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Good morning.

I came across the following integral in some field theory calculation:

$\int_0^\pi dx\,\log\left(a^2+b^2-2ab\cos x\right)=2\pi\log\left(\max\lbrace a,b\rbrace\right)$

for $0<a,b\in\mathbb{R}$. The notation has been adapated to make contact with Bronstein's integral 21.42.

Despite 6 pp.of calculations, I was not able to figure out a way to prove the identity (integration by parts, substitution, derivative trick, contour integration...). Does anyone have a proof?

Best regards, David

miracle173
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David Heider
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  • Seems to be tricky: Wolfram computes the integral, but the result does not look very pretty... – mkk030572 Apr 30 '16 at 06:23
  • Another reference where the integral occurs is Lamb - Hydrodynamics §163 - evaluation of self-propulsion velocity of inviscid axisymmetric vortex rings. There, the result given in the question is also stated without proof. – David Heider Apr 30 '16 at 06:27
  • This is something that I used to know, but the proof of it is escaping me at the moment. The result is sound, suppose a >> b, you collect a^2 which gives you 2 \pi log a + a "residual integral". Now you have to prove that the residual integral is really zero... I 'll continue to think about it. – Francesco Apr 30 '16 at 07:31
  • @mkk030572, at least we now know that that huge expression is equal to $2\pi\log(max{a,b})$ – Hasan Saad Apr 30 '16 at 08:28

2 Answers2

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Suppose WLOG $b > a$. Then with $c = b/a$ we have

$$\begin{align} I &= \int_0^\pi \log(a^2 + b^2 - 2ab \cos x) \, dx \\ &=\int_0^\pi \log a^2 \, dx + \int_0^\pi \log(1 + (b/a)^2 - 2(b/a) \cos x) \, dx \\ &= 2\pi \log a + \int_0^\pi \log(1 + c^2 - 2c \cos x) \, dx \end{align}$$

Now we can evaluate the second integral on the RHS as the limit of a Riemann sum

$$\begin{align}J &= \int_0^\pi \log(1 + c^2 - 2c \cos x) \, dx \\ &= \lim_{n \to \infty} \frac{\pi}{n}\sum_{j=1}^n\log(1 + c^2 - 2c\cos(\pi(j-1)/n)) \\ &= \lim_{n \to \infty} \frac{\pi}{n}\log\prod_{j=1}^n(1 + c^2 - 2c\cos(\pi(j-1)/n)) \\ &=\lim_{n \to \infty} \frac{\pi}{n}\log(1-c)^2+\lim_{n \to \infty} \frac{\pi}{n}\log\prod_{j=2}^{n}(1 + c^2 - 2c\cos(\pi(j-1)/n)) \\ &= \lim_{n \to \infty} \frac{\pi}{n}\log\prod_{j=1}^{n-1}(1 + c^2 - 2c\cos(\pi j/n))\end{align} .$$

Upon factoring, we obtain

$$1 + c^2 - 2c\cos(\pi j/n) = [c - \exp(i \pi j /n)][c - \exp(-i \pi j /n)],$$

and

$$c^{2n} -1 = (c-1)(c+1)\prod_{j=1}^{n-1}[c - \exp(i \pi j /n)][c - \exp(-i \pi j /n)].$$

Hence, recalling that $c = b/a > 1$, it follows that

$$\begin{align}J &= \lim_{n \to \infty} \frac{\pi}{n} \log\left(\frac{c^{2n}-1}{c^2-1} \right) \\ &= \pi \lim_{n \to \infty} \log\left(\frac{c^{2n}-1}{c^2-1} \right)^{1/n} \\ &= \pi \lim_{n \to \infty} \log\left[c^2 \left(\frac{1-c^{-2n}}{c^2-1} \right)^{1/n}\right] \\ &= \pi \log c^2 + \log \left[\lim_{n \to \infty} \left(\frac{1-c^{-2n}}{c^2-1} \right)^{1/n}\right]\\ &= \pi \log c^2\end{align}$$

Thus,

$$I = 2\pi \log a + \pi \log c^2 = 2\pi \log a + \pi \log (b/a)^2 \\ = 2\pi \log b $$

RRL
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  • Rechecking. The answer is right, expecpt the 4th line in the evaluation of J. i think the first term must be $\log(1-c)^2$ instead of $\log(1-c^2)$ because $\cos(\pi(j-1)/n)=\cos 0 = 1$ for $j=1$. In the limit $n\to\infty$ the contribution kicks nontheless. Thanks for the detailed solution! – David Heider Apr 30 '16 at 08:22
  • @DavidHeider: You're welcome. You are correct -- I'll correct it. – RRL Apr 30 '16 at 08:24
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Suppose that $b>a$. Then $$ \int_{0}^{\pi}\log(a^{2}+b^{2}-2ab\cos x)\, dx = 2\pi\log b + \int_{0}^{\pi}\log(c^{2}+ 1-2c\cos x)\, dx $$ where $c = \dfrac{a}{b}$ and $0 \le c < 1$.

We will prove that $$ I = \int_{0}^{\pi}\log(c^{2}+ 1-2c\cos x)\, dx = 0. $$ If $\log$ is the principal branch then $f(z) = \dfrac{\log(1-z)}{z}$ is an analytic function in |z| <1. Consequently $$ 0 = \oint_{|z|=c}f(z)\, dz = \int_{-\pi}^{\pi}\dfrac{\log(1-ce^{it})}{ce^{it}}ice^{it}\, dt = i\int_{-\pi}^{\pi}\log(1-ce^{it})\, dt $$ The imaginary part of the right hand side is $$ 0 = \int_{-\pi}^{\pi}\log|1-ce^{it}|\, dt = \int_{-\pi}^{\pi}\log(\sqrt{c^{2}+1-2c\cos t})\, dt = \int_{0}^{\pi}\log(c^{2}+1-2c\cos t)\, dt = I. $$

JanG
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