Question-152364

Question Number 152364 by mathdanisur last updated on 27/Aug/21

Answered by Kamel last updated on 28/Aug/21

Ω (a, b) = \int_{0}^{π} \frac{L n (t a n (a x))}{1 - 2 b c o s (x) + b^{2}} d x, ∣ b ∣< 1, 0 < a ⩽ \frac{1}{2} .

W e h a v e : L n (t a n (a x)) = - 2 \underset{n = 0}{\sum^{+ \infty}} \frac{c o s (2 (2 n + 1) a x)}{2 n + 1}

S o : Ω (a, b) = - \underset{n = 0}{\sum^{+ \infty}} \frac{1}{2 n + 1} \int_{0}^{2 π} \frac{c o s (2 (2 n + 1) a x)}{1 - 2 b c o s (x) + b^{2}} d x

I_{n} (a, b) = \int_{0}^{2 π} \frac{c o s (2 (2 n + 1) a x)}{1 - 2 b c o s (x) + b^{2}} d x = - \oint_{∣ z ∣= 1} \frac{z^{2 (2 n + 1) a}}{{b z}^{2} - (1 + b^{2}) z + b} \frac{d z}{i}

= - \oint_{∣ z ∣= 1} \frac{z^{2 (2 n + 1) a}}{b (z - b) (z - \frac{1}{b})} \frac{d z}{i}

= 2 π R e s [\frac{z^{\overset{}{2} (2 n + 1) a}}{b (z - b) (z - \frac{1}{b})}, z = b] = \frac{2 π}{1 - b^{2}} b^{2 (2 n + 1) a}

∴ Ω (a, b) = - \frac{2 π}{1 - b^{2}} \underset{n = 0}{\sum^{+ \infty}} \frac{b^{2 a (2 n + 1)}}{2 n + 1}

O r : \underset{n = 0}{\sum^{+ \infty}} \frac{x^{2 n + 1}}{2 n + 1} = \int_{0}^{x} \underset{n = 0}{\sum^{+ \infty}} t^{2 n} d t = \int_{0}^{x} \frac{d t}{1 - t^{2}} = \frac{1}{2} \int_{0}^{x} (\frac{1}{1 - t} + \frac{1}{1 + t}) d t = - \frac{1}{2} L n (\frac{1 - x}{1 + x})

T h e n :

\int_{0}^{π} \frac{L n (t a n (a x))}{1 - 2 b c o s (x) + b^{2}} d x = \frac{π}{1 - b^{2}} L n (\frac{1 - b^{2 a}}{1 + b^{2 a}})

K A M E L B E N A I C H A

Commented by puissant last updated on 28/Aug/21

M r K a m e l y o u a r e r e a l l y s t r o n g . .

Commented by Kamel last updated on 28/Aug/21

T h a n k y o u .

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