Evaluating the integral $\int_0^1 \frac{\cos bx}{\sqrt{x^2+s^2} }dx$

The asymptotic for large $n$ is determined by the behaviour of the integrand at $x = \pm 1$. Choosing a contour going in the direction $i$ from $x = -1$ and then in the direction $-i$ towards $x = 1$ and expanding the non-exponential part, we have $$\frac 1 {\sqrt {x^2 + s^2}} \bigg\rvert_{x = -1 + i \xi} = \frac 1 {\sqrt {1 + s^2}} + \frac {i \xi} {(1 + s^2)^{3/2}} + O(\xi^2), \\ \frac 1 {\sqrt {x^2 + s^2}} \bigg\rvert_{x = 1 + i \xi} = \frac 1 {\sqrt {1 + s^2}} - \frac {i \xi} {(1 + s^2)^{3/2}} + O(\xi^2).$$ The contributions from the first terms will cancel out, leaving $$I(\pi n, s) = \frac 1 2\int_{-1}^1 \frac {e^{i \pi n x}} {\sqrt {x^2 + s^2}} dx \sim i \int_0^\infty \frac {i \xi} {(1 + s^2)^{3/2}} e^{-i \pi n - \pi n \xi} d\xi = \\ \frac {(-1)^{n - 1}} {\pi^2 (1 + s^2)^{3/2} n^2}, \quad n \to \infty, \,n \in \mathbb N.$$

Getting a grasp of the original problem: we want to find the asymptotic behaviour of the Fourier coefficients of $\frac{1}{\sqrt{x^2+s^2}}\in L^2(0,1)$. Since the inverse Laplace transform of $\frac{1}{\sqrt{x^2+s^2}}$ is $J_0(as)$ and the Laplace transform of $\cos(\pi n x)\mathbb{1}_{(0,1)}(x)$ is $\frac{a}{a^2+n^2\pi^2}-\frac{a(-1)^n}{e^a(a^2+\pi^2 n^2)}$, we have

$$ \int_{0}^{1}\frac{\cos(\pi n x)}{\sqrt{x^2+s^2}}\,dx = \underbrace{\int_{0}^{+\infty}\frac{a J_0(as)}{a^2+\pi^2 n^2}\,da}_{K_0(\pi n s)} + (-1)^{n+1}\int_{0}^{+\infty}\frac{a J_0(as)}{e^a(a^2+\pi^2 n^2)}\,da $$ which is fairly easy to approximate numerically due to the known bounds for Bessel functions.
Close to the origin $J_0(a)$ can be approximated through its rapidly-convergent Maclaurin series, far from the origin we have Tricomi's $J_0(a)\approx\frac{\sin(a)+\cos(a)}{\sqrt{\pi a}}$. For $K_0$ we have Hankel's expansion.

In particular the Fourier coefficients of $\frac{1}{\sqrt{x^2+s^2}}$ decay like $\frac{1}{n^2 s^3}$.

It is also interesting to point out a curious inequality provided by Cauchy-Schwarz:

$$\begin{eqnarray*}\left|\int_{0}^{+\infty}\frac{a J_0(as)}{e^a(a^2+\pi^2 n^2)}\,da\right|^2&\leq& \int_{0}^{+\infty}\frac{J_0(as)^2}{e^{2a}}\,da\int_{0}^{+\infty}\frac{a^2}{(a^2+\pi^2 n^2)^2}\,da\\&=&\frac{1}{8n\,\text{AGM}(1,\sqrt{1+s^2})}\leq\frac{1}{8n}(1+s^2)^{-1/4}.\end{eqnarray*}$$