The integral $\int \ln{(1+m^{2} +2m\cos{x})}\,\mathrm{d}x?$

Question

I am trying to integrate the following: $$\int \ln{(1+m^{2} +2m\cos{x})}\,\mathrm{d}x.$$

Note that $m$ is a constant.

I tried using integration by parts but I didn't get the answer.

Answer 1

I can give you the answer Mathematica gave me:

$$\int \ln{(1+m^{2} +2m\cos{x})}\,\mathrm{d}x = \\\frac{1}{2} \, x \left(\mathrm{i}\,x-2\,\ln\left(\frac{\mathrm{e}^{\mathrm{i}\,x}+m}{m}\right)- 2\, \ln\left(1+\mathrm{e}^{\mathrm{i}\,x}\,m\right)+2\,\ln{(1+m^{2} +2m\cos{x})}\right)\\+\mathrm{i}\,\mathrm{Li}_2\left(-\frac{\mathrm{e}^{\mathrm{i}\,x}}{m}\right)+\mathrm{i}\,\mathrm{Li}_2\left(-\mathrm{e}^{\mathrm{i}\,x}\,m\right)$$

with $\mathrm{Li}_n()$ the polylogarithm function.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\cos\pars{x} = {z^{2} + 1 \over 2z}\,,\quad z \equiv \expo{\ic x}}$.

\begin{align} &\color{#f00}{\int\ln\pars{1+m^{2} + 2m\cos\pars{x}}\,\dd x} = \int\ln\pars{mz^{2} + \bracks{1 + m^{2}}z + m \over z}\,{\dd z \over \ic z} \\[5mm] = & -\ic\int\ln\pars{m\bracks{z + m}\bracks{z + 1/m} \over z}\,{\dd z \over z} = -\ic\int\ln\pars{m\bracks{1 + z/m}\bracks{1 + mz} \over z}\,{\dd z \over z} \\[1cm] = &\ -\ic\,\ln\pars{m}\int{\dd z \over z} - \ic\int{\ln\pars{1 - \bracks{-z/m}} \over -z/m}\,\dd\pars{-\,{z \over m}} - \ic\int{\ln\pars{1 - \bracks{-mz}} \over -mz}\,\dd\pars{-mz} \\[5mm] + &\ \ic\int{\ln\pars{z} \over z}\,\dd z \\[1cm] = &\ -\ic\ln\pars{m}\ln\pars{z} + \ic\int\Li{2}\, '\pars{-\,{z \over m}}\,\dd\pars{-\,{z \over m}} + \ic\int\Li{2}\, '\pars{-mz}\,\dd\pars{-mz} + \half\,\ic\ln^{2}\pars{z} \\[5mm] = &\ \ln\pars{m}x + \ic\,\Li{2}\pars{-\,{z \over m}} + \ic\,\Li{2}\pars{-mz} - \half\,x^{2}\,\ic \\[5mm] = &\ \color{#f00}{\ln\pars{m}x + \ic\,\Li{2}\pars{-\,{\expo{\ic x} \over m}} + \ic\,\Li{2}\pars{-m\expo{\ic x}} - \half\,x^{2}\,\ic} + \mbox{a constant} \end{align}

Answer 3

Wikipedia's page on Chebyshev polynomials includes the following generating function: $$\sum_{n=1}^\infty T_n(x)\frac{t^n}{n} = \ln\frac{1}{\sqrt{1-2xt+t^2}}.$$ Replacing $x\to \cos x$ and $t\to -m$, the above can be rearranged to $$ \ln (1+2m\cos x+m^2) = -2 \sum_{n=1}^\infty T_n(\cos x)\frac{(-m)^n}{n}=-2 \sum_{n=1}^\infty \cos(n x)\frac{(-m)^n}{n}.$$ Integrated term-by-term gives $$\int \ln (1+2m\cos x+m^2)\,dx=-2 \sum_{n=1}^\infty \sin(n x)\frac{(-m)^n}{n^2}+C$$ i.e. the Fourier series for this integral.

Answer 4

The indefinite integral has a rather simple anti-derivative given below

$$\int \ln{(1+m^{2} +2m\cos{x})}\,{d}x =-2\Im \text{Li}_2(-me^{i x})+C $$ which can be verified by taking the derivative of the RHS.