I would like to study the integral $I(L)$ (where $L$ is a positive integer): $$ I(L)= \int\limits_{{-}\pi L}^{{+}\pi L} \sqrt[4]{1-\cos\Big(\frac{q}{L}\Big)} \cos(q) dq $$ in the limit $L\to\infty$.
EDIT: From a numerical experiment in Wolfram alpha I expect this integral to go to zero as $\frac{1}{\sqrt{L}}$:
integrate_{-5*\pi}^{5*\pi}\sqrt[4]{1-cos(x/5)}*cos(x)dx = -0.47
integrate_{-50*\pi}^{50*\pi}\sqrt[4]{1-cos(x/50)}*cos(x)dx = -0.149
integrate_{-500*\pi}^{500*\pi}\sqrt[4]{1-cos(x/500)}*cos(x)dx = -0.047
integrate_{-1000*\pi}^{1000*\pi}\sqrt[4]{1-cos(x/1000)}*cos(x)dx = -0.033
I would like to prove analytically the result above, i.e. I would like to understand the origin of the trend $I(L)\sim_{\infty}\frac{1}{\sqrt{L}}$.
To this aim, I tried to expand the cosine and collect the first power $$ \begin{equation} \begin{split} \lim_{L\to\infty}I(L)&= \lim_{L\to\infty}\int\limits_{{-}\pi L}^{{+}\pi L} \sqrt[4]{1-\cos\Big(\frac{q}{L}\Big)} \cos(q) dq\\ &=\lim_{L\to\infty}\frac{1}{\sqrt{L}}\int\limits_{{-}\pi L}^{{+}\pi L} \sqrt[4]{({q})^2 \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2[n+1])!}} \Big(\frac{q}{L}\Big)^{2n}} \cos(q) dq=\\ &=\lim_{L\to\infty}\frac{1}{\sqrt{L}}\int\limits_{{-}\pi L}^{{+}\pi L} \sqrt[4]{({q})^2 \Big(\frac{1}{2!}-\frac{1}{3!}\Big(\frac{q}{L}\Big)^{2}+\frac{1}{4!}\Big(\frac{q}{L}\Big)^{4}+\dots} \cos(q) dq \end{split} \end{equation} $$ Now I was hoping to be able to show that $$ \lim_{L\to\infty} \left|\int_{-\pi L}^{+\pi L} \sqrt[4]{q^2 \left(\frac{1}{2!} - \frac{1}{3!} \left(\frac{q}{L}\right)^2 + \frac{1}{4!} \left(\frac{q}{L}\right)^4+\dots\right)} \cos(q) dq\right| \sim C\hspace{5mm} (*) $$ where $C$ is a positive constant. But this does not seem straightforward to me.
Does anyone see if there is any simple way to prove equation $(*)$? I expect $(*)$ to be true due to the numerical evidence listed above.
This is a case where the asymptotic of the Fourier coefficients of $f(u) = (1 - \cos u)^{1/4}$ is determined by the branch point of $f$ which is nearest to the real line. The integrals over the rays $[-\pi, -\pi + i \infty)$ and $[\pi, \pi + i \infty)$ are the same, thus $I(L)$ is equal to the integral along the banks of the branch cut $[0, i \infty)$. That integral can be estimated by applying Laplace's method. In the same way as here, we obtain $$f(i \xi \pm 0) \sim \frac {1 \pm i} {2^{3/4}} \sqrt \xi \quad\text{when } \xi \to 0^+, \\ I(L) = L \int_{-\pi}^\pi f(u) e^{i L u} du \sim i L \int_0^\infty \frac {2 i} {2^{3/4}} \sqrt \xi \,e^{-L \xi} d\xi = \\ -\sqrt {\frac \pi {2^{3/2} L}}, \quad L \to \infty, \;L \in \mathbb N.$$