What is the way to calculate $I_{t}^{+}\left(a;\lambda\right)=\int_{0}^{t}e^{a \times e^{-2\lambda s}}ds$

Question

Let $a$ and $\lambda$ be two strictly positive real numbers. How do you compute the following integral ?

$$I_{t}^{+}\left(a;\lambda\right)=\int_{0}^{t}e^{a \times e^{-2\lambda s}}ds$$

A priori, we cannot rely on the Exponential Integral as, given the value $a$ and $\lambda$, this means using diverging integrals...

Answer 1

Compute the following integral $$I_{t}^{+}\left(a;\lambda\right)=\int_{0}^{t}e^{ae^{-2\lambda s}}ds$$ where $a,\lambda\in\mathbb{R_+}$.

Substitute $ae^{-2\lambda s}=x$ so that \begin{align*} I_t^+(a;\lambda)&=-\frac1{2\lambda}\int_a^{ae^{-2\lambda t}} \frac{e^x}xdx\\ \end{align*} Let's integrate the more general \begin{align*} I&=\int\frac{e^x}xdx\\ &=\int\left(\frac1{x}-\frac1{x^2}+\frac1{x^2}-\frac2{x^3}+\frac2{x^3}-\frac6{x^4}+\cdots\right)e^xdx\\ &=\int\left(\frac1{x}-\frac1{x^2}\right)e^xdx+\int\left(\frac1{x^2}-\frac2{x^3}\right)e^xdx+\int\left(\frac2{x^3}-\frac6{x^4}\right)e^xdx+\cdots\\ &=e^x\left(\frac1{x}+\frac1{x^2}+\frac2{x^3}+\frac6{x^4}+\cdots\right)\\ \end{align*} Also, see here.