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!

• $\cos 2\pi=\cos 0=1$ so this means if we substitute with the cosine rule we get the same index ${\sqrt{a^2+b^2-2ab}}$ ,which implies our integral is zero if our anti derivative is even ,we shouls use this to show that $I=\in{ -\log a,-\log b}$... $I=\int_{\sqrt{a^2+b^2-2ab}}^{\sqrt{a^2+b^2-2ab}}\frac{c\log c}{ab\sqrt{1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}}} dc$ Nov 24 '13 at 17:01
• $I=\text{min} \{-\log a,-\log b \}$ Nov 24 '13 at 17:09
• This question lacks important context such as where you encountered the integral and why it is of interest. Nov 25 '13 at 4:15
• I tried to prove Newton's theorem in two dimensions, at some point I got stuck at this integral. Nov 25 '13 at 7:43

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 $$I(a) = \underbrace\int_0^{\pi} \ln \left(1+2a \cos(x) + a^2\right) dx}_{\spadesuit} = \overbrace{\dfrac12 \displaystyle \int_0^{2\pi} \ln \left(1-2a \cos(x) + a^2\right) dx}^{\clubsuit$$ $(\spadesuit)$ 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 $$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$
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 $2\pi$.