Integral $I_{t}^{-}\left(\alpha;\lambda\right)=\int_{0}^{t}\exp\left(-\frac{\alpha^{2}}{2\lambda}e^{-2\lambda s}\right)ds$ I need to compute the following integral of a double exponential function.
$$I_{t}^{-}\left(\alpha;\lambda\right)=\int_{0}^{t}\exp\left(-\frac{\alpha^{2}}{2\lambda}e^{-2\lambda s}\right)ds$$
The last integral may be computed using the incomplete Gamma function. Indeed, if we define $u=e^{-2\lambda s}$
, we have $\frac{du}{ds}=-2\lambda e^{-2\lambda s}=-2\lambda u$
and also $s=\frac{\ln\left(u\right)}{-2\lambda}$
so that $\frac{ds}{du}=\frac{1}{-2\lambda u}$:
$$I_{t}^{-}\left(\alpha;\lambda\right)=\int_{0}^{t}e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)e^{-2\lambda s}}ds=\int_{1}^{e^{-2\lambda t}}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{-2\lambda u}du=\frac{-1}{2\lambda}\int_{1}^{e^{-2\lambda t}}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{u}du
 $$
$$=\frac{1}{2\lambda}\int_{e^{-2\lambda t}}^{1}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{u}du$$
Using the integral exponential, after some straightforward developments, this leads:
$$I_{t}^{-}\left(\alpha;\lambda\right)=t+\frac{1}{2\lambda}\left(\sum_{k=1}^{+\infty}\frac{\left(-\frac{\alpha^{2}}{2\lambda}\right)^{k}}{k!k}-\sum_{k=1}^{+\infty}\frac{\left(-\frac{\alpha^{2}}{2\lambda}e^{-2\lambda t}\right)^{k}}{k!k}\right)$$
$$\Leftrightarrow I_{t}^{-}\left(\alpha;\lambda\right)=t+\frac{1}{2\lambda}\left(\sum_{k=1}^{+\infty}\frac{\left(-\frac{\alpha^{2}}{2\lambda}\right)^{k}\left(1-e^{-2k\lambda t}\right)}{k!k}\right)$$
This sum is finite, and it is also positive if the speed of mean reversion $\lambda$ stays strictly positive. Numerically speaking, we encounter difficulty to get convergence for a combination of a very low $\lambda$ (lower or equal to 0.01) combined to a very high $\alpha$ (Higher than 130%). In any other cases, we obtain the convergence and in the most common observed cases, a quick and convergent result is obtained with the first 15 elements of the sum above.
But is there another way to proceed ?
 A: Let $S\left(a,b\right)=\int_{b}^{+\infty}\frac{e^{-au}}{u}du$
with $a$  and $b$
two positive reals, then taking $v=u/b$, we have:
$$S\left(a,b\right)=\int_{b}^{+\infty}\frac{e^{-au}}{u}du=\int_{1}^{+\infty}\frac{e^{-abv}}{u}du=E_{1}\left(ab\right)$$
with $E_{n}\left(x\right)$ being the Generalized Exponential Integral function such that $E_{n}\left(x\right)=\int_{1}^{+\infty}\frac{e^{-xv}}{u^{n}}du$.
Then coming back $I_{t}^{-}\left(\alpha;\lambda\right)$, if we define $u=e^{-2\lambda s}$, we have $\frac{du}{ds}=-2\lambda e^{-2\lambda s}=-2\lambda u$
and also $s=\frac{\ln\left(u\right)}{-2\lambda}$
so that $\frac{ds}{du}=\frac{1}{-2\lambda u}$. This leads to:
$$I_{t}^{-}\left(\alpha;\lambda\right)=\int_{0}^{t}e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)e^{-2\lambda s}}ds=\int_{1}^{e^{-2\lambda t}}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{-2\lambda u}du=\frac{-1}{2\lambda}\int_{1}^{e^{-2\lambda t}}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{u}du$$
$$=\frac{1}{2\lambda}\int_{e^{-2\lambda t}}^{1}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{u}du=\frac{1}{2\lambda}\left(\int_{e^{-2\lambda t}}^{+\infty}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{u}du-\int_{1}^{+\infty}\frac{e^{\left(-\frac{\alpha^{2}}{2\lambda}\right)u}}{u}du\right)$$
$$=\frac{1}{2\lambda}\left(S\left(\frac{\alpha^{2}}{2\lambda},e^{-2\lambda t}\right)-S\left(\frac{\alpha^{2}}{2\lambda},1\right)\right)\\$$
$$=\frac{1}{2\lambda}\left(E_{1}\left(\frac{\alpha^{2}}{2\lambda}e^{-2\lambda t}\right)-E_{1}\left(\frac{\alpha^{2}}{2\lambda}\right)\right)$$
Using the development in serie of the exponential integral, after some straightforward developments, alternatively this leads:
$$I_{t}^{-}\left(\alpha;\lambda\right)=t+\frac{1}{2\lambda}\left(\sum_{k=1}^{+\infty}\frac{\left(-\frac{\alpha^{2}}{2\lambda}\right)^{k}}{k!k}-\sum_{k=1}^{+\infty}\frac{\left(-\frac{\alpha^{2}}{2\lambda}e^{-2\lambda t}\right)^{k}}{k!k}\right)$$
$$\Leftrightarrow I_{t}^{-}\left(\alpha;\lambda\right)=t+\frac{1}{2\lambda}\left(\sum_{k=1}^{+\infty}\frac{\left(-\frac{\alpha^{2}}{2\lambda}\right)^{k}\left(1-e^{-2k\lambda t}\right)}{k!k}\right)$$
