TITLE: Does this integral have a closed form?
QUESTION [12 upvotes]: I was working with this problem in an exam:
Given $\lambda\in(-1,1)\subset\Bbb R$, find 
$$f
   (\lambda)=\int_{0}^{\pi}\ln\left(1+\lambda \cos x\right)\mathrm{d}x
$$
My try: put $\delta\in (0,1)$ such that $\lambda\in(-\delta,\delta)$. Using the power series of $\ln(1+x)$ and uniform convergence to take $\sum_{n=0}^\infty$ outside $\int_{0}^{\pi}$, finally, I got this:
$$f(\lambda)=-\sum_{k=1}^\infty{(2k-1)!!\over (2k)!!}{\lambda^{2k-1}\over 2k}$$
But I still wonder if there is a closed form for $f(\lambda)?$ And is my solution above right or wrong?

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}%
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$\ds{\fermi\pars{\lambda}
=
\int_{0}^{\pi}\ln\pars{1 + \lambda\cos\pars{x}}\,\dd x:
\ {\large ? }.\qquad\lambda, \delta \in {\mathbb R}\,,
\quad\verts{\lambda} < \delta < 1.\quad}$ It's clear that $\ds{\fermi\pars{\lambda}}$ is an even function of $\ds{\lambda}$ and $\ds{\fermi\pars{0} = 0}$ such that
we calculate

\begin{align}
\fermi\pars{\lambda \not= 0}&=
\int_{0}^{\pi}\ln\pars{1 + \verts{\lambda}\cos\pars{x}}\,\dd x
=\half\int_{-\pi}^{\pi}\ln\pars{1 + \verts{\lambda}\cos\pars{x}}\,\dd x
\\[3mm]&=
\half\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}}
\ln\pars{1 + \verts{\lambda}\,{z^{2} + 1 \over 2z}}\,{\dd z \over \ic z}
\\[3mm]&=-\,\half\,\ic
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}}
\ln\pars{\verts{\lambda}z^{2} + 2z + \verts{\lambda} \over 2z}\,{\dd z \over z}
\end{align}
  The zeros $\ds{z_{\pm}}$ of $\ds{\verts{\lambda}z^{2} + 2z + \verts{\lambda}}$ are
  given by:
  $$
z_{\pm} \equiv {-1 \pm \root{1 - \lambda^{2}} \over \verts{\lambda}}\,,\qquad
z_{-} < -1\,,\quad -1 < z_{+} < 0
$$

Then,
$$
\fermi\pars{\lambda \not=0}=
-\,\half\,\ic
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}}
\ln\pars{\bracks{z - z_{-}}\bracks{z - z_{+}}}\,{\dd z \over z}
+\
\overbrace{\half\,\ic
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}}
\ln\pars{2z \over \verts{\lambda}}\,{\dd z \over z}}
^{\ds{\pi\ln\pars{\verts{\lambda} \over 2}}}
$$

\begin{align}
&\color{#c00000}{\fermi\pars{\lambda\not=0} - \pi\ln\pars{\verts{\lambda} \over 2}}
\\[3mm]&=\half\,\ic\int_{-1}^{z_{+}}{\ln\pars{\verts{x - z_{-}}\verts{x - z_{+}}} + \ic\pi
\over x + \ic 0^{+}}\,\dd x
+
\half\,\ic\int_{z_{+}}^{-1}{\ln\pars{\verts{x - z_{-}}\verts{x - z_{+}}} - \ic\pi
\over x - \ic 0^{+}}\,\dd x
\\[3mm]&=\half\,\ic\int_{-1}^{z_{+}}2\pi\ic\,{\dd x \over x}
=-\pi\ln\pars{\verts{z_{+}}}
=-\pi\ln\pars{1 - \root{1 - \lambda^{2}} \over \verts{\lambda}}
=\color{#c00000}{-\pi\ln\pars{\verts{\lambda} \over 1 + \root{1 - \lambda^{2}}}}
\end{align}

$$\color{#00f}{\large%
\fermi\pars{\lambda}
=
\int_{0}^{\pi}\ln\pars{1 + \lambda\cos\pars{x}}\,\dd x
=-\pi\ln\pars{2 \over 1 + \root{1 - \lambda^{2}}}}\,,\qquad
\lambda \in \pars{-1,1}
$$