Integral involving logarithm and cosine For $a,b>0$ I would like to compute the integral $$I=\int_0^{2\pi} -\log{\sqrt{a^2+b^2-2ab\cos{t}}}~dt.$$
Numerical computations suggest that $$I=\min\{-\log{a},-\log{b}\}.$$
How can I prove this? I tried to find the antiderivative using Mathematica, but the result looks awfull. 
Have you seen such an integral? Some reference would be great! 
 A: We have
$$I=\dfrac12\int_0^{2\pi} -\log(a^2+b^2-2ab\cos{t})dt=\int_0^{\pi} -\log(a^2+b^2-2ab\cos{t})dt$$
We then have
$$I=-2\pi\log(b) -\int_0^{\pi} \log((a/b)^2+1-2(a/b)\cos{t})dt$$
Let us call $a/b$ as $a$ from now on. Hence, we want to evaluate the integral
$$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$ Some preliminary results on $I(a)$. Note that we have $$I(a) = \underbrace{\displaystyle \int_0^{\pi} \ln \left(1+2a \cos(x) + a^2\right) dx}_{\spadesuit} = \overbrace{\dfrac12 \displaystyle \int_0^{2\pi} \ln \left(1-2a \cos(x) + a^2\right) dx}^{\clubsuit}$$ $(\spadesuit)$ can be seen by replacing $x \mapsto \pi-x$ and $(\clubsuit)$ can be obtained by splitting the integral from $0$ to $\pi$ and $\pi$ to $2 \pi$ and replacing $x$ by $\pi+x$ in the second integral.
Now let us move on to our computation of $I(a)$.
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx = \dfrac12 \int_0^{2\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-2a^2(1+ \cos(x))\right) dx = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-4a^2 \cos^2(x/2)\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left(1+a^2-2a \cos(x/2)\right) dx + \dfrac12 \int_0^{2\pi} \ln \left(1+a^2+2a \cos(x/2)\right) dx
\end{align}
Now replace $x/2=t$ in both integrals above to get
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1+a^2-2a \cos(t)\right) dt + \int_0^{\pi} \ln \left(1+a^2+2a \cos(t)\right) dt = 2I(a)
\end{align}
Now for $a \in [0,1)$, this gives us that $I(a) = 0$. This is because we have $I(0) = 0$ and $$I(a) = \dfrac{I(a^{2^n})}{2^n}$$ Now let $n \to \infty$ and use continuity to conclude that $I(a) = 0$ for $a \in [0,1)$. Now lets get back to our original problem. Consider $a>1$. We have
\begin{align*}
I(1/a) & = \int_0^{\pi} \ln \left(1-\dfrac2{a} \cos(x) + \dfrac1{a^2}\right)dx\\
& = \int_0^{\pi} \ln(1-2a \cos(x) + a^2) dx - 2\int_0^{\pi} \ln(a)dx\\
& = I(a) - 2 \pi \ln(a)\\
& = 0  & \text{(Since $1/a < 1$, we have $I(1/a) = 0$)}
\end{align*}
Hence, we get that
$$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$
Adapted from here
A: The term under the square root can be written as either $(ae^{it}-b)(ae^{-it}-b)$ or $(a-be^{it})(a-be^{-it})$.
You want to choose the one where neither factor crosses the negative real axis.  That is, you want to write
$$
a^2+b^2-2ab\cos t=(c-de^{it})(c-de^{-it}),
$$
where $c=\max(a,b)$ and $d=\min(a,b)$.  Then the integral becomes
$$
\int_{0}^{2\pi}-\log\sqrt{(c-de^{it})(c-de^{-it})}dt=-\frac{1}{2}\int_{0}^{2\pi}\log(c-de^{it})dt-\frac{1}{2}\int_{0}^{2\pi}\log(c-de^{-it})dt.
$$
Each term can be evaluated as a contour integral.  Letting $z=e^{it}$ in the first term, and $z=e^{-it}$ in the second, we have
$$
-\frac{1}{i}\int_{\Gamma}\frac{\log(c-dz)}{z}dz,
$$
where $\Gamma$ is the unit circle.  Because $c\ge d>0$, the contour avoids the branch cut, so the only contribution is from the pole at $z=0$:
$$
I=-{2\pi}\log c=-2\pi\log\max(a,b).
$$
This agrees with your result except for the factor of $2\pi$.
