I'd really love to evaluate this integral exactly in terms of known functions, because for large $b$ it becomes a pain numerically.
$$I(b,s)=\int_0^1 \frac{\cos bx}{\sqrt{x^2+s^2} }dx$$
Didn't get anywhere wth integration by parts, or series. I want an expression valid for both large and small values of the parameters.
Note: we have $b= \pi n$, where $n$ is an integer.
For $s \gg 1$ we can expand the root as a series and get a good approximation. However, this case is of small use to me, as in general $s$ is of the order of $1$ or smaller.
So here's my latest attempt:
Edited
$$x=s \sinh v$$
$$I(b,s)=\int_0^{\sinh^{-1} \frac{1}{s}} \cos \left(bs \sinh v \right) dv$$
Let's try integration by parts:
$$U=\cos \left(bs \sinh v \right) \\ dV=dv$$
$$dU=-bs\sin \left(bs \sinh v \right) \cosh v \\ V=v$$
$$I(b,s)=\cos b \sinh^{-1} \frac{1}{s}+bs \int_0^{\sinh^{-1} \frac{1}{s}} v \sin \left(bs \sinh v \right) \cosh v dv$$
$$I(b,s)=\cos b \sinh^{-1} \frac{1}{s}+bs \int_0^{\frac{1}{s}} \sinh^{-1} r \sin \left(bs r \right) dr$$
$$I(b,s)=\cos b \sinh^{-1} \frac{1}{s}+b \int_0^1 \sinh^{-1} \frac{x}{s} \sin \left(b x \right) dx$$
This seems a little better, at least we separated the special case $I(0,s)$, which is the first term. Not sure how to continue.
Using the fact that $b= \pi n$, we can transform the integral as:
$$b \int_0^1 \sinh^{-1} \frac{x}{s} \sin \left(b x \right) dx=\pi \sum_{k=0}^{n-1} (-1)^k \int_0^1 \sinh^{-1} \left( \frac{t+k}{ns} \right) \sin \pi t dt$$
The function $\sinh^{-1}$ is pretty nice, it has a logarithmic growth, and doesn't blow up anywhere. I guess, this could be the way to get a good approximation, especially since we have an alternating sum.
Best Answer
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 $$I(\pi n, s) = \frac 1 2\int_{-1}^1 \frac {e^{i \pi n x}} {\sqrt {x^2 + s^2}} dx \sim i \int_0^\infty \frac {i \xi} {(1 + s^2)^{3/2}} e^{-i \pi n - \pi n \xi} d\xi = \\ \frac {(-1)^{n - 1}} {\pi^2 (1 + s^2)^{3/2} n^2}, \quad n \to \infty, \,n \in \mathbb N.$$