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Question Number 36737 by abdo mathsup 649 cc last updated on 04/Jun/18
let g(θ) =∫_0 ^1 ln( 1−e^(iθ) x^2 )dx find a simple form of g(θ) .θ from R.
$${let}\:{g}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\:\mathrm{1}−{e}^{{i}\theta} {x}^{\mathrm{2}} \right){dx} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{g}\left(\theta\right)\:.\theta\:{from}\:{R}. \\ $$
Commented by math khazana by abdo last updated on 05/Aug/18
we have proved that for ∣z∣=1 ∫_0 ^1 ln(1−zx)dx =(1−(1/z))ln(1−z)−1 but g(θ) =∫_0 ^1 ln(1−(e^(i(θ/2)) x)^2 )dx =∫_0 ^1 ln(1−e^((iθ)/2) x)dx +∫_0 ^1 ln(1 +e^((iθ)/2) x)dx but ∫_0 ^1 ln(1−e^((iθ)/2) x)dx=(1−e^(−((iθ)/2)) )ln(1−e^((iθ)/2) )−1 let find f(z) =∫_0 ^1 ln(1+zx)dx we have for ∣u∣<1 ln^′ (1+u) =(1/(1+u))=Σ_(n=0) ^∞ (−1)^n u^n ⇒ ln(1+u) =Σ_(n=0) ^∞ (((−1)^n )/(n+1))u^(n+1) =Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ⇒ln(1+zx) =Σ_(n=1) ^∞ (−1)^(n−1) ((z^n x^n )/n) ⇒ f(z) =Σ_(n=1) ^∞ (−1)^(n−1) (z^n /n) ∫_0 ^1 x^n dx =Σ_(n=1) ^∞ (−1)^(n−1) (z^n /(n(n+1))) =Σ_(n=1) ^∞ (−1)^(n−1) ((1/n)−(1/(n+1)))z^n =Σ_(n=1) ^∞ (−1)^(n−1) (z^n /n) −Σ_(n=1) ^∞ (−1)^(n−1) (z^n /(n+1)) =ln(1+z) −Σ_(n=2) ^∞ (−1)^(n−2) (z^(n−1) /n) =ln(1+z) +(1/z)Σ_(n=2) ^∞ (−1)^(n−1) (z^n /n) =ln(1+z) +(1/z){Σ_(n=1) ^∞ (−1)^(n−1) (z^n /n) −z} =ln(1+z) +(1/z)ln(1+z)−1 =(1+(1/z))ln(1+z)−1⇒ ∫_0 ^1 ln(1+e^((iθ)/2) x)dx =(1+e^(−((iθ)/2)) )ln(1+e^((iθ)/2) )−1⇒ g(θ) =(1−e^((iπ)/2) )ln(1−e^((iθ)/2) ) +(1+e^(−((iθ)/2)) )ln(1+e^(−((iθ)/2)) )−2
$${we}\:{have}\:{proved}\:{that}\:{for}\:\mid{z}\mid=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{zx}\right){dx}\:=\left(\mathrm{1}−\frac{\mathrm{1}}{{z}}\right){ln}\left(\mathrm{1}−{z}\right)−\mathrm{1}\:{but} \\ $$$${g}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\left({e}^{{i}\frac{\theta}{\mathrm{2}}} {x}\right)^{\mathrm{2}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{e}^{\frac{{i}\theta}{\mathrm{2}}} {x}\right){dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}\:+{e}^{\frac{{i}\theta}{\mathrm{2}}} {x}\right){dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{e}^{\frac{{i}\theta}{\mathrm{2}}} {x}\right){dx}=\left(\mathrm{1}−{e}^{−\frac{{i}\theta}{\mathrm{2}}} \right){ln}\left(\mathrm{1}−{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)−\mathrm{1}\:{let} \\ $$$${find}\:{f}\left({z}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{zx}\right){dx}\:{we}\:{have}\:{for}\:\mid{u}\mid<\mathrm{1} \\ $$$${ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{u}^{{n}} }{{n}} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{zx}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{z}^{{n}} \:{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$${f}\left({z}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{z}^{{n}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{z}^{{n}} }{{n}\left({n}+\mathrm{1}\right)} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){z}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{z}^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{z}^{{n}} }{{n}+\mathrm{1}} \\ $$$$={ln}\left(\mathrm{1}+{z}\right)\:−\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \:\frac{{z}^{{n}−\mathrm{1}} }{{n}} \\ $$$$={ln}\left(\mathrm{1}+{z}\right)\:+\frac{\mathrm{1}}{{z}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{z}^{{n}} }{{n}} \\ $$$$={ln}\left(\mathrm{1}+{z}\right)\:+\frac{\mathrm{1}}{{z}}\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{z}^{{n}} }{{n}}\:−{z}\right\} \\ $$$$={ln}\left(\mathrm{1}+{z}\right)\:+\frac{\mathrm{1}}{{z}}{ln}\left(\mathrm{1}+{z}\right)−\mathrm{1} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right){ln}\left(\mathrm{1}+{z}\right)−\mathrm{1}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{e}^{\frac{{i}\theta}{\mathrm{2}}} {x}\right){dx}\:=\left(\mathrm{1}+{e}^{−\frac{{i}\theta}{\mathrm{2}}} \right){ln}\left(\mathrm{1}+{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)−\mathrm{1}\Rightarrow \\ $$$${g}\left(\theta\right)\:=\left(\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{2}}} \right){ln}\left(\mathrm{1}−{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)\:+\left(\mathrm{1}+{e}^{−\frac{{i}\theta}{\mathrm{2}}} \right){ln}\left(\mathrm{1}+{e}^{−\frac{{i}\theta}{\mathrm{2}}} \right)−\mathrm{2} \\ $$$$ \\ $$$$ \\ $$

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