Question Number 114526 by bobhans last updated on 19/Sep/20

(1/π)∫_0 ^π  e^(2cos θ)  dθ ?

Commented byJDamian last updated on 19/Sep/20

https://youtu.be/-UhFu0g9740

Commented bymathmax by abdo last updated on 19/Sep/20

the problem how to calculate Σ (1/((n!)^2 )) ...!

Answered by mathmax by abdo last updated on 19/Sep/20

I =(1/π) ∫_0 ^π  e^(2cosθ)  dθ ⇒πI =∫_0 ^(π/2)  e^(2cosθ)  dθ +∫_(π/2) ^π  e^(2cosθ)  dθ(→θ=(π/2)+u)  =∫_0 ^(π/2)  e^(2cosθ)  dθ  +∫_0 ^(π/2)  e^(−2sinu)  du =∫_0 ^(π/2) Σ_(n=0) ^∞  (((2cosθ)^n )/(n!)) dθ  +∫_0 ^(π/2) Σ_(n=0) ^∞  (((−2sinθ)^n )/(n!)) dθ =Σ_(n=0) ^∞  (2^n /(n!)) ∫_0 ^(π/2)  cos^n θ dθ  +Σ_(n=0) ^∞  (((−2)^n )/(n!)) ∫_0 ^(π/2)  sin^n θ dθ  changement  θ =(π/2)−t give  ∫_0 ^(π/2)  sin^n θ dθ =∫_0 ^(π/2) cos^n t dt ⇒  π I =Σ_(n=0) ^∞ {(2^n /(n!)) +(((−2)^n )/(n!))}∫_0 ^(π/2)  cos^n θ dθ  let w_n =∫_0 ^(π/2)  cos^n θ dθ (wallis integral)  πI =Σ_(n=0) ^∞  (2^n /(n!))(1+(−1)^n )W_n   =Σ_(n=0) ^∞  (2^(2n) /((2n)!))(2)W_(2n)  ⇒  I=(1/π)Σ_(n=0) ^∞  (2^(2n+1) /((2n)!)) W_(2n)      (w_(2n) is known by recurrence)