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Today I was discussing with a classmate about the sign of the integral

$$\int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy,$$ where $B(0,r)$ denotes the ball of center $0$ and radius $r$ in $\mathbb{R^2}$. My friend said that this integral is negative for every $r>0$ because the function $\ln |x-y|$ is "very negative" at $x=y$. I don't agree and I told him that I think there exists a critical $r$ from which the integral is positive. However, I don't know how to prove it. I rewrite it by using polar coordinates as $$\int_0^r\int_0^r\int_0^{2\pi}\int_0^{2\pi} \ln(\sqrt{r_1^2+r_2^2-2r_1r_2\cos(t_1-t_2)})r_1r_2dt_1dt_2dr_1dr_2.$$

I computed this integral with the software Mathematica and I obtained positive values with, for example, $r=2$. However, this proof is not valid for my friend. Does anyone know how to prove it rigorously?

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  • $\begingroup$ The double integral in the first expression and the quadruple integral in the second one don't make sense. If you're integrating on $\mathbb{R^2}$ you only need one and two integrals (respectively) in those expressions. $\endgroup$
    – Sam
    Commented Feb 25, 2022 at 11:26
  • $\begingroup$ In the first integral $x\in B(0,r)$ and $y\in B(0,r)$. I am integrating two times in $\mathbb{R^2}$ $\endgroup$
    – mejopa
    Commented Feb 25, 2022 at 11:29
  • $\begingroup$ Ahh....so $|x-y|$ is the $\mathbb{R^2}$ norm, gotcha. $\endgroup$
    – Sam
    Commented Feb 25, 2022 at 11:35
  • $\begingroup$ Yes, I edited that to avoid confusions again! $\endgroup$
    – mejopa
    Commented Feb 25, 2022 at 11:39

2 Answers 2

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First note that \begin{align} \int \limits_0^{2\pi} \int \limits_0^{2\pi} f(\cos(t_1 - t_2)) \, \mathrm{d} t_1 \mathrm{d} t_2 &= \int \limits_0^{2\pi} \int \limits_{-t_2}^{2\pi-t_2} f(\cos(s)) \, \mathrm{d} s \, \mathrm{d} t_2 = \int \limits_0^{2\pi} \int \limits_0^{2\pi} f(\cos(s)) \, \mathrm{d} s \, \mathrm{d} t_2 \\ &= 2 \pi \int \limits_0^{2\pi} f(\cos(s)) \, \mathrm{d} s = 4 \pi \int \limits_0^\pi f(\cos(s)) \, \mathrm{d} s \end{align} holds for any $f$ for which the integral exists. This allows us to rewrite your integral as \begin{align} I (r) &\equiv \int \limits_{\mathrm{B}(0,r)} ~ \int \limits_{\mathrm{B}(0,r)} \ln(\lVert x - y \rVert) \, \mathrm{d} x \, \mathrm{d} y = 2 \pi \int \limits_0^r \int \limits_0^r \int \limits_0^{\pi} \ln(r_1^2 + r_2^2 - 2 r_1 r_2 \cos(s)) r_1 r_2 \, \mathrm{d} s \, \mathrm{d} r_1 \mathrm{d} r_2 \\ &= 2 \pi \int \limits_0^r \int \limits_0^r \int \limits_0^{\pi} \left[2\ln(\max(r_1,r_2)) \vphantom{\frac{\min^2(r_1,r_2)}{\max^2(r_1,r_2)}} \right. \\ &\phantom{{} = 2 \pi \int \limits_0^r \int \limits_0^r \int \limits_0^{\pi} \left[\vphantom{\frac{\min^2(r_1,r_2)}{\max^2(r_1,r_2)}} \right.} \left. + \ln \left(1 + \frac{\min^2(r_1,r_2)}{\max^2(r_1,r_2)} - 2 \frac{\min(r_1,r_2)}{\max(r_1,r_2)} \cos(s) \right)\right] r_1 r_2 \, \mathrm{d} s \, \mathrm{d} r_1 \mathrm{d} r_2 \, . \end{align} The $s$-integral of the first term in the square brackets is trivial, the integral of the second term is zero (see this question). Therefore, \begin{align} I (r) &= 4 \pi^2 \int \limits_0^r \int \limits_0^r \ln(\max(r_1,r_2)) r_1 r_2 \, \mathrm{d} r_1 \mathrm{d} r_2 = 8 \pi^2 \int \limits_0^r \int \limits_0^{r_2} \ln(r_2) r_1 r_2 \, \mathrm{d} r_1 \mathrm{d} r_2\\ &= 4\pi^2 \int \limits_0^r \ln(r_2) r_2^3 \, \mathrm{d}r_2 = \pi^2 \left(r^4 \ln(r) - \int \limits_0^r r_2^3 \, \mathrm{d}r_2\right) = \pi^2 r^4 \left[\ln(r) - \frac{1}{4}\right] . \end{align} There is indeed a critical value $r_\text{c} = \mathrm{e}^{1/4}$ such that $I(r) < 0$ for $r < r_\text{c}$ and $I(r) > 0$ for $r > r_\text{c}$.

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  • $\begingroup$ Thanks for your answer. When you use polar coordinates, why don't you make the square root of the argument of $\ln$? Are your considering the norm $\|\|^2$ for any reason? $\endgroup$
    – mejopa
    Commented Feb 26, 2022 at 9:49
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    $\begingroup$ @mejopa Using basic properties of logarithms, $\ln{(\sqrt{x})}=\frac12\ln{(x)}$. $\endgroup$
    – David H
    Commented Feb 26, 2022 at 11:20
  • $\begingroup$ Of course! Sorry $\endgroup$
    – mejopa
    Commented Feb 26, 2022 at 12:04
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Let $A = \int_{B(0,1)} \ln(\|x\|)dx$ and choose $r$ so that $\pi r^2 \gt 4\pi + \frac{|A|}{\ln 2}$.

For any fixed $y \in B(0,r)$, write $$B_1= \{x \in B(0,r):\|x-y\|\le 1 \}$$ and $$B_2= \{x \in B(0,r):\|x-y\|\ge 2 \}$$ Then we have:

$$\int_{B(0,r)} \ln(\|x-y\|)dx \ge \int_{B_1} \ln(\|x-y\|)dx + \int_{B_2} \ln(\|x-y\|)dx $$ $$\ge \int_{B(0,1)} \ln(\|x|)dx + \int_{B_2}(\ln 2) dx$$ $$\ge A + (\pi r^2 - 4\pi)(\ln 2) \gt 0$$

With $C = A + (\pi r^2 - 4\pi)(\ln 2)$, this implies: $$\int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy \gt \int_{B(0,r)} C dy = \pi r^2 C \gt 0$$

Now observe that the function $$f(r) = \int_{B(0,r)} \int_{B(0,r)} \ln(\|x-y\|)dxdy$$

is continuous, $f(1) \lt 0$ and, by the above calculations, $f(r) \gt 0$ for large enough $r$, so there is a minimum value $r_0$ where $f(r_0) = 0$ and it's easy to see that $f(r) \gt 0$ for all $r \gt r_0$

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  • $\begingroup$ Thanks for your answer! Really helpful $\endgroup$
    – mejopa
    Commented Feb 26, 2022 at 17:29

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