Frullani version of the classic $\log \left( 1 + 2\alpha \cos px + \alpha^2\right)$ integral.

Question

I read in a paper about Frullani integrals the following claim $$ \begin{align*} I & :=\int_0^\infty \frac{1}{x}\log\left(\frac{1 + 2\alpha \cos px + \alpha^2}{1 + 2\alpha \cos qx + \alpha^2}\right)\,\mathrm{d}x = \left\{ \begin{array}{lrr} \log \frac{p}{q} \log\left( 1 + \alpha^2\right) & \text{if} & \alpha^2>1\\ \log \frac{p}{q} \log\left( 1 + \frac{1}{\alpha}\right) & \text{if} & \alpha^2<1 \end{array} \right. \end{align*} $$ Where the author points at the dangers of using this method. The problem lies in the fact that the function $f(n,r) := \log(1+2n \cos rx + n^2)$ does not have a limit as $x\to \infty$. And hence the Frullani identity can not be used.

A very similar integral $f(n,-r)$ has been evaluated several times before. See for an example A question in Complex Analysis $\int_0^{2\pi}\log(1-2r\cos x +r^2)\,dx$ and can also be solved by differentiating under the integral sign . Even $f(n,-r)^2$ has been evaluated.

My question is: Can any of this be used to prove the claim above?

Answer 1

I think there was a typo in that paper you read.

Entry 4.324.2 in Gradshteyn and Ryzhik says that the value of $I$ is $ \log(1+\alpha) \log \left(\frac{q^{2}}{p^{2}} \right)$ when $-1 <\alpha \le 1$, and $\log \left(1+ \frac{1}{\alpha} \right)\log \left(\frac{q^{2}}{p^{2}} \right) $ when $\alpha^{2} >1$.

I will show the first case.

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Let $$I(s) = \int_{0}^{\infty}x^{s-1} \log \left(\frac{1+2\alpha \cos(px)+\alpha^{2}}{1+2 \alpha \cos(qx) + \alpha^{2}} \right) \, \mathrm dx,$$ where $-2 < s < 0$, $\alpha^{2} <1$, and $p$ and $q$ are positive real numbers.

The integral also converges for $0 \le s < 1$, but those values are purposely not included so that we can use Fubini's theorem to justify interchanging the order of summation and integration below.

It then follows from a property of the Mellin transform that $$I = \int_{0}^{\infty} \frac{1}{x} \, \log \left(\frac{1+2\alpha \cos(px)+\alpha^{2}}{1+2 \alpha \cos(qx) + \alpha^{2}} \right) \, \mathrm dx = \lim_{s \to 0^{-}} I(s).$$

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Using the Fourier series $$\log(1+2\alpha \cos (\theta) + \alpha^{2}) = -2 \sum_{k=1} \frac{(-\alpha)^{k}\cos(k \theta)}{k}, \quad \alpha^{2}<1,$$

and the Mellin transform $$\int_{0}^{\infty}x^{s-1} \left( \cos(px) - \cos(qx) \right)\, \mathrm dx = \Gamma(s) \cos \left(\frac{\pi s}{2} \right) \left(p^{-s}-q^{-s} \right), \quad -2 <s <1, $$ we have

$$ \begin{align} I(s) &= -2 \int_{0}^{\infty} x^{s-1}\sum_{k=1}^{\infty} \left( \frac{(-\alpha)^{k} \cos(kpx)}{k} - \frac{(-\alpha)^{k}\cos(kqx)}{k} \right)\, \mathrm dx \\ &= -2 \sum_{k=1}^{\infty} \frac{(-\alpha)^{k}}{k} \int_{0}^{\infty} x^{s-1} \left(\cos(kpx)-\cos(kqx) \right) \, \mathrm dx \\ &= - 2 \sum_{k=1}^{\infty} \frac{(-\alpha)^k}{k} \Gamma(s) \cos \left(\frac{\pi s}{2} \right)\left((kp)^{-s}-(kq)^{-s} \right) \\ &=-2 \, \Gamma(s) \cos \left(\frac{\pi s}{2} \right) \left(p^{-s}-q^{-s} \right) \sum_{k=1}^{\infty} \frac{(-\alpha)^{k}}{k^{s+1}} \\ &= 2 \, \Gamma(s) \cos \left(\frac{\pi s}{2} \right) \left(q^{-s}-p^{-s} \right) \operatorname{Li}_{s+1}(-\alpha). \end{align}$$

Therefore, $$ \begin{align} I &= \lim_{s \to 0^{-}} I(s) \\ &= \lim_{s \to 0^{-}} 2 \left(\frac{1}{s} + \mathcal{O}(s) \right)\cos \left(\frac{\pi s}{2} \right) \left(s (\log p -\log q) + \mathcal{O}(s^{2}) \right)\operatorname{Li}_{s+1}(-\alpha) \\ &= 2( \log p - \log q) \operatorname{Li}_{1}(-\alpha) \\ &= -2 ( \log p - \log q) \log(1+\alpha) \\ &= \log(1+\alpha) \log \left(\frac{q^{2}}{p^{2}} \right). \end{align}$$

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UPDATE:

If $\alpha^{2} <1$, then $$\int_{0}^{n \pi/p} \log \left(1+2 \alpha \cos(px)+ \alpha^{2}\right) =0, \quad n \in \mathbb{N}_{\ge 0},$$ which means that $I$ converges by Dirichlet's test.

The case $\alpha^{2}>1$ comes from replacing $\alpha$ with $\frac{1}{\alpha}$ in the Fourier series.

Answer 2

Untilize the Frullani-like integral $\int_0^\infty \frac{\cos px -\cos qx}xdx =\ln\frac qp$ to obtain \begin{align} I_{a^2<1}= &\int_{0}^{\infty} \frac{1}{x} \, \ln \frac{1+2a\cos px+a^{2}}{1+2 a\cos qx + a^{2}} \ dx\\ =&\> 2 \int_{0}^{\infty} \sum_{k=1}^\infty \frac{(-a)^{k+1}}{k}\frac{\cos(k px)-\cos(k qx)}x \ dx\\ = &\>2\ln \frac qp \sum_{k=1}^\infty \frac{(-a)^{k+1}}{k} = 2\ln \frac qp \ \ln(1+a) \\ \\ I_{a^2>1}=&\int_{0}^{\infty} \frac{1}{x} \, \ln \frac{1+\frac2a\cos px+\frac1{a^2}}{1+\frac2a\cos qx+\frac1{a^2}} \ dx = 2\ln \frac qp \ \ln(1+\frac1a) \end{align}