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let-f-0-1-ln-1-e-i-x-dx-find-a-simple-form-of-f-

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Question Number 36736 by abdo mathsup 649 cc last updated on 04/Jun/18
let f(θ) = ∫_0 ^1 ln(1−e^(iθ) x)dx find a simple form of f(θ)
letf(θ)=∫01ln(1−eiθx)dxfindasimpleformoff(θ)
Commented by prof Abdo imad last updated on 06/Jun/18
for that let find ϕ(z) =∫_0 ^1 ln(1−zx)dx if ∣z∣=1 we have ∣zx∣≤1 and ln^′ (1−u)=((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ ln(1−u)=−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒ ϕ(z) =−∫_0 ^1 {Σ_(n=1) ^∞ ((z^n x^n )/n)}dx =−Σ_(n=1) ^∞ (z^n /n) ∫_0 ^1 x^n dx =−Σ_(n=1) ^∞ (z^n /(n(n+1))) =−Σ_(n=1) ^∞ ( (1/n) −(1/(n+1)))z^n =−Σ_(n=1) ^∞ (z^n /n) +Σ_(n=1) ^∞ (z^n /(n+1)) but −Σ_(n=1) ^∞ (z^n /n) =ln(1−z) and Σ_(n=1) ^∞ (z^n /(n+1)) = Σ_(n=2) ^∞ (z^(n−1) /n) =(1/z) Σ_(n=2) ^∞ (z^n /n) = (1/z){ Σ_(n=1) ^∞ (z^n /n) −z} =(1/z){−ln(1−z) −z} =−((ln(1−z))/z) −1 ⇒ ϕ(z)= ln(1−z) −((ln(1−z))/z) −1 f(θ) =ϕ(e^(iθ) ) = (1−(1/e^(iθ) ))ln(1−e^(iθ) ) −1 =(1−e^(−iθ) ) ln(1−e^(iθ) ) −1 .
forthatletfindφ(z)=∫01ln(1−zx)dxif∣z∣=1wehave∣zx∣⩽1andln′(1−u)=−11−u=−∑n=0∞un⇒ln(1−u)=−∑n=0∞un+1n+1=−∑n=1∞unn⇒φ(z)=−∫01{∑n=1∞znxnn}dx=−∑n=1∞znn∫01xndx=−∑n=1∞znn(n+1)=−∑n=1∞(1n−1n+1)zn=−∑n=1∞znn+∑n=1∞znn+1but−∑n=1∞znn=ln(1−z)and∑n=1∞znn+1=∑n=2∞zn−1n=1z∑n=2∞znn=1z{∑n=1∞znn−z}=1z{−ln(1−z)−z}=−ln(1−z)z−1⇒φ(z)=ln(1−z)−ln(1−z)z−1f(θ)=φ(eiθ)=(1−1eiθ)ln(1−eiθ)−1=(1−e−iθ)ln(1−eiθ)−1.

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