Is it a coincidence that the $x^2$, and $x^3$ of $\cos$ and sinusoidal integrals relate to Gamma functions? Is there any generalization of
$$
\int_0^\infty\cos(\xi^k)d\xi=\frac{\Gamma\left(\frac 1 k\right)}{2\sqrt k}\quad?
$$
It is surprisingly valid for $k=2,3$ and $\sin$ as well, but not for other $k$.
 A: Using a change of contour for the last integral you have something like:
$$ \int_0^\infty e^{i \xi^k} d\xi =  \frac1k \int_0^\infty e^{it} t^{1/k-1} dt =\frac1k  i^{1/k} \int_0^\infty e^{-u} u^{1/k-1} du =\frac1k i^{1/k} \Gamma(1/k). $$
A: Broadly speaking (ignoring branches), let $x= \xi^k$, $x^{1/k} = \xi$, $\frac{d\xi}{dx} = \frac{1}{k}x^{\frac{1}{k}-1}$
$$
I = \int_0^\infty  \frac{1}{k}x^{\frac{1}{k}-1} \cos(x) \; dx
$$
compare this to a Mellin transform. Because $\cos$ and $\sin$ are made out of exponential functions, also consider that
$$
\Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \; dx
$$
so we can see the origin of $\Gamma(1/k)$, in fact, many functions are related to gamma functions through their Mellin transforms.
For Mellin transforms in general we have
$$
\mathcal{M}[f(x^k)](s) = \frac{1}{k}\mathcal{M}[f(x)](\frac{s}{k})
$$
we have the Mellin transform of $\cos(x)$
$$
\mathcal{M}[\cos(x)](s) = \int_0^\infty x^{s-1} \cos(x) \; dx = \cos \left(\frac{\pi  s}{2}\right) \Gamma (s)
$$
$$
\mathcal{M}[\cos(x^k)](s) = \int_0^\infty x^{s-1} \cos(x^k) \; dx = \frac{1}{k}\cos \left(\frac{\pi  s}{2k}\right) \Gamma \left(\frac{s}{k}\right)
$$
your integral is for $s=1$
$$
\int_0^\infty\cos(x^k) \; dx = \frac{1}{k}\cos \left(\frac{\pi }{2k}\right) \Gamma (\frac{1}{k})
$$
your example works because $\frac{1}{3}\cos(\frac{\pi}{2\cdot 3})=\frac{\sqrt{3}}{2 \cdot 3} = \frac{1}{2 \sqrt{3}}$.
In terms of a generalisation for any random function one chooses, consider a very general series, e.g. a hypergeometric function. which has Mellin transform
$$
\mathcal{M}[_2F_1(a,b,c,-d x)](s) =  \frac{\Gamma (c) d^{-s} \Gamma (s) \Gamma (a-s) \Gamma (b-s)}{\Gamma (a) \Gamma (b) \Gamma (c-s)}
$$
we have
$$
\mathcal{M}[_2F_1(a,b,c,-d x^k)](s) =  \frac{\Gamma (c) d^{-\frac{s}{k}} \Gamma (\frac{s}{k}) \Gamma (a-\frac{s}{k}) \Gamma (b-\frac{s}{k})}{k\Gamma (a) \Gamma (b) \Gamma (c-\frac{s}{k})}
$$
so for example, in principle using $\;_2F_1(\frac{1}{2},\frac{1}{2};1;x) = \frac{2}{\pi} K(x)$, with elliptic $K$ function, we have in analogy
$$
\frac{2}{\pi}\int_0^\infty x^{s-1} K(-\eta x^k) \; dx = \frac{\Gamma (1) \eta^{-\frac{s}{k}} \Gamma (\frac{s}{k}) \Gamma (\frac{1}{2}-\frac{s}{k}) \Gamma (\frac{1}{2}-\frac{s}{k})}{k\Gamma (\frac{1}{2}) \Gamma (\frac{1}{2}) \Gamma (1-\frac{s}{k})}
$$
whether this converges or not will depend on various technical details. This last example is just to show that the gamma functions do not relate in particular to sinusoidal functions, but most analytic functions, which are definable using a generalized contour integral, along with the integration region $[0,\infty)$. In general, wherever gamma functions are found, some kind of relationship to a Mellin transform is not far away.
