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Poisson like Integral

Let a \in [-\pi, \pi]. Prove that:

    \[\bigintsss_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x = \frac{\pi^2}{2} - \pi \left| a \right|\]

Solution

We recall the following Fourier series.

Lemma 1: Let x \in [-\pi, \pi] then

(1)   \begin{equation*} \sum_{n=1}^{\infty} \frac{\sin nx}{n} = \frac{\pi-x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{\pi^2}{6} -\frac{\pi x}{2} + \frac{x^2}{4} \end{equation*}

Lemma 2: Let x \in [-\pi, \pi] then

(2)   \begin{equation*} \sum_{n=1}^{\infty} \frac{(-1)^n \sin nx}{n} = - \frac{x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^n \cos nx}{n^2} = \frac{x^2}{4} - \frac{\pi^2}{12} \end{equation*}

Lemma 3: It holds that:

(3)   \begin{equation*} \ln \left ( 1 - 2x \cos a + x^2 \right ) = -2\sum_{n=1}^{\infty} \frac{x^n \cos na}{n} \end{equation*}

Hence for a \in [0, \pi]

    \begin{align*} \ln \left ( \frac{1+2x \cos a + x^2}{1-2x \cos a + x^2} \right ) &= \ln \left ( 1 + 2x \cos a + x^2 \right ) - \ln \left ( 1 - 2x \cos a + x^2 \right ) \\ &=-2 \sum_{n=1}^{\infty} \frac{(-x)^n \cos na}{n} + 2 \sum_{n=1}^{\infty} \frac{x^n \cos na}{n} \\ &= -2 \sum_{n=1}^{\infty} \frac{\left ( (-x)^n - x^n \right ) \cos na}{n} \\ &= 4 \sum_{n=1}^{\infty} \frac{x^{2n+1} \cos (2n+1)a}{2n+1} \end{align*}

since (-x)^n - x^n is 0 whenever n is even. Hence,

    \begin{align*} \int_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x &= 4\int_{0}^{1} \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{2n+1} \cos (2n+1)a}{2n+1} \, \mathrm{d}x \\ &=4 \sum_{n=0}^{\infty} \frac{\cos(2n+1)a}{2n+1} \int_{0}^{1} x^{2n} \, \mathrm{d}x \\ &=4 \sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{(2n+1)^2} \end{align*}

This is nothing else than \chi_2 Legendre function directly associated with the series in (1), (2). Hence,

    \begin{align*} \sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{(2n+1)^2} & = \mathfrak{Re} \left ( \sum_{n=0}^{\infty} \frac{e^{i(2n+1)a}}{(2n+1)^2} \right ) \\ &=\frac{1}{2}\mathfrak{Re} \left ( \mathrm{Li}_2 \left ( e^{ia} \right ) - \mathrm{Li}_2 \left ( - e^{ia} \right ) \right ) \end{align*}

The real part of \mathrm{Li}_2(e^{ix}) is equation (1) whereas the real part of \mathrm{Li}_2(-e^{ix}) is equation (2). Thus,

    \begin{align*} 4\sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{\left ( 2n+1 \right )^2} &= 2 \left ( \frac{\pi^2}{6} - \frac{\pi a}{2} + \frac{a^2}{4} - \frac{a^2}{4} + \frac{\pi^2}{12} \right ) \\ &=\frac{\pi^2}{2} - \pi a \end{align*}

Because the LHS is even so must be the RHS. So, the result can be extended for a \in [-\pi, 0]. Therefore,

    \[\bigintsss_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x = \frac{\pi^2}{2} - \pi \left| a \right|\]

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