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Consider the integral $$ f(\alpha,\beta)= \int_0^{2\pi}\,dx \sqrt{1- \cos(\alpha x ) \cos(\beta x)}$$ as a function of the two parameters $\alpha,\beta$. I am interested in the asymptotic behavior for $\alpha, \beta \gg 1$.

For $\alpha = \beta$ the integral can be evaluated explicitly with the result $$ f(\alpha , \alpha) = \frac{2}{\alpha} \left[ \lfloor 2 \alpha\rfloor + \sin^2 \left(\frac\pi2 \{ 2 \alpha \} \right) \right]$$ with $\{ x \} = x - \lfloor x \rfloor$. For large $\alpha$ the function $f(\alpha, \alpha)$ thus approaches $4$.

If we see what happens if we keep $\beta$ large but fixed and vary $\alpha$, we see that $\beta \approx \alpha$ with $f(\alpha,\beta) \approx 4$ looks like a minimum of the function and it quickly approaches $2\pi$ (at least for $\alpha$ large) for $\beta$ sufficiently different from $\alpha$. However, there are oscillations on top of the mean value $2\pi$. In the figure you see a numerical evaluation of the integral for $\beta=20$ and $\alpha$ between 0 and 40.

numerical evaluation of f as a function of alpha with beta=20 fixed

  • Is the value of $f(\alpha,\beta)$ for $\alpha,\beta \gg 1$ and $\alpha \neq \beta$ indeed $2\pi$?
  • Why the value $f=4$ for $\alpha \approx \beta$ is lower than the generic value $2\pi$ (for $\alpha$, $\beta$ large)?
  • What is the period of the (fast) oscillations as a function of $\alpha$ with $\beta$ fixed which are visible in the plot?
  • What is the shape of the envelope? (it is a peaked function -> Lorentzian, or Gaussian, or ...?)
  • Does anybody know how to obtain a good asymptotic expression for $f(\alpha, \beta)$?
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Regarding your second question: when $\alpha=\beta$, the argument $1-\cos^2(\alpha x)$ of the square root is always at most 1, while it sometimes exceeds 1 when $\alpha\ne\beta$. To me that explains why the special value is less than the average value. – Greg Martin Mar 26 '12 at 7:56
up vote 5 down vote accepted

Set $\alpha-\beta = r$ and $\alpha+\beta =s$. So our integral is $$f(r,s) := \int_0^{2 \pi} \sqrt{1-(\cos (rx) + \cos (sx))/2} dx.$$ It looks like, for $r$ constant and $s$ large, a pretty good approximation is just to ignore the $s$ term. Set $$g(r) := \int_0^{2 \pi} \sqrt{1-\cos (rx)/2} dx.$$ The figure below shows $f$ (in red) and $g$ (in blue) for $s=40$. enter image description here

An even better approximation seems to be using $3$ terms of the Taylor series for the square root: $$\sqrt{1-\cos(rx)/2 - \cos(sx)/2} \approx$$ $$ \left(1-\cos(rx)\right)^{1/2} - \frac{\cos{sx}}{4} \left(1-\cos(rx)\right)^{-1/2} - \frac{\cos^2{sx}}{32} \left(1-\cos(rx)\right)^{-3/2}=$$ $$\left(1-\cos(rx)\right)^{1/2} - \frac{1}{64} \left(1-\cos(rx)\right)^{-3/2} + \cos(sx) (\mbox{something}) + \cos(2x)(\mbox{something}).$$ Here I have replaced $\cos^2(sx)$ by $(\cos(2sx)+1)/2$.

So $$f(r,s) \approx \int_0^{2 \pi} \left( \left(1-\cos(rx)\right)^{1/2} - \frac{1}{64} \left(1-\cos(rx)\right)^{-3/2} \right) dx +$$ $$\int_0^{2 \pi} \cos(sx) (\mbox{something}) dx + \int_0^{2 \pi} \cos(2sx) (\mbox{something}) dx.$$ The two terms below the line break will go to zero as $s \to \infty$, by the Riemann-Lebesgue lemma. (As a general rule of thumb, anytime that you have a highly oscillatory integral, try to use Riemann-Lebesgue.)

Set $$h(r) := \int_0^{2 \pi} \left( \left(1-\cos(rx)\right)^{1/2} - \frac{1}{64} \left(1-\cos(rx)\right)^{-3/2} \right) dx.$$ Here is the above plot with $h$ added on (in green) enter image description here

Here is a plot of $f(r,40)-h(r)$ (note the small vertical range). I wouldn't trust this data too much -- the jaggedness is often a sign that we are getting close to Mathematica's numerical tolerance: enter image description here

I suspect that one should be able to show that there exists a function $F(y)$ such that $$\lim_{s \to \infty} f(r,s) = \int_{0}^{2 \pi} F(1-\cos(rx)/2) dx,$$ given by a convergent power series which starts $F(y) = y^{1/2} - y^{-3/2}/64+\cdots$.

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Does this mean that the limit $\lim_{t\to\infty}[\lim_{s\to\infty} f(r,s)]$ does not exist? – Fabian Mar 29 '12 at 14:47
I assume that $t$ is meant to be $r$. I don't know, but I would guess that that limit DOES exist. Looking at $g(r)=\int_{0}^{2 \pi} \sqrt{1-\cos(rx)/2} dx$, we have $g(r)=2 \pi \frac{1}{2 \pi r} \int_0^{2 \pi r} \sqrt{1 - \cos(u)/2}$ so $\lim_{r \to \infty} g(r)$ is $2 \pi$ times the average value of $\sqrt{1-\cos(u)/2}$, which exists. I would guess that $f(r,s)$ behaves the same way. – David Speyer Mar 29 '12 at 15:17
According to Mathematica, the limit thus would be $2\sqrt{2} E(i \sqrt{2}) \approx 6.178$ which is different from @joriki's answer. – Fabian Mar 29 '12 at 16:53
Sorry, I was probably unclear. I suspect that $\lim_{s \to \infty} f(r,s)$ exists; call that limit $\ell(r)$. It looks like $g$ is a pretty good approximation to that limit and that $h$ is a better one. I don't think that either of them are equal to $\ell$, or even that $\lim_{r \to \infty} g(r)$ is equal to $\lim_{r \to \infty} \ell(r)$. – David Speyer Mar 29 '12 at 18:11
@Fabian: I don't think that "the" limit exists. It depends very much on how you go to $\infty$. My answer shows, that for $\alpha=r\beta$ there are solutions depending on $r$. Davids solution shows, that for $\alpha = \beta + c$ there are different solutions depending on $c$. You could think of other "paths" to $\infty$ like $\alpha=c \beta^2$ and they will probably result in other answers (most of which will at least be close to $2\pi \langle\sqrt{1-\cos(x)\cos(y)}\rangle_{x,y}$). – example Mar 30 '12 at 17:04

Greg has already indicated in a comment why the value for $\alpha=\beta$ is lower. In this case the phases of the two cosines are maximally correlated and their product is non-negative; in fact the integrand simplifies to $|\sin\alpha x|$ in this case. For $\alpha,\beta,|\alpha-\beta|\gg1$, on the other hand, the phases of the cosines are approximately uncorrelated (in fact, for incommensurable $\alpha,\beta$ they would come arbitrarily close to every pair of phases if we extend the integral to infinity). I don't think the asymptotic value is $2\pi$; it should be $2\pi$ times the average of $\sqrt{1-\cos x\cos y}$ over full periods of $x$ and $y$. This is approximately $6.01987$, which seems to agree with your image.

Regarding the frequency of the oscillations, it looks like it's simply $1$, which would make sense, since we add one full period of $\cos\alpha x$ when we increase $\alpha$ by $1$. The integrand can be written as $\sqrt{1-(\cos(\alpha+\beta)x+\cos(\alpha-\beta)x)/2}$, and both of these cosines are at their maximum at $2\pi$ when $\alpha$ is an integer (because you've chosen $\beta$ as an integer). On that basis, my guess for the shape of the envelope would be $I\pm c/|\alpha-\beta|$, where $I$ is the average value and $c$ some constant.

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Do you expect a difference between $\alpha/\beta \in \mathbb{Q}$ and $\alpha/\beta \notin \mathbb{Q}$? – Fabian Mar 26 '12 at 17:33
@Fabian: No, that was just a side comment; the integral should depend continously on $\alpha$ and $\beta$, so there can't be such a difference. – joriki Mar 27 '12 at 2:13
@Fabian: User example makes a valid point about the limit for fixed $\alpha/\beta$; my comment on the integral depending continuously on $\alpha$ and $\beta$ doesn't preclude that limit from behaving differently for rational or irrational $\alpha/\beta$. – joriki Mar 28 '12 at 14:23
I wish a bounty could be shared. Your answer helped a lot but David's answer was more useful for me... – Fabian Apr 2 '12 at 14:54

Let's look at some special cases:

  1. b constant, $a\rightarrow\infty$: The limit is roughly $6.01987$. As $a$ becomes large, the number of different phases between the two cosines increases so that it approaches the average of $\sqrt{1-\cos(x)\cos(y)}$.

  2. $a=b\rightarrow\infty$: You already concluded correctly, that this converges to 4.

  3. (this one is almost visible in your plot - if it were of a higher resolution) $a=n\cdot b$ with $b\rightarrow\infty$ and $n \in \mathbb{N}$. These have unique values that differ from $6.01987...$.

    Eg. $n=2$ converges to $f(a,2a)=\frac{4\sqrt{10}}{3}\approx 4.22$

    $n=3$ converges to $f(a,3a)=2\sqrt{5}+\text{ArSinh}(2)\approx 5.91577$

It appears to me, that the right way to go to $\infty$ is by fixing a ratio $a/b=:r$. Once this ratio is set, any increase in a (and thus b) can be neglected after reaching the least common multiple (if this LCM is $c$ then clearly $f(c,rc)=f(2c,2rc)=f(3c,3rc)=\dots$ ). Such numbers have a finite limit value $\neq 6.0198...$ .

Any two numbers with $\text{LCM}(a,b)=\infty\,\Leftrightarrow\,r\notin\mathbb{Q}$ converge to the abovementioned $6.01987...$ by the same argument. To see this lets define a new function $$ f'(a,b,\phi) = \int_0^{2\pi}\!\!\!\!\!\sqrt{1-\cos(a x+\phi)\cos(b x)}\,\text{d}x $$ $$ \Rightarrow f'(a+1,r(a+1),\phi) = \frac{a}{a+1}f'(a,ra,\phi) + \frac{1}{a+1}f'(1,r,\phi'(\phi,r,a))$$ for fitting $\phi'$. The series of all $\phi,\phi',\phi'',\dots$ repeats for $r\in\mathbb{Q}$ after $c$ terms, so that the resulting limit is just equal to the average of these $c$ (different) integrals over one period of $\cos(a x)$. The larger the LCM is, the closer we come to the actual average of $6.01987...$, especially for $r\notin\mathbb{Q}$ the limit is identical to this number.

Sidenote: The limit need not depend continuously on $a/b$ even though $f(a,b)$ is continuous in both arguments.

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You posted this answer after my answer, but you didn't comment on why you disagree with it. Why do you think that my argument against $2\pi$ is incorrect; and does it seem to you in the image that the asymptotic value is closer to $2\pi$ than to $6.01987$? I don't find your "on average so to say" argument for $2\pi$ convincing; the average of a function is generally not the function value at the average argument. – joriki Mar 28 '12 at 12:35
Ok, sorry, I actually missed that (and misunderstood that point in your answer). When I considered this I just approximated $\sqrt{1-x}\approx 1-0.5 x+\dots$ which would have made the average of the function roughly the function value at the average argument. My point about the difference of $a/b\in\mathbb{Q}$ vs. $a/b\notin\mathbb{Q}$ remains valid though. One should just add, that the limit in the latter case is not $2\pi$ but rather 6.01987... – example Mar 28 '12 at 12:52
I only got notified of your comment because my comment was the only one so far under this answer; usually you have to use the @username idiom to ping someone if you want them to get notified of a comment that isn't under one of their posts. It's not clear to me what you mean by "The limit value need not depend continuously on $a$ and $b$" -- if $a$ and $b$ are still free to vary, then what limit is this? – joriki Mar 28 '12 at 13:07
@joriki I just edited my answer to give the correct average in the case of $a/b\notin\mathbb{Q}$. It now looks much more like your answer, sry again that I didn't see that earlier (or I might just have commented you answer instead of writing a new one) When talking about the difference between $a/b\in\mathbb{Q}$ or $\notin\mathbb{Q}$ you argued with continuity (comments under your answer). This is correct for finite $a$ and $b$ but not as you take the limit $\rightarrow\infty$. In fact I calculated some examples of function values for ratios $a/b$ that do not converge to anything near 6.01987 – example Mar 28 '12 at 13:35
I didn't (or at least didn't intend to) claim that the limit for fixed ratio $\alpha/\beta$ (not sure why you renamed them $a$ and $b$) is always $6.01987$. This is just the limit as $\alpha\to\infty$ for fixed $\beta$, and it seems we agree on that. Regarding the continuity argument, I still don't understand what you mean because you didn't answer the question in my last comment: What do you mean by "The limit value need not depend continuously on a and b" -- if a and b are still free to vary, then what limit is this? – joriki Mar 28 '12 at 13:54

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