Definite integral of exponential of square root

Assuming $a\geq 0$, $$I_2(a) \stackrel{\text{parity}}{=} 2\int_{0}^{+\infty}e^{-\sqrt{t^2+a^2}}\, \mathrm dt \stackrel{t\mapsto a\sinh(z)}{=} 2a\int_{0}^{+\infty}\cosh(z)\,e^{-a\cosh z}\,\mathrm dz\tag{1}$$ leads to: $$ I_2(a) = 2a\int_{1}^{+\infty}\frac{z}{\sqrt{z^2-1}} e^{-az}\, \mathrm dz = 2a\cdot K_1(a)\tag{2}$$ where $K_1$ is a modified Bessel function of the second kind.
$I_1(a)$ can be computed from Simply Beautiful Art's comment above.