Would a series representation work for integrating this natural log?

Question

I have the following troublesome integral:

$$\int_0^{2\pi} \ln\Bigg(\frac{\color{red}{r^2}+\color{red}{r_0^2}-2rr_0\cos(\theta_0)}{r^2+\frac{a^4}{r_0^2}-2a^2\frac{r}{r_0}\cos(\theta_0)}\Bigg)d\theta_0$$

Where $r,r_0,a$ are all unspecified constants.

We could also dumb it down to a more general form: $$\int_0^{2\pi} \ln\Bigg(\frac{a-\color{red}c\cos(\theta_0)}{b-\color{red}d\cos(\theta_0)}\Bigg)d\theta_0$$ if anyone is more comfortable with this representation instead, it shouldn't change how it's solved.

The series representation of $\ln(x) = -\sum_{n=1}^\infty \frac{(-1)^n(x-1)^n}{n}$ which has a radius of convergence of $x=[0,2]$. I'll approximate this to be $\ln(x) \approx (x-1)$, so our integral would then become:

$$\int_0^{2\pi} \Bigg(\frac{\color{red}{r^2}+\color{red}{r_0^2}-2rr_0\cos(\theta_0)}{r^2+\frac{a^4}{r_0^2}-2a^2\frac{r}{r_0}\cos(\theta_0)} -1\Bigg) d\theta_0$$, right? I'm thinking that because $x = \cos(\theta_0)$, then $x = [-1,1]$ and our radius of convergence is $x=[0,2]$, that this might not work.

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Main question is, is the series representation the right way to go?

Subsequent question is, is there anything wrong with my series representation?

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EDIT: I forgot to change the integrand to what I had originally intended for it to be. I made the correction in red.

Answer 1

For simplicity, $$ \begin{aligned} \int_0^{2\pi} \ln \left(\frac{b-\cos x}{a-\cos x}\right) d x = & \int_0^ {2\pi}\left(\int_a^b \frac{1}{y-\cos x} d y\right) d x \\ = & 2\int_a^b\left( \underbrace{ \int_0^ {\pi}\frac{d x}{y-\cos x}}_{J} \right) d y \end{aligned} $$ Noticing that when $ x\mapsto \pi-x, J= \int_0^ {\pi}\frac{d x}{y+\cos x}$ and averaging the two versions yields $$ \begin{aligned} J& =\frac{1}{2} \int_0^\pi\left(\frac{1}{y-\cos x}+\frac{1}{y+\cos x}\right) d x \\ & =2 y \int_0^{\frac{\pi}{2}}\left(\frac{1}{y^2-\cos ^2 x}\right) d x \\ & =2 y \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{y^2 \sec ^2 x-1} d x \\& =2 y \int_0^{\infty} \frac{d t}{\left(y^2-1\right)+y^2 t^2}\\&=\frac{2}{\sqrt{y^2-1}}\left[\tan ^{-1}\left(\frac{y t}{\sqrt{y^2-1}}\right)\right]_0^{\infty}\\&= \frac{\pi}{\sqrt{y^2-1}} \end{aligned} $$ Plugging back gives $$\int_0^{2\pi} \ln \left(\frac{b-\cos x}{a-\cos x}\right) d x =2\int_a^b \frac{\pi}{\sqrt{y^2-1}} d y$$ Putting $y=\cosh \theta$, we have \begin{aligned} \int_0^{2\pi} \ln \left(\frac{b-\cos x}{a-\cos x}\right) d x = & 2 \pi \int_{\cosh ^{-1} a}^{\cosh ^{-1}b} \frac{1}{\sinh \theta} \sinh \theta d \theta \\ & =2 \pi\left(\cosh ^{-1}b-\cosh ^{-1}a\right) \\ & =2 \pi\left[\ln \left(b+\sqrt{b^2-1}\right)-\ln \left(a+\sqrt{a^2-1}\right)\right] \\ & =2 \pi \ln \left(\frac{b+\sqrt{b^2-1}}{a+\sqrt{a^2-1}}\right) \end{aligned} As @Researcher R noted, replacing $a,b$ by $\frac{a}{c}$ and $\frac{b}{c}$ respectively yields generally $$\int_0^{2\pi} \ln \left(\frac{b-c\cos x}{a-c\cos x}\right) d x = 2 \pi \ln \left(\frac{b+\sqrt{b^2-c^2}}{a+\sqrt{a^2-c^2}}\right) $$

Answer 2

Hint

Consider $$I(c)=\int_0^{2\pi} \log(a-c\cos(x))\,dx \qquad \text{with}\qquad a>1$$ $$I'(c)=-\int_0^{2\pi} \frac{\cos (x)}{a-c \cos (x)}\,dx=\frac{2 \pi }{c} \left(1-\frac{a}{\sqrt{a^2-c^2}}\right)$$ $$\int_0^{2\pi} \log(a-\cos(x))\,dx =\int_0^1 I'(c)\,dc=-2 \pi \log \left(\frac{2 a}{a+\sqrt{a^2-1}}\right)$$

Do the same for the second logarithm and recombine the results in a single logarithm.

Answer 3

Putting together what I've seen across different answer, I decided to piece together my own answer using an amalgamation of different workings I've seen.

Splitting up the natural log:

$$\frac{1}{4\pi}\int_0^{2\pi}\ln\Big(\frac{a^2}{r_0^2}\frac{r^2+r_0^2-2rr_0\cos\theta_0}{r^2+\frac{a^4}{r_0^2}-2\frac{r}{r_0}a^2\cos\theta_0}\Big) = \frac{1}{4\pi}\int_0^{2\pi}\ln\Big(\frac{a_1}{a_2}\frac{b_1-c_1\cos\theta_0}{b_2-c_2\cos\theta_0}\Big)d\theta_0 \\ = \frac{1}{4\pi}\Bigg[\underbrace{\int_0^{2\pi}\Big(\ln(a_1)+\ln(b_1-c_1\cos\theta_0)\Big)d\theta_0}_{(1)}-\underbrace{\int_0^{2\pi}\Big(\ln(a_2)-\ln(b_2-c_2\cos\theta_0)\Big)d\theta_0}_{(2)}\Bigg]$$

Starting with (1) by following from Claude's setup:

$$\cases{I(b_1,c_1) = \int_0^{2\pi}\ln(b_1-c_1\cos\theta_0)d\theta_0 \\ I'(b_1,c_1) = \frac{\partial}{\partial c_1} I(b_1,c_1)} \\ \text{We'll also use: }\int_0^c I'(b_1,c_1)dc_1 = I(b_1,c_1) - I(b_1,0) \rightarrow I(b_1,c_1) = I(b_1,0) + \int_0^c I'(b_1,c_1)dc \\ \text{And we can change the order of integration if we replace $c_1$ with a dummy variable : } \\I(b_1,c_1) = I(b_1,0) + \int_0^{2\pi} \Big(\int_0^c \frac{\partial}{\partial x'} f(b_1,x',\theta_0) dx' \Big)d\theta_0 = I(b_1,0) + \int_0^c \int_0^{2\pi} f(b_1,x',\theta_0)_{x'} d\theta_0 dx'$$

Solving the double integral can be done via half-angle tan substitution:

$$I(b_1,c_1) = \int_0^c \int_{-\pi}^{\pi}\frac{\cos\theta_0}{x'\cos\theta_0-b_1}d\theta_0 dx' \\ = \int_0^c\Bigg(-\frac{b_1}{x'}\int_{-\infty}^{\infty}\frac{dt}{(b_1-x')+(b_1+x')t^2}dt + \frac{2\pi}{x'}\Bigg)dx' \\ = \int_0^c\Bigg(-\frac{b_1}{x'(b_1+x')}\int_{-\infty}^{\infty}\frac{2dt}{\frac{b_1-x'}{b_1+x'}+t^2}dt + \frac{2\pi}{x'}\Bigg)dx' \\ = \int_0^c\Bigg(-\frac{2b_1}{x'\sqrt{b_1^2-x'^2}}\int_{-\infty}^{\infty}\frac{dt}{1+\Bigg(\frac{t}{\sqrt{\frac{b_1-x'}{b_1+x'}}}\Bigg)^2}dt + \frac{2\pi}{x'}\Bigg)dx' \\ =\boxed{2\pi\int_0^c \frac{1}{x'}-\frac{b_1}{x'\sqrt{b_1^2-x'^2}}dx'}$$

where we've used polynomial division to turn $\frac{\cos\theta_0}{c_1\cos\theta_0-b_1} = \frac{1}{x'}-\frac{b_1}{x'(b_1-x'\cos\theta_0)}$ prior to integrating to make the integrand a little easier to work with.

The next step is to then integrate $2\pi\int_0^c \frac{1}{x'}-\frac{b_1}{x'\sqrt{b_1^2-x'^2}} dx'$ which can be done using simple substitution:

$$2\pi\int_0^c \frac{1}{x'^2}\Bigg(1-\frac{b_1}{\sqrt{b_1^2-x'^2}}\Bigg) x'dx' \\ u^2 = b_1^2-x^2, -udu = x'dx \\ = -2\pi\int_{b_1}^\sqrt{b_1^2-c^2} \frac{1}{(b_1^2-u^2)}\Bigg(1-\frac{b_1}{u}\Bigg) udu \\ = 2\pi\int_{b_1}^\sqrt{b_1^2-c^2} \frac{1}{(b_1+u)(b_1-u)}\Bigg(b_1-u\Bigg) udu \\ v = b_1+x', dv = dx' \\ 2\pi\int_{2b_1}^{b_1+\sqrt{b_1^2-c^2}} \frac{dv}{v} \\ = 2\pi\ln\Bigg(\frac{b_1+\sqrt{b_1^2-c^2}}{2b_1}\Bigg)$$

Now subtract off $I(b_1,0) = \int_0^{2\pi} ln(b_1) d\theta_0$

$$\boxed{I(b_1,c_1) = 2\pi\ln\Bigg(\frac{b_1+\sqrt{b_1^2-c_1^2}}{2}\Bigg)} \\ \boxed{I(b_2,c_2) = 2\pi\ln\Bigg(\frac{b_2+\sqrt{b_2^2-c_2^2}}{2}\Bigg)}$$

And combining the natural logs together:

$$\frac{1}{2}\Bigg(\frac{a_1}{a_2}\frac{\frac{b_1+\sqrt{b_1^2-c_1^2}}{2}}{\frac{b_2+\sqrt{b_2^2-c_2^2}}{2}}\Bigg)$$

Just cancel out the 2s and:

$$I(b_1,c_1,b_2,c_2) = \frac{1}{2}\ln\Bigg(\frac{a_1}{a_2}\frac{b_1+\sqrt{b_1^2-c_1^2}}{b_2+\sqrt{b_2^2-c_2^2}}\Bigg)$$

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Below is going to be a way to rewrite this result so that it becomes significantly easier for subsequent integration:

Rewriting this is in terms of $r$, $a$, and $r_0$:

$$-\frac{1}{2}\ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+\frac{a^4}{r_0^2}+\sqrt{\Big(r^2+\frac{a^4}{r_0^2}\Big)^2-4\Big(\frac{r}{r_0}a^2\Big)^2}}{r^2+r_0^2+\sqrt{(r^2+r_0^2)^2-4(rr_0)^2}}\Bigg) \\ = -\frac{1}{2}\ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+\frac{a^4}{r_0^2}+\sqrt{\Big(r^4+2r^2\frac{a^4}{r_0^2}+a^4(\frac{a}{r_0})^4\Big)-4\Big(\frac{r}{r_0}a^2\Big)^2}}{r^2+r_0^2+\sqrt{(r^4+2r^2r_0^2+r_0^4)-4(rr_0)^2}}\Bigg) \\ = -\frac{1}{2}\ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+\frac{a^4}{r_0^2}+\sqrt{r^4+a^4(\frac{a}{r_0})^4-2\Big(\frac{r}{r_0}a^2\Big)^2}}{r^2+r_0^2+\sqrt{r^4+r_0^4-2(rr_0)^2}}\Bigg) \\ =-\frac{1}{2}\ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+\frac{a^4}{r_0^2}+\sqrt{\Big(r^2-a^2(\frac{a}{r_0})^2\Big)^2}}{r^2+r_0^2+\sqrt{\Big(r^2-r_0^2\Big)^2}}\Bigg) \\ =-\frac{1}{2}\ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+\frac{a^4}{r_0^2}+|(r)^2-\Big(\frac{a^2}{r_0}\Big)^2|}{r^2+r_0^2+|(r)^2-(r_0)^2|}\Bigg)$$

$$\boxed{I(r,r_0) = -\frac{1}{2}\ln\Bigg(\frac{r_0^2}{a^2}\frac{r^2+\frac{a^4}{r_0^2}+|r^2-\Big(\frac{a^2}{r_0}\Big)^2|}{r^2+r_0^2+|r^2-(r_0)^2|}\Bigg)}$$

Answer 4

The integral vanishes per $\int_0^\pi \ln\left(1+2b\cos \theta+b^2\right)d\theta=0$

\begin{align} I =\int_0^{2\pi} \ln\frac{1-2\frac{r_0}{r}\cos\theta+ \frac{r^2_0}{r^2}}{1-2\frac{a^2}{r r_0}\cos\theta+ \frac{a^4}{r^2 r^2_0}}d\theta =0 \end{align}

Answer 5

Answer to the question:

No, replacing $\log(x)$ by $x-1$ does not work at all even if you remain in the interval of convergence, because this approximation is gross, you neglect all other terms.

Illustration:

$$\int_1^2\log(x)\,dx=\left.(x\log x-x)\right|_1^2=2\log2-1$$

while

$$\int_1^2(x-1)\,dx=\left.(\tfrac{x^2}2-x)\right|_1^2=-\frac12.$$

Answer 6

@ResearcherR I'll provide a straightforward computation by using $1$st kind Chebyshev polynomials. This'll be a complementary explanation of what @Quanto means. Note the property $$\log(1-2xz+z^2)=-2\sum_{n\geq1}\dfrac{\mathrm T_n(x)}{n}z^n,\enspace -1<x<1;\enspace |z|<1$$ In our case $x=\cos\theta_0$ and so $$\begin{aligned} \int_0^{2\pi}\log(1-2z\cos\theta_0+z^2)\,\mathrm d\theta_0&=-2\sum_{n\geq1}\int_0^{2\pi}\dfrac{\mathrm T_n(\cos\theta_0)}{n}z^n\,\mathrm d\theta_0\\ &=-2\sum_{n\geq1}\int_0^{2\pi}\dfrac{\cos(n\theta_0)}{n}z^n\,\mathrm d\theta_0=0 \end{aligned} $$

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In light of this, we can now compute your integral easily without recurring to differentiation under the integral sign. This is because your goal integral looks like $$\int_0^{2\pi}\ln\left(\dfrac{a^2+b^2-2ab\cos\theta_0}{c^2+d^2-2cd\cos\theta_0}\right)\,\mathrm d\theta_0$$ which can be separated into $$\int_0^{2\pi}\ln\left(a^2+b^2-2ab\cos\theta_0\right)\,\mathrm d\theta_0-\int_0^{2\pi}\ln\left(c^2+d^2-2cd\cos\theta_0\right)\,\mathrm d\theta_0$$ And by factoring either $a^2$ or $b^2$ in the first integral or $c^2$ or $d^2$ in the second such that $|z_1|<1$ and $|z_2|<1$ in $$\small\int_0^{2\pi}\ln(C_1)\,\mathrm d\theta_0+\int_0^{2\pi}\ln\left(1+z_1^2-2z_1\cos\theta_0\right)\,\mathrm d\theta_0-\int_0^{2\pi}\ln(C_2)\,\mathrm d\theta_0-\int_0^{2\pi}\ln\left(1+z_2^2-2z_2\cos\theta_0\right)\,\mathrm d\theta_0,$$ then it so follows that $$\int_0^{2\pi}\ln\left(\dfrac{a^2+b^2-2ab\cos\theta_0}{c^2+d^2-2cd\cos\theta_0}\right)\,\mathrm d\theta_0=2\pi\ln\left(\dfrac{C_1}{C_2}\right)$$ That being, there are $4$ situations: $\small \mathbf{1)}\ r>r_0\wedge r>\dfrac{a^2}{r_0}$, $\small \mathbf{2)}\ r>r_0\wedge r<\dfrac{a^2}{r_0}$, $\small \mathbf{3)}\ r<r_0\wedge r>\dfrac{a^2}{r_0}$ and $\small \mathbf{4)}\ r<r_0\wedge r<\dfrac{a^2}{r_0}$. We'll treat the cases where they are equal later.

In situation $\mathbf{1)}$ your goal integral is $$\small\int_0^{2\pi}\ln\left(\dfrac{r^2}{r^2}\cdot\dfrac{1+\dfrac{r_0^2}{r^2}-2\dfrac{r_0}{r}\cos\theta_0}{1+\dfrac{a^4}{r^2r_0^2}-2\dfrac{a^2}{rr_0}\cos\theta_0}\right)\,\mathrm d\theta=0$$ In $\mathbf{2)}$ $$\small\int_0^{2\pi}\ln\left(\dfrac{r^2}{\dfrac{a^4}{r_0^2}}\cdot\dfrac{1+\dfrac{r_0^2}{r^2}-2\dfrac{r_0}{r}\cos\theta_0}{1+\dfrac{r^2r_0^2}{a^4}-2\dfrac{rr_0}{a^2}\cos\theta_0}\right)\,\mathrm d\theta=4\pi\ln\left(\dfrac{rr_0}{a^2}\right)$$ In $\mathbf{3)}$ $$\small\int_0^{2\pi}\ln\left(\dfrac{r_0^2}{r^2}\cdot\dfrac{1+\dfrac{r^2}{r_0^2}-2\dfrac{r}{r_0}\cos\theta_0}{1+\dfrac{r^2r_0^2}{a^4}-2\dfrac{rr_0}{a^2}\cos\theta_0}\right)\,\mathrm d\theta=4\pi\ln\left(\dfrac{r_0}{r}\right)$$ And in $\mathbf{4)}$ $$\small\int_0^{2\pi}\ln\left(\dfrac{r_0^2}{\dfrac{a^4}{r_0^2}}\cdot\dfrac{1+\dfrac{r^2}{r_0^2}-2\dfrac{r}{r_0}\cos\theta_0}{1+\dfrac{r^2r_0^2}{a^4}-2\dfrac{rr_0}{a^2}\cos\theta_0}\right)\,\mathrm d\theta=8\pi\ln\left(\dfrac{r_0}{a}\right)$$ As for the equal cases, we cannot apply the same technique as the series expansion is not analytical when $z=1$. Your integral may then reduce to $$\small\int_0^{2\pi}\ln(C_1')\,\mathrm d\theta_0+\int_0^{2\pi}\ln\left(1-\cos\theta_0\right)\,\mathrm d\theta_0-\int_0^{2\pi}\ln(C_2)\,\mathrm d\theta_0,$$ $$\small\int_0^{2\pi}\ln(C_1)\,\mathrm d\theta_0-\int_0^{2\pi}\ln(C_2')\,\mathrm d\theta_0-\int_0^{2\pi}\ln\left(1-\cos\theta_0\right)\,\mathrm d\theta_0,$$ $$\text{or }\small\int_0^{2\pi}\ln(C_1')\,\mathrm d\theta_0-\int_0^{2\pi}\ln(C_2')\,\mathrm d\theta_0.$$ What's clear is that now we need to compute $$\int_0^{2\pi}\ln\left(1-\cos\theta_0\right)\,\mathrm d\theta_0$$ without employing that series expansion. Well, it turns out it isn't too cumbersome since $$\begin{aligned} \small\int_0^{2\pi}\ln\left(1-\cos\theta_0\right)\,\mathrm d\theta_0&\small=\int_0^{2\pi}\ln\left(2\sin^2\left(\frac{\theta_0}{2}\right)\right)\,\mathrm d\theta_0\\ &\small=2\pi\ln(2)+4\int_0^{\pi}\ln(\sin(\theta_0))\,\mathrm d\theta_0\\ &\small=2\pi\ln(2)+4\int_0^{\pi}\left(-\ln(2)-\sum_{k\geq 1}\dfrac{\cos(2k\theta_0)}{k}\right)\,\mathrm d\theta_0\\ &\small=-2\pi\ln(2) \end{aligned}$$ With that done let's enumerate the rest of possibilities: $\small \mathbf{5)}\ r=r_0\wedge r>\dfrac{a^2}{r_0}$, $\small \mathbf{6)}\ r=r_0\wedge r<\dfrac{a^2}{r_0}$, $\small \mathbf{7)}\ r>r_0\wedge r=\dfrac{a^2}{r_0}$ and $\small \mathbf{8)}\ r<r_0\wedge r=\dfrac{a^2}{r_0}$ and $\small \mathbf{9)}\ r=r_0\wedge r=\dfrac{a^2}{r_0}$.

For $\mathbf{5)}$ we've got $$\small\int_0^{2\pi}\ln\left(\dfrac{2r^2}{r^2}\cdot\dfrac{1-\cos\theta_0}{1+\dfrac{a^4}{r^2r_0^2}-2\dfrac{a^2}{rr_0}\cos\theta_0}\right)\,\mathrm d\theta=0$$ For $\mathbf{6)}$ $$\small\int_0^{2\pi}\ln\left(\dfrac{2r_0^2}{\dfrac{a^4}{r_0^2}}\cdot\dfrac{1-\cos\theta_0}{1+\dfrac{r^2r_0^2}{a^4}-2\dfrac{rr_0}{a^2}\cos\theta_0}\right)\,\mathrm d\theta=8\pi\ln\left(\dfrac{r_0}{a}\right)$$ For $\mathbf{7)}$ $$\small\int_0^{2\pi}\ln\left(\dfrac{r^2}{2r^2}\cdot\dfrac{1+\dfrac{r_0^2}{r^2}-2\dfrac{r_0}{r}\cos\theta_0}{1-\cos\theta_0}\right)\,\mathrm d\theta=0$$ For $\mathbf{8)}$ $$\small\int_0^{2\pi}\ln\left(\dfrac{r_0^2}{2\dfrac{a^4}{r_0^2}}\cdot\dfrac{1+\dfrac{r^2}{r_0^2}-2\dfrac{r}{r_0}\cos\theta_0}{1-\cos\theta_0}\right)\,\mathrm d\theta=8\pi\ln\left(\dfrac{r_0}{a}\right)$$ And finally for $\mathbf{9)}$, $$\small\int_0^{2\pi}\ln\left(\dfrac{2r_0^2}{2r_0^2}\cdot\dfrac{1-\cos\theta_0}{1-\cos\theta_0}\right)\,\mathrm d\theta=0$$ To sum up, $$\scriptsize\int_0^{2\pi}\ln\left(\dfrac{r^2+r_0^2-2rr_0\cos\theta_0}{r^2+\dfrac{a^4}{r_0^2}-2a^2\dfrac{r}{r_0}\cos\theta_0}\right)\,\mathrm d\theta_0= \left\{\begin{aligned} & r>r_0: \left[\begin{aligned} & 0,\enspace r>\dfrac{a^2}{r_0}\\ & 0,\enspace r=\dfrac{a^2}{r_0}\\ & 4\pi\ln\left(\dfrac{rr_0}{a^2}\right),\enspace r<\dfrac{a^2}{r_0}\\ \end{aligned}\right. \\ & r=r_0: \left[\begin{aligned} & 0,\enspace r>\dfrac{a^2}{r_0}\\ & 0,\enspace r=\dfrac{a^2}{r_0}\\ & 8\pi\ln\left(\dfrac{r_0}{a}\right),\enspace r<\dfrac{a^2}{r_0}\\ \end{aligned}\right. \\ & r<r_0: \left[\begin{aligned} & 4\pi\ln\left(\dfrac{r_0}{r}\right),\enspace r>\dfrac{a^2}{r_0}\\ & 8\pi\ln\left(\dfrac{r_0}{a}\right),\enspace r=\dfrac{a^2}{r_0}\\ & 8\pi\ln\left(\dfrac{r_0}{a}\right),\enspace r<\dfrac{a^2}{r_0}\\ \end{aligned}\right. \end{aligned}\right\} =4\pi\max\{\ln(r),\ln(r_0)\}-4\pi\max\{\ln(r),\ln\left(\frac{a^2}{r_0}\right)\} $$