Need help evaluating the complex integral $\int_{-\pi}^{\pi} \dfrac{x \sin x\ dx}{1- 2a \cos x + a^2}$ where $0 < a < 1$.
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I have tried this: \begin{align}\int_{-\pi}^{\pi} \dfrac{x \sin x\ dx}{1- 2a \cos x + a^2} &= \Im \left [ \int_{-\pi}^{\pi} \dfrac{x e^{ix}\ dx}{(a - e^{ix})(a - e^{-ix})} \right ] \ &= \Im \left [ \int_{-\pi}^{\pi} \dfrac{-i log(z) z\ dx}{(a - z)(a - z^{-1})} \frac{dz}{iz} \right ] \ &= \Im \left [ \int_{C_1} \dfrac{- log(z) z\ dx}{(a-z)(az-1)}\ dz \right ] \ &= \Im \left [ \int_{C_1} \dfrac{- log(z) z\ dx}{(a-z)(az-1)}\ dz \right ] \ \end{align} – May 13 '16 at 04:16
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Almost surely a duplicate.... – user_of_math May 13 '16 at 05:08
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4Here goes. Your answer is exactly twice of this integral http://math.stackexchange.com/questions/822484/a-difficult-integral-evaluation-problem – user_of_math May 13 '16 at 05:10