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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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marked as duplicate by Claude Leibovici, Ramiro, choco_addicted, colormegone, JonMark Perry May 13 '16 at 8:20

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  • $\begingroup$ 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*} $\endgroup$ – Gorgio May 13 '16 at 4:16
  • $\begingroup$ Almost surely a duplicate.... $\endgroup$ – user_of_math May 13 '16 at 5:08
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    $\begingroup$ Here goes. Your answer is exactly twice of this integral math.stackexchange.com/questions/822484/… $\endgroup$ – user_of_math May 13 '16 at 5:10