Let $((-\pi,\pi], \mathcal{B}((-\pi,\pi]), \mu)$ be a measure space with Lebesgue measure $\mu$. Show that for $m, d\in\mathbb{N}$,
$$
\int_{-\pi}^\pi \left|\lim_{m\to\infty}\lim_{d\to \infty} \sqrt{d}\cos(m!x)^{2d}\right|dx=0
$$
and
$$
\lim_{m\to\infty}\lim_{d\to \infty}\int_{-\pi}^\pi \left| \sqrt{d}\cos(m!x)^{2d}\right|dx=1/\sqrt{\pi}
$$
It seems that we can use this result: Show that $ \frac{1}{n}\sum_{k=1}^n[\cos (x+k)]^d\to (d-1)!!/d!! \mbox{ for $d$ is even } $
for $m$-a.e. all $x\in (-\pi, \pi]$ that
$$
\frac{1}{n}\sum_{k=1}^n[\cos (x+k)]^d\to (d-1)!!/d!!\sim \frac{1}{\sqrt{d\pi}} \mbox{ for $d$ is even }
$$
otherwise it is $0$, as $n\to \infty$.
Best Answer
For the first one, for a fixed $m \in \mathbb{N}$ and for any $x \neq \frac{k\pi}{m!}$, $|\cos{(m!x)}|<1$, so
$$ \lim_{d \to \infty} \sqrt{d} \left(\cos{(m!x)}\right)^{2d} = 0 $$
so if you take $m$ to infinity avoiding those points the limit is always zero, since the set
$$ \left\{x \in [-\pi,\pi] \colon x=\frac{k\pi}{m!} \ \text{for some} \ k \in \mathbb{Z} \ \text{and} \ m \in \mathbb{N}\right\} $$ is countable, the function that you are integrating is zero a.e.
Now, for the second equality, we use that
$$ \int_{-\pi}^{\pi} \left(\cos{x}\right)^{2n} dx = 2\pi\frac{(2n)!}{(2^n n!)^2} $$
notice that, by the formula above and Stirling Formula:
\begin{align*} \int_{-\pi}^{\pi} |\sqrt{d} \left(\cos{(m!x)}\right)^{2d}| dx &= \sqrt{d}\int_{-\pi}^{\pi} \left(\cos{(m!x)}\right)^{2d} dx = \frac{\sqrt{d}}{m!}\int_{-m!\pi}^{m!\pi} \left(\cos{x}\right)^{2d} dx \\ &= \frac{\sqrt{d}}{m!}m!\int_{-\pi}^{\pi} \left(\cos{x}\right)^{2d} dx = \sqrt{d}\frac{2\pi(2d)!}{(2^d d!)^2} \\ &= \frac{2\sqrt{d}\pi}{2^{2d}}\frac{\sqrt{2\pi(2d)}((2d)e^{-1})^{2d}(1+O(d^{-1}))}{(\sqrt{2\pi d}(de^{-1})^{d}(1+O(d^{-1})))^2} \\ &= 2\sqrt{\pi}\frac{1+O(d^{-1})}{(1+O(d^{-1}))^2} = 2\sqrt{\pi} + o(1) \end{align*}
hence
$$ \lim_{m \to \infty} \lim_{d \to \infty} \int_{-\pi}^{\pi} |\sqrt{d} \left(\cos{(m!x)}\right)^{2d}| dx = \lim_{m \to \infty} 2\sqrt{\pi} = 2\sqrt{\pi}. $$
Maybe I made some mistake in my computation because the result in the end was different but I think this idea should work.