# $I(r) = \int_0^{2\pi}\frac{\cos(t) - r}{1 - 2r\cos t + r^2}\,dt$ is always zero for $r\in[0,1)$. Why?

For $$r\in [0,1)$$ define $$I(r) = \int_0^{2\pi}\frac{\cos(t) - r}{1 - 2r\cos t + r^2}\,dt.$$ Numerical experiments hint at $$I(r) = 0$$ for all $$r\in [0,1)$$ but I can't show this analytically.

This integral appears when you compute the Cauchy transform of $$\overline z$$ over the unit circle. The latter therefore seems to be constantly zero.

• @Dr.SonnhardGraubner Edited – amsmath Jan 26 at 17:23
• I think your integral isn't zero – Dr. Sonnhard Graubner Jan 26 at 17:31
• @Dr.SonnhardGraubner Well, that was helpful... Why? – amsmath Jan 26 at 17:33
• @Dr.SonnhardGraubner: I think the integral is zero. – Larry Jan 26 at 18:01
• @Dr.SonnhardGraubner You gave us the proof below. ;-) – amsmath Jan 26 at 18:08

An alternative proof using complex methods:

For $$0< r < 1$$ let $$f_r \colon B_\frac{1}{r} (0) \to \mathbb{C} \, , \, f_r(z) = \frac{- \ln(1-rz)}{z} \, ,$$ where $$f_r(0) = r$$ . Then $$f_r$$ is holomorphic, so $$I(r) \equiv - \int \limits_0^{2\pi} \ln(1-r \mathrm{e}^{\mathrm{i}t}) \, \mathrm{d} t = - \mathrm{i} \int \limits_{S^1} f_r(z) \, \mathrm{d} z = 0$$ holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm: $$I(r) = \sum \limits_{n=1}^\infty \frac{r^n}{n} \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i} n t} \, \mathrm{d} t = 0 \, .$$ This implies \begin{align} \int \limits_0^{2\pi} \frac{\cos(t) - r}{1 - 2 r \cos(t) + r^2} \, \mathrm{d} t &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln(1 - 2 r \cos(t) + r^2) \, \mathrm{d} t \\ &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln[(1 -r \mathrm{e}^{\mathrm{i} t})(1 -r \mathrm{e}^{-\mathrm{i} t})] \, \mathrm{d} t \\ &= \frac{\mathrm{d}}{\mathrm{d} r} I(r) = \frac{\mathrm{d}}{\mathrm{d} r} 0 = 0 \end{align} as desired.

(Since it is not mentioned yet) The function $$P_r(t) = \sum_{n=-\infty}^\infty r^{|n|} e^{int} =\frac{1-r^2}{1-2r\cos t +t^2},\quad0\le r<1, 0\le t\le 2\pi$$ is called the Poisson kernel. For any continuous function $$f:[0,2\pi]\to \Bbb C$$ the Poisson integral $$u(r,\theta) = \frac{1}{2\pi i}\int_0^{2\pi} f(t)P_r(\theta-t)\mathrm{d}t\tag{*}$$ gives the unique solution of the Dirichlet problem $$\triangle u =0, \lim\limits_{r\to 1^-}u(r,\theta)=f(\theta)$$.

Let $$u(r,\theta)=r\cos \theta$$. Since it is a real part of the analytic function $$z=r e^{i\theta}$$, we find that $$\triangle u=0,\ \ \ \lim\limits_{r\to 1^-}u(r,\theta)=\cos \theta.$$ By $$\text{(*)}$$ it follows that $$r\cos \theta =\frac{1}{2\pi}\int_0^{2\pi} \frac{\cos t(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t$$ for all $$0\le r<1$$ and $$0\le \theta\le 2\pi$$. Using the fact $$\frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t=\frac{1}{2\pi}\sum_{n=-\infty}^\infty r^{|n|} \int_0^{2\pi}e^{in(t-\theta)}\mathrm{d}t =1,$$ it follows $$\frac{1}{2\pi}\int_0^{2\pi} \frac{\left(\cos t-r\cos \theta\right)(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t =0.$$ By letting $$\theta =0$$, we obtain $$\frac{1-r^2}{2\pi }\int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0,$$ or equivalently $$\int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0.$$

• That is also very nice. Thank you. In fact, I tried to fuddle around with the Poisson kernel, but I did not see things as clear as you, obviously. ;-) – amsmath Jan 26 at 19:18
• @amsmath I hope this will help :-) – Song Jan 26 at 21:31

It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.

Here, consider the integral $$\vec I(M) = \int_{X\in C}\frac{\vec{MX}}{MX^2}\mathrm ds$$ where $$M$$ is a point at a distance $$r<1$$ away from the center $$O$$ of the unit circle $$C$$, and $$\mathrm ds$$ is a length element of the circle. What you are asking for is a component of $$\vec I$$, the one directed along $$\vec{OM}$$. In fact, we have $$\vec I=\vec 0$$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.

• interesting physical interpretation (+1) – G Cab Jan 26 at 19:43

With $$\gamma$$ being the counter-clockwise unit circle and $$\bar\gamma$$ being the clockwise unit circle, \begin{align} \int_0^{2\pi}\frac{r-\cos(t)}{1-2r\cos(t)+r^2}\,\mathrm{d}t &=\int_0^{2\pi}\frac12\left(\frac1{r-e^{it}}+\frac1{r-e^{-it}}\right)\mathrm{d}t\tag1\\ &=\frac1{2i}\int_\gamma\frac{\,\mathrm{d}z}{z(r-z)}-\frac1{2i}\int_{\bar\gamma}\frac{\,\mathrm{d}z}{z(r-z)}\tag2\\ &=\frac1i\int_\gamma\frac{\,\mathrm{d}z}{z(r-z)}\tag3\\ &=\frac1{ir}\int_\gamma\left(\frac1z-\frac1{z-r}\right)\mathrm{d}z\tag4\\ &=\left\{\begin{array}{} 0&\text{if }|r|\lt1\\ \frac{2\pi}r&\text{if }|r|\gt1 \end{array}\right.\tag5 \end{align} Explanation:
$$(1)$$: partial fractions
$$(2)$$: $$z=e^{it}$$ in the left integral and $$z=e^{-it}$$ in the right integral
$$(3)$$: $$\bar\gamma$$ is in the opposite direction from $$\gamma$$
$$(4)$$: partial fractions
$$(5)$$: $$2\pi i$$ times the sum of the residues inside $$\gamma$$

Here is yet another slightly different approach.

Let $$P_r(t) = \frac{\cos t - r}{1 - 2r \cos t + r^2}, \qquad r \in [0,1).$$ Using $$\cos t = (e^{it} + e^{-it})/2$$, we can rewrite $$P_r(t)$$ as \begin{align} P_r (t) &= \frac{1}{2} \frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\\ &= -\frac{1}{2} \frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\\ &= -\frac{1}{2} \left [\frac{1}{r - e^{it}} + \frac{1}{r - e^{-it}} \right ]\\ &= \frac{1}{2} \left [\frac{e^{-it}}{1 - r e^{-it}} + \frac{e^{it}}{1 - r e^{it}} \right ]\\ &= \frac{1}{2} \left [e^{-it} \sum_{n = 0}^\infty r^n e^{-int} + e^{it} \sum_{n = 0}^\infty r^n e^{int} \right ]\\ &= \sum_{n = 0}^\infty r^n \left (\frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} \right )\\ &= \sum_{n = 0}^\infty r^n \cos (n + 1)t. \end{align}

Now \begin{align} \int^{2\pi}_0 P_r (t) \, dt = \int^{2 \pi}_0 \sum_{n = 0}^\infty r^n \cos (n + 1)t \, dt = \sum_{n = 0}^\infty r^n \int^{2\pi}_0 \cos (n + 1)t \, dt = 0. \end{align}

Note that $$I(r) =\frac{-1}{2r}\int_{0}^{2\pi}\frac{1-2r\cos t+r^2+r^2-1}{1-2r\cos t+r^2}\,dt$$ which is same as $$-\frac{\pi} {r} +\frac{1-r^2}{r}\int_{0}^{\pi}\frac{dt}{1+r^2-2r\cos t}$$ The integral above is equal to $$\frac{\pi} {\sqrt{(1+r^2)^2-4r^2}}=\frac{\pi}{1-r^2}$$ and hence $$I(r) =0$$.

We have used the formula $$\int_{0}^{\pi}\frac{dt}{a+b\cos t} =\frac{\pi} {\sqrt{a^2-b^2}},\,a>|b|$$ which can be proved using the substitution $$(a+b\cos t) (a-b\cos u) =a^2-b^2$$

Hint: $$\cos(t)=\frac{1-a^2}{1+a^2}$$ and $$dt=\frac{2da}{1+a^2}$$ Now your indefinite integral is given by $$\int-\frac{2 \left(a^2 r+a^2+r-1\right)}{\left(a^2+1\right) \left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1\right)}da$$ and this is rational The solution of this integral is given by $$-2 \left(\frac{\tan ^{-1}(a)}{2 r}-\frac{\tan ^{-1}\left(\frac{a (r+1)}{1-r}\right)}{4 r}+\frac{\tan ^{-1}\left(\frac{a (r+1)}{r-1}\right)}{4 r}\right)$$

• Thank you very much. So my integral is indeed zero. You can fuse the second and the third term to one term: $-\tfrac 1{2r}\arctan(\tfrac{1+r}{1-r}a)$. The boundaries after substitution are zero and $\infty$. – amsmath Jan 26 at 18:04

Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $$r$$:

\begin{align} I'(r) &= \int_0^{2\pi}\frac{d}{dr}\left(\frac{\cos t - r}{1 - 2r\cos t + r^2}\right)\,dt\\ &= \int_0^{2\pi} \frac{(r^2-1)-2r\cos t + 2\cos^2t}{(1 - 2r\cos t + r^2)^2} \\ &= \frac1{i}\int_{|z|=1} \frac{(r^2-1)-r(z+z^{-1}) + \frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}\cdot\frac{dz}{z}\\ &= \frac1{i}\int_{|z|=1} \frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2\left(z-\frac1r\right)^2}\,dz\\ \end{align}

Since $$r \in [0,1\rangle$$, the relevant residues are $$\operatorname{Res}(f,0) = \frac1{2r^2}$$ $$\operatorname{Res}(f,r) = -\frac1{2r^2}$$ so $$I'(r) = 0$$ for $$r \in [0,1\rangle$$.

We conclude that $$I$$ is constant on $$[0,1\rangle$$. Hence $$I(r)=I(0) = \int_0^{2\pi}\cos t\,dt = 0$$