My problem is to integrate this expression: $$\int_0^{2\pi}\log(1-2r\cos x +r^2)dx.$$ where $r$ is any constant in $[0,1]$.

I know the answer is zero. Can you explain you idea to me or just prove that? Maybe you will use the "Cauchy integral theorem ".


Edit: I have proved a little more. Namely, this integral is $0$ for $| r|\leq 1$, and $4\pi\log|r|$ for $|r|\geq 1$.

1) If $r=0$, this is $0$.

2) Now assume $0< r\leq 1$. We have $$ 1-2r\cos x+r^2=(1-re^{ix})(1-re^{-ix})=|1-re^{ix}|^2. $$ Hence the integral can be written $$ 2\int_0^{2\pi}\log |1-re^{ix}|dx=4\pi\cdot \frac{1}{2\pi} \int_0^{2\pi}\log |1-re^{ix}|dx= \frac{2}{i}\int_{\gamma_r}\frac{\log|1-z|}{z}dz. $$

where $\gamma_r(x)=re^{ix}$ over $[0,2\pi]$, i.e. $\gamma_r$ is the circle of radius $r$ centered at $0$.

Shortcut: as pointed out by robjohn, $\log|1-z|=\mbox{Re}\;\log(1-z)$, so we can conclude by Cauchy's integral formula. I'll leave my harmonic argument below.

Recall that if $f$ is holomorphic, then $\log|f|$ is harmonic on its domain. So $$g:z\longmapsto \log|1-z|$$ is harmonic on the open unit disk $D=\{z\in\mathbb{C}\;;\;|z|<1\}$.

If $0<r<1$, then the circle $\gamma_r$ is contained in $D$ and contains $0$, so $$ 0=g(0)=\frac{1}{2\pi} \int_0^{2\pi}g(re^{ix})dx=\frac{1}{2\pi} \int_0^{2\pi}\log |1-re^{ix}|dx $$ by the mean value property of harmonic functions.

3) For $r=1$ now. Fix $x$ and study the variations of $r\longmapsto \log(1-2r\cos x+r^2)$ on $[0,1]$. If $\cos x\leq 0$, this is increasing and nonnegative, so $|\log(1-2r\cos x+r^2)|\leq \log(1-2\cos x+1)$. And if $\cos x\geq 0$, it is nonpositive with a minimum at $r=\cos x$, so $|\log(1-2r\cos x+r^2)|\leq| \log(1-2\cos^2 x+\cos^2 x)|=| \log(\sin^2 x)|$. In any case, we see that the integrand is uniformly bounded by an integrable function. It follows that $$ \int_0^{2\pi}\log(1-2\cos x+1)dx=\lim_{r\rightarrow 1^-} \int_0^{2\pi}\log(1-2r\cos x+r^2)dx=0 $$ by Lebesgue dominated convergence.

So these integrals are $0$ for every $0\leq r\leq 1$.

4) By change of variable and periodicity, we get $0$ for $-1\leq r\leq 1$.

5) Finally, for $|r|>1$, we get $$ \int_0^{2\pi}\log(1-2r\cos x+r^2)dx=\int_0^{2\pi}\log\left(r^2\left(\frac{1}{r^2}-2\frac{1}{r}\cos x+1\right)\right)dx $$ $$ =\int_0^{2\pi}\log r^2dx+ \int_0^{2\pi}\log\left(\frac{1}{r^2}-2\frac{1}{r}\cos x+1\right)dx=4\pi \log |r|+0. $$

  • $\begingroup$ Thank U very much!and I've mistaken its range (0,1) for [0,1].Nevertheless,I'll be curious about whether this still holds when r=1. $\endgroup$ – sopin Apr 5 '13 at 12:47
  • $\begingroup$ one more question,does this proof hold if x∈[0,π]?I had typed this,but it turned out to be [0,2π],and maybe it's the system who changed my input...= = $\endgroup$ – sopin Apr 5 '13 at 12:57
  • $\begingroup$ well,I think its integration on [0,π] equals that on [π,2π],if I change the variant x into -x,then add 2π onto the integral interval.am I right? $\endgroup$ – sopin Apr 5 '13 at 13:12
  • $\begingroup$ @sopin The change of variable $u=\pi+x$ is what you want. ANd you're right, then it works. So the integral over $[0,\pi]$ is half the integral over $[0,2\pi]$. $\endgroup$ – Julien Apr 5 '13 at 13:15
  • $\begingroup$ Thank you again for bothering to answer my question patiently. $\endgroup$ – sopin Apr 5 '13 at 13:18

Copying from my blog.

Let $$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$ 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 \tag{Since $1/a < 1$, we have $I(1/a) = 0$} \end{align*} Hence, we get that $$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$

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  • $\begingroup$ I would have never thought of doing it like this, funny...+1. $\endgroup$ – Julien Apr 5 '13 at 16:54
  • $\begingroup$ Thank u for ur answer.It is good to find many different ways of solving the same question. $\endgroup$ – sopin Apr 6 '13 at 6:01

Note that $$ \begin{align} \int_0^{2\pi}\log(1-2r\cos(x)+r^2)\,\mathrm{d}x &=2\,\mathrm{Re}\left(\oint_\gamma\log(1-z)\frac{\mathrm{d}z}{iz}\right) \end{align} $$ where the contour of integration is $\gamma(x)=re^{ix}$.

Since there is no singularity inside the contour, the integral is $0$.

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  • $\begingroup$ Nice to invoke the (obvious) harmonic conjugate, +1. As lots of people seem to like holomorphic functions better than harmonic ones...Note this is true for $0<r<1$ only. $\endgroup$ – Julien Apr 5 '13 at 13:53
  • $\begingroup$ well,seems great...but can you explain further why the equation holds? $\endgroup$ – sopin Apr 5 '13 at 13:55
  • $\begingroup$ my book writes:log(z)=log|z|+iargz,with 0<argz<2π. $\endgroup$ – sopin Apr 5 '13 at 14:03
  • $\begingroup$ @sopin Yes, sorry, typo. So you get the formula when you take the real part. $\endgroup$ – Julien Apr 5 '13 at 14:15
  • $\begingroup$ @julien oh,I see.thank you!and thank robjohn! $\endgroup$ – sopin Apr 5 '13 at 14:22

An idea: putting

$$z:=re^{ix}:=r\cos x+i\,r\sin x\;,\;\;r,x\in\Bbb R\implies dz=rie^{ix}dx\;,\;\; 1-2r\cos x+r^2=|z-1|^2$$

so by an extension to Cauchy's Integral Formula from the MVHF (see Note and comments below) we get $$\int\limits_0^{2\pi}\log(1-2r\cos x+r^2)dx=\frac{2}{i}\int\limits_{S^1}\frac{\log|z-1|}{z}dz=4\pi\left(\log|z-1|\right)_{z=0}=0$$

Note: In fact the result follows at once from the Mean Value Property for Harmonic Functions (see comments below and Julien's answer), which can be seen as a "special" kind of extension or, perhaps more accurate, special case of the CIF. Follow the links below and above

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  • $\begingroup$ Cauchy's integral formula is for holomorphic functions. This one is harmonic. Also, this only works for $0<r<1$. And finally, this formula is exactly the one I used. So we agree on this. $\endgroup$ – Julien Apr 5 '13 at 12:49
  • $\begingroup$ First, yes: this only works for $\,0\le r<1\,$ as we don't want the logarithm to go around (the/a) point where its argument vanishes. Second, because the same reason, I think this function is holomorphic in the domain limited by the unit circle as we'll always be away from the problematic point $\,z=1\,$ and thus CIF is appliable...unless I'm missing something basic here. $\endgroup$ – DonAntonio Apr 5 '13 at 13:06
  • $\begingroup$ See my answer. We can compute this for every $r$, actually. The intergral always converges. Now $\log|z-1|$ is real-valued, harmonic, not holomorphic. The only real-valued holomorphic functions are constant. $\endgroup$ – Julien Apr 5 '13 at 13:12
  • $\begingroup$ See the open mapping theorem to see why real valued holomorphic implies constant. Although with such a strong assumption as real-valued, it can be done with more elementary tools. $\endgroup$ – Julien Apr 5 '13 at 13:22
  • $\begingroup$ Oh, I know we can compute for every $\,r\,$ yet we'll have to add multiples of $\,2\pi\,$ to the argument and/or choose branch cuts and it all becomes another mess. $\endgroup$ – DonAntonio Apr 5 '13 at 13:24

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