Integrating an exponential of an exponential I want to integrate integral $A$, $$A=\int_{-\frac{\mu}{\beta}}^{\infty}\exp\left[-\left(\frac{\beta}{2}+1\right)z-e^{-z}\right]\textrm{d}z,$$ where $\mu,\beta,z>0$. Unfortunately, so far I have been unable to find a close form solution.
So far I have tried substituting $t=z+\frac{\mu}{\beta}$, which yields, $$A=\int_{0}^{\infty}\exp\left[-\left(\frac{\beta}{2}+1\right)t+\frac{\mu}{2}+\frac{\beta}{2}-e^{-t+\frac{\mu}{\beta}}\right]\textrm{d}t.$$ 
Again, I could not find a closed form solution.
Any ideas on how I could proceed with this? Could you suggest another substitution perhaps?

The equation can be rearranged as, $$A=\exp\left(\frac{\mu}{2}+\frac{\mu}{\beta}\right)\int_{0}^{\infty}\exp\left[-\left(\frac{\beta}{2}+1\right)t-e^{\frac{\mu}{\beta}}e^{-t}\right]\textrm{d}t.$$ The integral is in the form, $\int_{0}^{\infty}\exp\left(-Ax-Be^{-x}\right)dx$, where the integral solution can be found in [Gradshteyn2007, eq. (3.331-1)], in the form, $B^{-A}\gamma\left(A,B\right)$ conditioned on $\left[\textrm{Re}A>0\right]$. 
Hence the closed-form solution can be given as,
$$A=\exp\left(\frac{\mu}{2}+\frac{\mu}{\beta}\right)\exp\left(\frac{\mu_{Q}}{\beta_{Q}}\right)^{-\left(\frac{\beta_{Q}}{2}+1\right)}\gamma\left(\frac{\beta_{Q}}{2}+1,\exp\left(\frac{\mu_{Q}}{\beta_{Q}}\right)\right).$$
