6
$\begingroup$

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 ".

$\endgroup$
5
$\begingroup$

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 $$ \int_0^{2\pi}\log(1-2r\cos x+r^2)dx=\int_0^{2\pi}\log\left(r^2\left(\frac{1}{r^2}-2\frac{1}{r}\cos x+1\right)\right)dx $$ $$ =\int_0^{2\pi}\log r^2dx+ \int_0^{2\pi}\log\left(\frac{1}{r^2}-2\frac{1}{r}\cos x+1\right)dx=4\pi \log |r|+0. $$

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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 $$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.