Asymptotic expression of an oscillatory integral 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.



*

*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 $f(\alpha, \beta)$?

 A: 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.
A: 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 $s=40$.

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)

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:

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 $F(y) = y^{1/2} - y^{-3/2}/64+\cdots$.
A: Let's look at some special cases:


*

*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 $\sqrt{1-\cos(x)\cos(y)}$.

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

*(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.


