Evaluate $\int \log(1+2m \cos x+m^2) dx$

Question

The question is to evaluate $$\int \log(1+2m \cos x+m^2) dx$$ when $m > 0$

I noticed that $(1+2m \cos x+m^2)$ can be written as $(1+me^{-x})(1+me^{ix})$ and then I can apply the Taylor expansion of $\log(x)$ but I am not sure if the approach is correct.

Answer 1

$$\int \ln{\left(1+2m\cos{x}+2m^2\right)}dx=x\ln{(1+2m^2)}+\int\ln{\left(1+\frac{2m}{1+2m^2}\cos{x}\right)}dx=$$ where $\displaystyle \left|\frac{2m\cos{x}}{1+2m^2}\right|<1$

$\displaystyle =x\ln{(1+2m^2)}+\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left(\frac{2m}{1+2m^2}\right)^n\int\cos^n{x}dx$

$\displaystyle =x\ln{(1+2m^2)}+\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^nn}\left(\frac{2m}{1+2m^2}\right)^n\sum\limits_{k=0}^n\binom{n}{k}\frac{e^{i(n-2k)x}}{i(n-2k)}+C$

Answer 2

Mathematica Output

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Mathematica instantly produces the following: $$\frac{1}{2} x \left(2 \log \left(m^2+2 m \cos (x)+1\right)-2 \log \left(\frac{m+e^{i x}}{m}\right)-2 \log \left(1+m e^{i x}\right)+i x\right)+i \operatorname{Li}_2\left(\frac{-e^{i x}}{m}\right)+i \operatorname{Li}_2\left(-e^{i x} m\right)+C$$ I am working on a way to prove this by expanding and greatly improving on Durgesh's answer, but I am not there yet. Should all else fail, converting each of the terms to integral form and combining should work. There are definitely patterns emerging in the output; for example, the derivatives of the two Dilogarithm terms in the output are the latter two logarithm terms present in the output, i.e. $$\frac{d}{dx} i \operatorname{Li}_2\left(\frac{-e^{i x}}{m}\right) = \log \left(\frac{m+e^{i x}}{m}\right)$$ $$\frac{d}{dx} i \operatorname{Li}_2\left(-e^{i x} m\right) = \log \left(1+m e^{i x}\right)$$

Answer 3

I'm not so sure of indefinite integration, but if we add bounds, it comes out beautifully, as shown here.

Take the derivative with respect to $m$ to get

$$\begin{align}-I'(-m)&=\int_0^\pi\frac{2m-2\cos(x)}{1-2m\cos(x)+m^2}\ dx\\&=\frac1m\int_0^\pi1-\frac{1-m^2}{1-2m\cos(x)+m^2}\ dx\\&=\frac\pi m-\frac2m\tan^{-1}\left(\frac{1+m}{1-m}\tan(x/2)\right)\bigg|_{x=0}^{x=\pi}\\&=\begin{cases}0&|m|<1\\\frac{2\pi}m&|m|>1\end{cases}\end{align}$$

It thus becomes clear that

$$I(-m)=\begin{cases}C_1&|m|<1\\2\pi\ln|m|+C_2&|m|>1\end{cases}$$

By substituting in a few values, one may deduce that $C_1=C_2=0$, thus

$$\int_0^\pi\ln(1+2m\cos(x)+m^2)\ dx=\begin{cases}0&|m|<1\\2\pi\ln|m|&|m|>1\end{cases}$$

Answer 4

This is a well-known problem about the Poisson kernel.
Since $\log\|z\|=\text{Re}\log(z)$ and for every $n\in\mathbb{Z}$ we have $\int_{0}^{2\pi}e^{ni\theta}\,d\theta=2\pi\,\delta(n)$,

$$\forall r\in\mathbb{R},\quad \int_{0}^{2\pi}\log\|1-r e^{i\theta}\|\,d\theta = 2\pi \log\max(1,|r|),\tag{1}$$ $$\forall r\in\mathbb{R},\quad \int_{0}^{2\pi}\log(1+r^2-2r\cos\theta)\,d\theta = 4\pi\log\max(1,|r|).\tag{2}$$

The indefinite integral is related with the dilogarithm function, but I doubt you really need it.