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Question Number 156362 by MathSh last updated on 10/Oct/21

Answered by mindispower last updated on 10/Oct/21

∣Ω_1 +iΩ_2 ∣^2 =Ω_1 ^2 +Ω_2 ^2   Ω_1 +iΩ_2 =∫_(−∞) ^∞ e^(i(e^(2x) +e^(−2x) )) e^(x−sh^2 (x))   =∫_0 ^∞ e^(i(e^(2x) +e^(−2x) )) e^(−sh^2 (x)) (e^x +e^(−x) )dx  let u=sh(t)  =du=ch(t)dt  =2∫_0 ^∞ e^(i(4sh^2 (x)+1)) e^(−u^2 ) du  =2e^i ∫_0 ^∞ e^((−1+4i)u^2 ) du  ∫_0 ^∞ e^(au^2 ) du=f(a),Re(a)<0  f(a)=((√π)/( 2(√a)))  Ω_1 +iΩ_2 =(e^i /((−1+4i)^(1/2) ))(√π)  ∣Ω_1 +iΩ_2 ∣=((√π)/( (√(√(17)))))⇒Ω_1 ^2 +Ω_2 ^2 =(π/( (√(17))))

Ω1+iΩ22=Ω12+Ω22Ω1+iΩ2=ei(e2x+e2x)exsh2(x)=0ei(e2x+e2x)esh2(x)(ex+ex)dxletu=sh(t)=du=ch(t)dt=20ei(4sh2(x)+1)eu2du=2ei0e(1+4i)u2du0eau2du=f(a),Re(a)<0f(a)=π2aΩ1+iΩ2=ei(1+4i)12πΩ1+iΩ2∣=π17Ω12+Ω22=π17

Commented by MathSh last updated on 10/Oct/21

Perfect my dear Ser, thank you

PerfectmydearSer,thankyou

Commented by mindispower last updated on 11/Oct/21

withe pleasur

withepleasur

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