Computation of the integral of the product of a double exponential and an exponential I would like to compute the following integral:
$$\int_0^t\exp\left(\frac{\alpha^2}{2\lambda}e^{-2\lambda s}-\lambda s\right)ds\space\space\space(1)$$
An integral near from this one is:
$$\int_0^t\exp\left(\frac{\alpha^2}{2\lambda}e^{-2\lambda s}\right)ds\space\space\space   (2)$$
and may be computed by setting the following change of variables $u=e^{-2\lambda s}$. This leads to a integral of the following kind:
$$\int_c^d \frac{e^{au}}{u}du$$
By using the integral serie development of the exponential, a calculation of the integral (2) may be achieved. The remaining issue is the speed of convergence of the calculus but that is another issue that is a numeric one.
Unfortunately, if i am not wrong, this trick is not  usable to compute the integral (1). Any idea to obtain an usable calculus of the integral (1)?
 A: Considering $$I=\int\exp\left(\frac{\alpha^2}{2\lambda}e^{-2\lambda s}-\lambda s\right)\,ds$$ let (not so obvious, I agree but quite close to your $u$)
$$s=-\frac 1{\lambda }{\log \left(\frac{ x}{\sqrt{\frac{\alpha ^2}{2\lambda
   }}}\right)}\implies ds=-\frac{dx}{\lambda  x}$$
$$I=-{\lambda  \sqrt{\frac{2\alpha ^2}{\lambda }}}\int e^{x^2}\,dx=-{\lambda  \sqrt{\frac{2\alpha ^2\pi}{2\lambda }}}  \text{erfi}(x)$$
Back to $s$, this should give
$$I=-\frac{\sqrt{\frac{\pi\alpha ^2}{2\lambda }}}{\alpha ^2}\text{erfi}\left(\sqrt{\frac{\alpha ^2}{2\lambda }} e^{-\lambda  s}\right)$$ and simplifying
$$J=\int_0^t\exp\left(\frac{\alpha^2}{2\lambda}e^{-2\lambda s}-\lambda s\right)\,ds=\frac 1 \alpha \sqrt{\frac \pi {2\lambda}}\left(\text{erfi}\left(\frac{\alpha }{ \sqrt{2\lambda
   }}\right)-\text{erfi}\left(\frac{\alpha  }{
   \sqrt{2\lambda }}e^{-\lambda  t}\right) \right)$$
Edit
Making, as you thought about it, $u=e^{-2\lambda s}$ is good and leads for the antiderivative to $$I=\int -\frac{e^{\frac{\alpha ^2 u}{2 \lambda }}}{2 \lambda  \sqrt{u}}\,du$$ Now
$${\frac{\alpha ^2 u}{2 \lambda }}=x^2\implies u=\frac{2 \lambda  x^2}{\alpha ^2}\implies du=\frac{4 \lambda  x}{\alpha ^2}\,dx$$
$$I=-\frac{ \sqrt{\frac{2\lambda }{\alpha ^2}}}{\lambda }\int e^{x^2} \,dx=-\frac{ \sqrt{\frac{2\lambda }{\alpha ^2}}}{\lambda }\frac{1}{2} \sqrt{\pi } \text{erfi}(x)$$
Assuming $\alpha >0$ and $\lambda>0$, the definite integral is
$$\color{blue}{\int_0^t\exp\left(\frac{\alpha^2}{2\lambda}e^{-2\lambda s}-\lambda s\right)\,ds=\frac 1 \alpha \sqrt{\frac{\pi }{2\lambda}}\Big[ \text{erfi}\left(\frac{\alpha }{ \sqrt{2\lambda
   }}\right)-\text{erfi}\left(\frac{\alpha  }{
   \sqrt{2\lambda }}e^{-\lambda  t}\right)\Big]}$$
