Consider
$$\tag{1} I(x)=\int\limits_{-\infty}^\infty du \int\limits_{-1}^1dy' \ \frac{e^{-x\sqrt{u^2+a^2}}}{u^2+a^2}\cos(u(y-y')) $$
For fixed $a>0$ and real $|y|<2$. I would like to find an approximation for $I$ as $x \to 0$ (or an expression in terms of 'common' non-elementary functions, but I suspect that is not likely).
My attempt
I expand the exponential around $x=0$. This of course renders the resulting series divergent, but does facilitate using Basset's integral to perform each $u$ integral, which is of the form
$$\tag{2} \int du \ (u^2+a^2)^{-1+n/2}\cos(u(y-y')) $$
The first few terms of the resulting series are
$$\tag{3} I\stackrel{?}{\sim}\int\limits_{-1}^1dy' \ \left[\frac{\pi}{a}e^{-a(y-y')}-2xK_0(a(y-y'))+\frac{ax^3}{3(y-y')}K_1(a(y-y'))+\cdots \right] $$
Where $K$ is a modified Bessel function. The $y'$ integral for the first two terms can now be done using the relations here. Since the argument of $K_0$ changes from positive to negative over the integral, I split up the integration and use the continuation of $K_0$ for negative argument
$$\tag{4} \int\limits_{-1}^1 dy' \ K_0(a(y-y')=\int\limits_{-1}^y dy' K_0(a(y-y'))+\int\limits_y^1 dy' K_0(a(y-y')) $$
In the second term on the RHS of (4) I express $K_0$ with negative argument using
$$\tag{5} K_0(-z)=K_0(z)-i\pi I_0(z) $$
The resulting expression is lengthy, and features Bessel and Struve functions. Here is a plot of $I$ as a function of $y$ using the first two terms of (3) for the approximation versus numerical integration of (1)
On the left $x=0$, and on the right $x\approx 0.5$. The approximation is good when $y>1$, but becomes complex (for nonzero $x$) when $y<1$. This makes me suspect that the method could be fruitful, but somehow one of the manipulations was not quite right. On the other hand, I am not attached to this method, so any other methods are welcome.
Background The integral arises when solving a Laplacian boundary value problem in three dimensions. The $y$ dependence of the result is quite important, while the $x$ dependence less so.