# Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$

For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.

My attempt: Let

\begin{align*} f_n(x) &= \frac{\ln\left(1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\ &=\frac{\ln\left(\displaystyle\frac{1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\ &=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}. \end{align*} Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But $$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider $$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$ So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$

It looks too easy, is there any crucial lack?

$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.

• What is this problem from? Commented Jan 25, 2014 at 2:34
• The problem comes from my today calculus exam. Commented Jan 25, 2014 at 2:36
• I ask this question because of the level of the rest of the questions on the exam. This one doesn't seem to require anything more than bunching skills, unlike the rest. I do not exclude the possibility that proof is OK (modulo some constants etc) but I'm a little afraid that there is some fundamental lack in my reasoning and I want just be sure about that. Commented Jan 25, 2014 at 2:42
• I think there may be a problem with the equality right after "But" and before "Consider." In the numerator you have a term of $-2 \cos x$ and in the denominator you have a term of $-2 a \cos x$; note the factor of $a$. I can't see how the equality follows. Commented Jan 25, 2014 at 3:07
• – Ben
Commented Jan 15, 2016 at 5:41

Here is an elementary way to compute the integral.

First, let us prove some initial results.

1. Making the substitution $x \mapsto \pi - x$ yields $I(a) = \int^\pi_0 \log \left (1 + 2a\cos x + a^2 \right ) \, dx = I(-a)$ so that $$I(a) = I(-a). \tag{\dagger}$$

2. Then, consider \begin{align*} I(a) + I(-a) &= \int^{\pi}_{0}\log \! \Big ( \left (1 - 2a\cos x + a^2 \right ) \left (1 + 2a\cos x + a^2 \right ) \Big) \> dx\\ &= \int^{\pi}_{0}\log \! \Big ( \left (1 + a^2 \right )^2 - \left (2a\cos x \right )^2 \Big) \> dx.\\ \end{align*} Using double angle formulae produces \begin{align*} I(a) + I(-a)&= \int^{\pi}_{0}\log \left ( 1 + 2a^2 + a^4 - 2a^2 \left ( 1 + \cos 2x \right ) \right) \, dx\\ &= \int^{\pi}_{0}\log \left ( 1 - 2a^2\cos 2x + a^4 \right) \, dx,\\ \end{align*} so we may let $x \mapsto \frac{1}{2}x$ to give \begin{align*} I(a) + I(-a) &= \frac{1}{2}\int^{2\pi}_{0}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, dx.\\ \end{align*} We can then split the integral at $\pi$ and set $x \mapsto 2\pi - x$ for the second integral: \begin{align*} I(a) + I(-a) &= \frac{1}{2} I(a^2) + \frac{1}{2}\int^{2\pi}_{\pi}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, dx\\ &= \frac{1}{2} I(a^2) + \frac{1}{2}\int^{\pi}_{0}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, dx\\ &= I(a^2). \end{align*} We thus have (applying $(\dagger)$) $$I(a)= \frac{1}{2}I(a^2). \tag{\star}$$

It follows from $(\star)$ that $I(0) = 0$ and $I(1) = 0$.

Consider the case when $0 \le a < 1$. We may use $(\star)$ iteratively $n$ times to write

$$I(a) = \frac{1}{2^n} I \left ( a^{2^{n}} \right ).$$

Setting $n \to \infty$ allows $\frac{1}{2^n} \to 0$ and $a^{2^{n}} \to 0$ so that $I \left ( a^{2^{n}} \right ) \to 0$ which gives the result

$$I(a) = 0.$$

When $a > 1$, it follows that $0 < 1/a < 1$ and consequently $I(1/a) = 0$. We have

\begin{align*} I(a) &= \int^\pi_0 \log \! \Big ( a^2 \left ((1/a)^2 + (1/a)\cos x + 1 \right ) \Big ) \> dx\\ &= 2\pi\log(a) + I(1/a)\\ &= 2\pi\log\left(a\right). \end{align*}

We could use $(\dagger)$ to extend the result to negative $a$, obtaining the final solution valid for all real $a$,

$$I(a) = \begin{cases} 0 &\text{if } |a| \le 1;\\ 2\pi\log|a| &\text{otherwise}. \end{cases}$$

• Beautiful use of symmetries! +1 Commented Oct 1, 2015 at 17:01
• Nice answer! +1. Is $I(a)$ is continuous function? You used that fact when you deal with $0<a<1$
– RFZ
Commented May 27, 2016 at 11:22
• $\displaystyle +1$. Nice. Commented Oct 16, 2022 at 23:02

Considering the following diagram:

$\hspace{2cm}$

we get that \begin{align} \int_0^\pi\log\left(1-2r\cos(\theta)+r^2\right)\,\mathrm{d}\theta &=\int_0^{2\pi}\log\sqrt{1-2r\cos(\theta)+r^2}\,\mathrm{d}\theta\\ &=\mathrm{Re}\left(\int_\gamma\log(z-1)\frac{\mathrm{d}z}{iz}\right) \end{align} along the path $\gamma=r\,e^{i[0,2\pi]}$.

If $r\le1$, the singularity at $z=0$ has residue $0$. Thus, the integral is $0$.

If $r\gt1$, then we need to modify the path to avoid the branch cut for $\log(z-1)$ along $\{t\in\mathbb{R}:t\ge1\}$. That is, in addition to the circular contour $\gamma=r\,e^{i[0,2\pi]}$, we need to follow the two contours $[r,1]$ below the real axis and $[1,r]$ above the real axis. The sum of the integrals along these two contours is $$\int_r^12\pi i\frac{\mathrm{d}z}{iz}+\int_1^r0\frac{\mathrm{d}z}{iz}=-2\pi\log(r)$$ Since the integral along all three contours is $0$, the integral along the circular part must be $2\pi\log(r)$.

Putting these two cases together, we get $$\int_0^\pi\log\left(1-2r\cos(\theta)+r^2\right)\,\mathrm{d}\theta =\left\{\begin{array}{l} 0&\text{if }r\le1\\ 2\pi\log(r)&\text{if }r\gt1 \end{array}\right.$$

• I know this is old but can you explain why $\int_0^{2\pi}\log\sqrt{1-2r\cos(\theta)+r^2}\,\mathrm{d}\theta =\mathrm{Re}\int_\gamma\log(z-1)\frac{\mathrm{d}z}{iz}$? Commented Mar 29, 2020 at 21:04
• Note that on the circle $|z|=1$, $\mathrm{d}\theta=\frac{\mathrm{d}z}{iz}$, and by the Law of Cosines, $|z-1|=\sqrt{1-2r\cos(\theta)+r^2}$.
– robjohn
Commented Mar 29, 2020 at 21:18

If $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x,$$ then \begin{align} I'(a)&=\int_{0}^{\pi}\frac{2a-2\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x,\\ I'(a) & = \frac{1}{a}\int_{0}^{\pi}\frac{2a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\ & = \frac{1}{a}\int_{0}^{\pi}\frac{1-1+a^2+a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\ & = {\pi \over a} + \frac{1}{a}\int_{0}^{\pi}\frac{a^2-1}{1-2a\cos x+a^2} \ \mathrm{d}x \end{align} and making the Weierstrass substitution, $$\cos x = \frac{1 - t^2}{1 + t^2},$$ $$\mathrm{d}x = \frac{2 \,\mathrm{d}t}{1 + t^2}.$$ $$I'(a)={\pi \over a} + \frac{2}{a}\int_{0}^{\infty}\frac{a^2-1}{(1+a^2)(1+t^2)-2a(1-t^2)}\ \mathrm{d}t$$ $$I'(a)={\pi \over a} + \frac{2}{a}\int_{0}^{\infty}\frac{a^2-1}{(1-a)^2 + (1+a)^2t^2} \ \mathrm{d}t$$ $$I'(a)={\pi \over a} + \frac{\pi}{a} \operatorname{sgn} (a^2-1),$$ so for $a > 1$, $$I'(a)={2\pi \over a},$$ $$I(a)={2 \pi \log a},$$ given that $\displaystyle \lim_{a \rightarrow 1^+} I(a)=0$. You have to be careful to show this last statement I believe, but you can see the result here - this integral is easy to evaluate:

\begin{align} \int_0^{\pi} \log(2-2\cos x) \ \mathrm{d}x &= \int_0^{\pi} \log(4 \sin^2 x) \ \mathrm{d}x \\ & = \pi \log 4 + 2\int_0^{\pi} \log(\sin x) \ \mathrm{d}x \\ & = 2\pi \log 2 + 4\int_0^{\pi/2} \log(\sin x) \ \mathrm{d}x \\ \end{align} That final integral can be found here, which gives us the final result.

• Feynman comes to the rescue! +1 Commented Oct 1, 2015 at 17:01
• Just wondering, how do you get that sgn function? I tried to 'brute force' it with arctan and got an undefined answer $\int_{0}^{\infty}{\frac{a^2-1}{(1-a)^2+(1+a)^2t^2}}dt = -\tan^{-1}(\frac{a-1}{(a+1)\tan{(\frac{x}{2}})})$ Using standard arctan results. Commented May 27, 2018 at 8:36
• @GarethMa That's actually where $\text{sgn}$ function comes into the game. You see, when $x->0^+$ the thing inside the $\tan^{-1}$ becomes unbounded, and as a consequence $\tan^{-1}$ approaches $\pi/2$ or $-\pi/2$ depending on the sign of a thing inside the $\tan^{-1}$. And this sign is equal to the sign of $(a-1)/(a+1)$(since $x$ is positive) which is equal to the sign of $(a-1)(a+1)=a^2-1$ for $a \ne 1$ Commented May 12, 2021 at 14:56

For some reason, this problem appeared on the main page for me even though the problem was posted quite a while ago. Let me present an alternative approach which does not really require any integrating.

$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos(x)+a^2\right)\ \text{d}x,\;\ a>1$$

Consider a triangle $$ABC$$ with angles $$\alpha,\beta,\gamma$$ opposite to the sides $$a,b,c$$, respectively:

The law of cosines states $$c^2=a^2+b^2-2ab\cos(\gamma)$$ To keep with this notation, let $$b=1$$ and $$\gamma=x$$.

Our integral becomes \begin{align} I(a) & = \int_{0}^{\pi}\ln\left(c^2\right)\ \text{d}\gamma \\ & = 2\int_{0}^{\pi}\ln\left(c\right)\ \text{d}\gamma \\ \end{align} The law of sines states $$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$$ Our integral becomes \begin{align} I(a) & = 2\int_{0}^{\pi}\ln\left(a\ \frac{\sin(\gamma)}{\sin(\alpha)}\right)\ \text{d}\gamma \\ & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\gamma \\ \end{align} Consider the right-hand integral. In a triangle, the three angles $$\alpha,\beta,\gamma$$ add up to $$\pi$$ radians. Let $$\gamma=\pi-\alpha-\beta$$ such that $$\text{d}\gamma=-\text{d}\alpha-\text{d}\beta$$.

When $$\gamma\rightarrow 0:\alpha\rightarrow \pi,\beta\rightarrow 0$$. When $$\gamma\rightarrow \pi:\alpha\rightarrow 0,\ \beta\rightarrow 0$$.

Our integral becomes \begin{align} I(a) & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma\\ & -2\left[\int_{\pi}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\alpha)+\int_{0}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\beta)\right] \\ & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\alpha \\ \end{align} Recognize that the integrals evaluate to the same value and therefore will cancel out each other.

$$I(a)=2\pi\ln(a),\;\ a>1$$

• Could you please explain how you get the new integration boundaries in $-2\left[\int_{\pi}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\alpha)+\int_{0}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\beta)\right]$? Commented Jan 22, 2020 at 22:42
• This method fascinated me a lot, but I think it is flawed. $\int_0^\pi\ln(\sin\alpha)\cdot\frac{d\beta}{d\gamma}\ d\gamma$ happens to be $0$ in this case, but in general this is not true. Take for example $\int_0^\pi\frac{d\beta}{d\gamma}\cdot\frac{d\beta}{d\gamma}\ d\gamma$, which is strictly greater than 0. Perhaps with more work and splitting up the integral it is possible to show $\int_0^\pi\ln(\sin\alpha)\cdot\frac{d\beta}{d\gamma}\ d\gamma=0$, but it seems difficult. Commented Jan 14, 2021 at 23:57
• @Alexander51413 where does $\int_0^\pi \ln(\sin \alpha) d\beta = 0$ follow from? The boundaries in the solution are different, namely 0 and 0. Commented Apr 13, 2023 at 7:32

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm I}\pars{a} = \int_{0}^{\pi}\ln\pars{1 - 2a\cos\pars{x} + a^{2}}\, \dd x:\ {\large ?}.\qquad a \geq 0}$.

\begin{align} {\rm I}\pars{a}&=\half\int_{-\pi}^{\pi}\ln\pars{1 - 2a\cos\pars{x} + a^{2}}\,\dd x =\half\int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{1 - 2a\,{z^{2} + 1 \over 2z} + a^{2}}\,{\dd z \over \ic z} \\[3mm]&=-\,\half\ic \int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{-az^{2} + \bracks{a^{2} + 1}z - a \over z}\,{\dd z \over z} \\[3mm]&=-\,\half\ic \int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{-a\bracks{z - a}\bracks{z - 1/a} \over z}\,{\dd z \over z} \\[3mm]&=-\,\half\ic \int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{-a\bracks{z - \mu_{<}}\bracks{z - \mu_{>}} \over z}\,{\dd z \over z} \quad\mbox{where}\quad \mu_{< \atop >} \equiv {\min \atop \max}\braces{a, {1 \over a}} \\[3mm]&\mbox{and}\quad 0\ \leq\ \mu_{<} < 1\,,\quad \mu_{>}\ >\ 1 \end{align}

$${\rm I}\pars{a}= -\,\half\ic\int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{\bracks{z - \mu_{<}}\bracks{\mu_{>} - z}}\,{\dd z \over z} +\half\ic\int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{z \over a}\,{\dd z \over z}$$

$$\half\ic\int_{\verts{z}=1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}} < \pi}} \ln\pars{z \over a}\,{\dd z \over z} =\half\ic\int_{-\pi}^{\pi}\ln\pars{\expo{\ic\theta} \over a}\,{\expo{\ic\theta}\ic\,\dd\theta \over \expo{\ic\theta}} = \pi\ln\pars{a}$$

\begin{align} &\color{#c00000}{\large{\rm I}\pars{a} - \pi\ln\pars{a}} =\half\,\ic\int_{-1}^{\mu_{<}}{\ln\pars{\bracks{\mu_{<} - x}\bracks{\mu_{>} - x}} + \ic\pi \over x + \ic 0^{+}}\,\dd x \\[3mm]&\mbox{} + \half\,\ic\int_{\mu_{<}}^{-1} {\ln\pars{\bracks{\mu_{<} - x}\bracks{\mu_{>} - x}} - \ic\pi \over x - \ic 0^{+}}\,\dd x \\[3mm]&=\half\,\ic\pars{\int_{-1}^{\mu_{<}}{\braces{\ln\pars{\bracks{\mu_{<} - x}\bracks{\mu_{>} - x}} + \ic\pi}}\bracks{\pp{1 \over x} - \ic\pi\delta\pars{x}} \,\dd x} \\[3mm]&\phantom{=}\mbox{} -\half\,\ic\pars{\int_{-1}^{\mu_{<}}{\braces{\ln\pars{\bracks{\mu_{<} - x}\bracks{\mu_{>} - x}} - \ic\pi}}\bracks{\pp{1 \over x} + \ic\pi\delta\pars{x}}\,\dd x} \\[3mm]&=\half\ic\pp\int_{-1}^{\mu_{<}}2\pi\ic\,{\dd x \over x} =-\pi\lim_{\epsilon \to 0^{+}}\pars{\int_{-1}^{-\epsilon}{\dd x \over x} +\int_{\epsilon}^{\mu_{<}}{\dd x \over x}}=-\pi\ln\pars{\mu_{<}} \\[3mm]&=\color{#c00000}{\large -\pi\Theta\pars{1 - a}\ln\pars{a} -\pi\Theta\pars{a - 1}\ln\pars{1 \over a}} \end{align}

$$\color{#00f}{\large% {\rm I}\pars{a} = \Theta\pars{a - 1}\bracks{2\pi\ln\pars{a}}}$$ which was calculated for $\ds{\color{#c00000}{a > 0}}$.

From the $\ds{{\rm I}\pars{a}}$ original definition it's clear that $\ds{\color{#c00000}{{\rm I}\pars{a}\ \mbox{is an}\ \ul{\mbox{even}}\ \mbox{function of}\ a}}$ and that $\ds{\color{#c00000}{{\rm I}\pars{0} = 0}}$. Then, the solution $\ds{\color{#c00000}{\forall\ a \in {\mathbb R}}}$ is given by: $$\color{#00f}{\large% {\rm I}\pars{a} = \Theta\pars{\verts{a} - 1}\bracks{2\pi\ln\pars{\verts{a}}}}$$

Old thread, but this question came up again and was marked a duplicate, so I have to post my physics lecture here.

You know the symmetries of the problem that the electric field of a line charge must be purely radial and that its magnitude has no axial or azimuthal dependence $\vec E=E(r)\hat e_r$. Thus we may apply Gauss's law over a cylinder or radius $r$ and length $\ell$ to find the magnitude of the electric field of a line charge $\lambda$ coincident with the $z$-axis: $$\oint\vec E\cdot d^2\vec A=E(r)A_{\text{curvy}}=2\pi rE(r)=\int\frac{\rho}{\epsilon_0}d^3V=\frac{\lambda\ell}{\epsilon_0}$$ So $\vec E=\frac{\lambda}{2\pi\epsilon_0r}\hat e_r=-\vec\nabla V$. So we can get the potential $V=-\frac{\lambda}{2\pi\epsilon_0}\ln r+C$. Now that constant of integration is a bit problematic because $V$ is unbounded at both $r=0$ and $r=\infty$, but if we have a line charge $\lambda$ at $(x,y)=(b\cos\theta,b\sin\theta)$ and another charge $-\lambda$ on the $z$-axis, then the potential of the combination is $$V(\theta)=-\frac{\lambda}{2\pi\epsilon_0}\ln\sqrt{1-2\frac br\cos\theta+\frac{b^2}{r^2}}$$ And this potential does go to $0$ as $r$ goes to $\infty$. Now we can create the potential for a cylinder centered on the $z$-axis with radius $b$ and surface charge density $\sigma$ with a line charge $-2\pi b\sigma$ running along the $z$-axis by superposition: $$V=\int_0^{2\pi}-\frac{a\sigma}{4\pi\epsilon_0}\ln\left(1-2\frac br\cos\theta+\frac{b^2}{r^2}\right)d\theta$$ For $r>b$, applying Gauss's law over a cylinder of radius $r$ and length $\ell$ this time shows that $\vec E=\vec0$ because the cylinder contains no net charge. So this time $V=C$, a constant for $r>b$. Since $$\lim_{r\rightarrow\infty}V(r)=0$$ by construction, so we have shown that for $0<a<1$, $$\int_0^{2\pi}\ln(1-2a\cos\theta+a^2)d\theta=0$$ If $a>1$, then \begin{align}\int_0^{2\pi}\ln(1-2a\cos\theta+a^2)d\theta&=\int_0^{2\pi}\left[2\ln a+\ln(1-2a^{-1}\cos\theta+a^{-2})\right]d\theta\\ &=4\pi\ln a+0=4\pi\ln a\end{align}

By my post, I had found $$\int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right) \tag*{(*)}$$ for any $$c\neq 0$$ and $$-1\leq \frac{b}{c} \leq 1.$$

Putting $$b=-2 a$$ and $$c=a^{2}+1$$ into $$(*)$$, we have \begin{aligned} \int_{0}^{\pi} \ln \left(1-2 a \cos x+a^{2}\right) d x =& \pi \ln \left(\frac{a^{2}+1+\sqrt{\left(a^{2}+1\right)^{2}-4 a^{2}}}{2}\right) \\ =& \pi \ln \left(\frac{a^{2}+1+\sqrt{\left(a^{2}-1\right)^{2}}}{2}\right) \\ =& \pi \ln \left(\frac{a^{2}+1+\left|a^{2}-1\right|}{2}\right) \end{aligned} We can now conclude that

For any $$|a| \leqslant 1$$, $$I=\pi \ln \left(\frac{\left.a^{2}+1+1-a^{2}\right)}{2}\right)=0$$ for any $$|a|>1$$, $$I=\pi \ln \left(\frac{a^{2}+1+a^{2}-1}{2}\right)=2 \ln |a|$$