How to integrate $\int_{-\infty}^{+\infty} \exp(-\sqrt{a^2+z^2}) dz$ 
$$\int_{-\infty}^{+\infty} \exp(-\sqrt{a^2+z^2}) dz$$

According to info provided by @skbmoore:
Gradshteyn, Izrail Solomonovich, and Iosif Moiseevich Ryzhik. Table of integrals, series, and products. Academic press, 2014. Eq.3.914.1 and 3.914.4 
 A: Gradshteyn and Ryzhik have several integrals of this type.  For instance, eq. 3.914.1 has 
$$ \int_0^\infty \exp{(-\beta\,\sqrt{a^2+z^2}\,)} \cos{(b\,z)} \, dz = \frac{\beta \, a}{\sqrt{\beta^2+b^2}} K_1\big(a \sqrt{\beta^2+b^2} \, \big.)$$
Set $\beta=1$ and $b=0$ to get the answer you need.  References to this formula are given, so if you want to know how it was derived, you might examine those.
A: By assuming $a>0$ and letting $x=a\sinh z$ we have
$$ I(a)=2\int_{0}^{+\infty}e^{-\sqrt{a^2+x^2}}\,dx = 2a \int_{0}^{+\infty}\cosh(z) e^{-a\cosh z}\,dz \tag{1}$$
and
$$ I(a)=2a\int_{1}^{+\infty}\frac{u e^{-au}}{\sqrt{u^2-1}}\,du =2a e^{-a}\int_{0}^{+\infty}\frac{v+1}{\sqrt{v^2+2v}} e^{-av}\,dv\tag{2}$$
such that $I(a)$ is clearly related to the Laplace transform of $\frac{v+1}{\sqrt{v^2+2v}}$ and it is given by $2a e^{-a}$ times a log-convex function. Equivalently
$$ I(a) = 2e^{-a}+2ae^{-a}\int_{0}^{+\infty}\frac{e^{-av}}{\sqrt{v^2+2v}(v+1+\sqrt{v^2+2v})}\,dv\tag{3}$$
which in combination with the Cauchy-Schwarz inequality allows tight approximations for $a\approx 0$ or $a\gg 0$.
By the very definition of $K_1$ we have $I(a)= 2a\,K_1(a)$ and tight lower bounds can be derived from $\frac{u}{\sqrt{u^2-1}}\geq 1+\frac{1}{2u^2}$ and Cauchy-Schwarz, again. Unsurprisingly, $I(a)$ behaves like the density of a normal distribution close to the origin and like the density of a Laplace distribution far from the origin.
