Integral of $\log|e^{it}-1|$

Question

Does there exist an elegant proof of $$ \int_0^{2\pi}\log|e^{it}-1|\,dt=0 \ ? \label{1}\tag1 $$ Of course, one can do some $\varepsilon$-$\delta$ stuff and get it, but I look for a nice proof. In the literature, it is reduced to $$ \int_0^\pi \log(2\sin t)\,dt=0, \label{2}\tag2 $$ but if you want to find a proof of \eqref{2}, then it is usually said that it follows from \eqref{1}.

Answer 1

Here's another elementary proof, again assuming one has checked that the integral converges absolutely despite the singularity at $t=0$.

Let $I$ be the average of the $2\pi$-periodic function $\log \left| e^{it}-1 \right|$. This is the same as the average of $\log \left| e^{it}+1 \right|$ (translate $t$ by $\pi$). It's also the average of $\log \left| e^{2it} - 1 \right|$. But $$ e^{2it} - 1 = (e^{it}-1) (e^{it}+1), $$ so $$ \log\left|e^{2it} - 1\right| = \log\left|e^{it}-1\right| + \log\left|e^{it}+1\right|. $$ Hence $I=I+I$, so $I=0$.

If we prefer trig functions to complex exponentials, we can use the identity $2 \cos t \sin t = \sin 2t$ to the same effect: $\log\left|2\cos t\right|$ averages to zero (because it's the difference between the averages of $\log\left|2\sin t\right|$ and $\log\left|2\sin 2t\right|$, which are the same); by shifting $t$, the same is true of the average of $\log\left|2\sin t\right|$.

Answer 2

I like the proof with Riemann sums along the arithmetic progressions: since the integral converges and the function $\log |e^{it}-1|$ is piecewise monotone, it equals $$\int_0^{2\pi} \log|e^{it}-1|=\lim_{n\to\infty}\frac{2\pi}n\sum_{k=1}^{n-1}\log|e^{i(2\pi k/n)}-1|\\=\lim_{n\to\infty}\frac{2\pi}n\log \left|\prod_{\xi:1+\xi+\ldots+\xi^{n-1}=1}(1-\xi)\right|=\lim_{n\to\infty}\frac{2\pi}n\log n=0.$$

Answer 3

I once traced references for the proofs by Fourier expansion (Euler 1770), differentiation under an integral sign (Poisson 1823), contour integration (Cauchy 1825), Riemann sums (Ellis 1841), $\underline {I+I=I}$ (Goodwin 1843).

Addendum:
Coxeter (1935, 5.5) notes that Goodwin’s $I+I=I$ (or really $J+J = J+\tfrac\pi2\log\tfrac12$, where $J$ is the integral from $0$ to $\frac\pi2$) is already in Lobachevsky’s Application of Imaginary Geometry to Certain Integrals (Russian, 1836): set $x=0$ in his formula (14) (translation, p. 54).

Answer 4

Well, one definitely needs to work a bit to justify the convergence of the integral at the endpoints. Apart from that, the comment by mathworker21 is on point. Namely, let us consider the holomorphic function $f(z):=\log(1-z)$ on $\mathbb{C}\setminus[1,\infty)$, where $\log$ is the principal logarithm on $\mathbb{C}\setminus(-\infty,0]$. Then $f(z)$ is holomorphic, hence $g(z):=\Re f(z)=\log|1-z|$ is harmonic. In particular, $$\int_0^{2\pi}g(re^{it})\,dt=2\pi g(0)=0,\qquad r\in (0,1),$$ and the same also holds for $r=1$ by Lebesgue's dominated convergence theorem (say).

Answer 5

Start with the identity: $$ \sin(x)=\frac{\exp(ix)-\exp(- ix)}{2 i}. $$ Some motivation, informal stuff. If $x$ is $0$ then the difference is $0$. This is valid without an absolute sign. This is valid for $x \in \mathbb R$.

It is straightforward to show that the subtraction cancels out the imaginary unit complete in the form of the quotient the right hand is real too.

If You have done this and comprehend it then the identity is nearly easy.

Since the question starts with $\exp( i x)-1$ some additional work is needed. So factor out $\exp\left( \frac{i x}{2}\right)$. Then take then absolute value of both factors for implications. See that $$ \begin{split} \left|\exp\Big( \frac{i x}{2}\Big)\right| &= 1 \\ \left|\exp\Big( \frac{i x}{2}\Big)\right|\left|\left[\exp( \frac{i x}{2})-(\exp( \frac{-i x}{2})\right]\right| &= \left|\exp\Big( \frac{i x}{2}\Big)-\exp\Big( -\frac{ x}{2}\Big)\right| \end{split} $$

We to compensate the $2$ in the demoninator not the imaginare unit $i$.

Now the identity is cleared by a sequence of complex numbers absolute value identities.

Now remember $\sin(x)$ is positive right of $0$, the origin. We are allowed to leave away the absolute value taking till the first zero crossing of the graph of the function. That is $\pi$ .

By the symmetry argument we are allowed to cut the interval down to $\pi$ since we have to the take the absolute value.

This argument is rather fast-stepped. But it is easily convincing since it is school knowledge.

The logarithmic function for a real-valued argument in this case a real-valued, absolute-valued function, is changing nothing. The logarithm is defined only for positive real arguments and the value interval on the output is the complete reals. Logarithm does as natural as it is not change the monotone at all.

So that is it in informal words but with formal intentions in mind.

The last argument is, that taking the half wave once or twice does not matter, because both are zero.

In graphical analysis, the result expresses that the area above the real axis in compensated by that below the real axis between zero and $\pi$. That is surprising and untrue from intuition unless one is firm with the natural logarithm and the exponential function.