{.\pi}{\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right){sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}{cos}\left(\frac{\pi}{\mathrm{4}{n}}\right),\frac{\mathrm{1}}{{n}}={t} \\ $$$$\Rightarrow{A}_{{n}} =\frac{\pi{t}}{\Gamma\left(\mathrm{1}−\frac{{t}}{\mathrm{2}}\right){sin}\left(\frac{\pi{t}}{\mathrm{2}}\right)}{cos}\left(\frac{\pi}{\mathrm{4}}{t}\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\pi{t}}{{sin}\left(\frac{\pi{t}}{\mathrm{2}}\right)}=\frac{\mathrm{2}}{\pi},{we}\:{get}\:\frac{\mathrm{2}}{\pi\Gamma\left(\mathrm{1}\right)}{cos}\left(\mathrm{0}\right)=\frac{\mathrm{2}}{\pi} \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 27/Jun/20

Are you talking to yourself, Sir ?��

Commented by mathmax by abdo last updated on 27/Jun/20

i talk to sir mind what happen to you...

$$\mathrm{i}\:\mathrm{talk}\:\mathrm{to}\:\mathrm{sir}\:\mathrm{mind}\:\mathrm{what}\:\mathrm{happen}\:\mathrm{to}\:\mathrm{you}... \\ $$

Commented by Ar Brandon last updated on 27/Jun/20

Oh sorry ! �� I just feel you both are alike. ����

Commented by maths mind last updated on 27/Jun/20

withe pleasur sir

$${withe}\:{pleasur}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com