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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

Ω(a,b)=∫0πLn(tan(ax))1−2bcos(x)+b2dx,∣b∣<1,0<a⩽12.Wehave:Ln(tan(ax))=−2∑+∞n=0cos(2(2n+1)ax)2n+1So:Ω(a,b)=−∑+∞n=012n+1∫02πcos(2(2n+1)ax)1−2bcos(x)+b2dxIn(a,b)=∫02πcos(2(2n+1)ax)1−2bcos(x)+b2dx=−∮∣z∣=1z2(2n+1)abz2−(1+b2)z+bdzi=−∮∣z∣=1z2(2n+1)ab(z−b)(z−1b)dzi=2πRes[z2(2n+1)ab(z−b)(z−1b),z=b]=2π1−b2b2(2n+1)a∴Ω(a,b)=−2π1−b2∑+∞n=0b2a(2n+1)2n+1Or:∑+∞n=0x2n+12n+1=∫0x∑+∞n=0t2ndt=∫0xdt1−t2=12∫0x(11−t+11+t)dt=−12Ln(1−x1+x)Then:∫0πLn(tan(ax))1−2bcos(x)+b2dx=π1−b2Ln(1−b2a1+b2a)KAMELBENAICHA

Commented by puissant last updated on 28/Aug/21

Mr Kamel you are really strong..

MrKamelyouarereallystrong..

Commented by Kamel last updated on 28/Aug/21

Thank you.

Thankyou.

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