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

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

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

thnk you sir.

Commented byAr Brandon last updated on 26/Jun/20

wow, cool !

Commented bymaths 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)

Commented byAr Brandon last updated on 27/Jun/20

Are you talking to yourself, Sir ?😅

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

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

Commented byAr Brandon last updated on 27/Jun/20

Oh sorry ! 😹 I just feel you both are alike. 😂😂

Commented bymaths mind last updated on 27/Jun/20

withe pleasur sir