Question Number 152364 by mathdanisur last updated on 27/Aug/21
Answered by Kamel last updated on 28/Aug/21
![Ω(a,b)=∫_0 ^π ((Ln(tan(ax)))/(1−2bcos(x)+b^2 ))dx , ∣b∣<1, 0<a≤(1/2). We have: Ln(tan(ax))=−2Σ_(n=0) ^(+∞) ((cos(2(2n+1)ax))/(2n+1)) So: Ω(a,b)=−Σ_(n=0) ^(+∞) (1/(2n+1))∫_0 ^(2π) ((cos(2(2n+1)ax))/(1−2bcos(x)+b^2 ))dx I_n (a,b)=∫_0 ^(2π) ((cos(2(2n+1)ax))/(1−2bcos(x)+b^2 ))dx=−∮_(∣z∣=1) (z^(2(2n+1)a) /(bz^2 −(1+b^2 )z+b)) (dz/i) =−∮_(∣z∣=1) (z^(2(2n+1)a) /(b(z−b)(z−(1/b)))) (dz/i) =2πRes[(z^(2^ (2n+1)a) /(b(z−b)(z−(1/b)))),z=b]=((2π)/(1−b^2 )) b^(2(2n+1)a) ∴ Ω(a,b)=−((2π)/(1−b^2 ))Σ_(n=0) ^(+∞) (b^(2a(2n+1)) /(2n+1)) Or: Σ_(n=0) ^(+∞) (x^(2n+1) /(2n+1))=∫_0 ^x Σ_(n=0) ^(+∞) t^(2n) dt=∫_0 ^x (dt/(1−t^2 ))=(1/2)∫_0 ^x ((1/(1−t))+(1/(1+t)))dt=−(1/2)Ln(((1−x)/(1+x))) Then: ∫_0 ^𝛑 ((Ln(tan(ax)))/(1−2bcos(x)+b^2 ))dx=(𝛑/(1−b^2 ))Ln(((1−b^(2a) )/(1+b^(2a) ))) KAMEL BENAICHA](https://www.tinkutara.com/question/Q152403.png)
$$ \\ $$$$\Omega\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{Ln}\left({tan}\left({ax}\right)\right)}{\mathrm{1}−\mathrm{2}{bcos}\left({x}\right)+{b}^{\mathrm{2}} }{dx}\:,\:\mid{b}\mid<\mathrm{1},\:\mathrm{0}<{a}\leqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$${We}\:{have}:\:{Ln}\left({tan}\left({ax}\right)\right)=−\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){ax}\right)}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${So}:\:\Omega\left({a},{b}\right)=−\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){ax}\right)}{\mathrm{1}−\mathrm{2}{bcos}\left({x}\right)+{b}^{\mathrm{2}} }{dx} \\ $$$${I}_{{n}} \left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){ax}\right)}{\mathrm{1}−\mathrm{2}{bcos}\left({x}\right)+{b}^{\mathrm{2}} }{dx}=−\underset{\mid{z}\mid=\mathrm{1}} {\oint}\frac{{z}^{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){a}} }{{bz}^{\mathrm{2}} −\left(\mathrm{1}+{b}^{\mathrm{2}} \right){z}+{b}}\:\frac{{dz}}{{i}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\underset{\mid{z}\mid=\mathrm{1}} {\oint}\frac{{z}^{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){a}} }{{b}\left({z}−{b}\right)\left({z}−\frac{\mathrm{1}}{{b}}\right)}\:\frac{{dz}}{{i}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\pi{Res}\left[\frac{{z}^{\overset{} {\mathrm{2}}\left(\mathrm{2}{n}+\mathrm{1}\right){a}} }{{b}\left({z}−{b}\right)\left({z}−\frac{\mathrm{1}}{{b}}\right)},{z}={b}\right]=\frac{\mathrm{2}\pi}{\mathrm{1}−{b}^{\mathrm{2}} }\:{b}^{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right){a}} \\ $$$$\therefore\:\Omega\left({a},{b}\right)=−\frac{\mathrm{2}\pi}{\mathrm{1}−{b}^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{{b}^{\mathrm{2}{a}\left(\mathrm{2}{n}+\mathrm{1}\right)} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\:\:\:{Or}:\:\:\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}=\int_{\mathrm{0}} ^{{x}} \underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{t}^{\mathrm{2}{n}} {dt}=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \left(\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt}=−\frac{\mathrm{1}}{\mathrm{2}}{Ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$\:\:\:\:\:\:\:{Then}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{{Ln}}\left(\boldsymbol{{tan}}\left(\boldsymbol{{ax}}\right)\right)}{\mathrm{1}−\mathrm{2}\boldsymbol{{bcos}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{b}}^{\mathrm{2}} }\boldsymbol{{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{1}−\boldsymbol{{b}}^{\mathrm{2}} }\boldsymbol{{Ln}}\left(\frac{\mathrm{1}−\boldsymbol{{b}}^{\mathrm{2}\boldsymbol{{a}}} }{\mathrm{1}+\boldsymbol{{b}}^{\mathrm{2}\boldsymbol{{a}}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\ $$
Commented by puissant last updated on 28/Aug/21
$${Mr}\:{Kamel}\:{you}\:{are}\:{really}\:{strong}.. \\ $$
Commented by Kamel last updated on 28/Aug/21
$${Thank}\:{you}. \\ $$