Closed form expression for $e^{-ne^{-at}}$

Question

I am solving an equation, and I have come across the following expression which I am unable to solve. I will really appreciate if someone can help me with this.

$$\int_{0}^{\infty} e^{-ne^{-at}}dt$$

The original expression was...

$$\int_{0}^{\infty} t e^{-ne^{-at}}dt$$

Solving by by-part method, $u = t$, $v=-ne^{-at}$.

$$\int u,v \ dt= u\int v\ dt- \int u' (\int v \ dt) \ dt$$

Now the second part of by-part equation i.e. $$\int v \ dt= \int -ne^{-at} \ dt= \frac{1}{na}e^{-ne^{-at}} $$

At this point I am stuck and need some hint on how to solve this

Answer 1

$$I = \int_{0}^{\infty} e^{-ne^{-at}}dt$$

Let $$y =e^{-ne^{-at}} \Rightarrow I = \int_{e^{-n}}^{1}y\cdot dt$$ $$\Rightarrow \ln y =-ne^{-at}$$ $$\Rightarrow \frac{dy}{y} = nae^{-at} \cdot dt = -a\,(-ne^{-at}) \cdot dt = -a\,(\ln{y}) \cdot dt$$ $$\Rightarrow -\frac{dy}{na \ln{y}} = y \cdot dt$$

$$\Rightarrow I = -\frac{1}{n \cdot a} \int_{e^{-n}}^{1} \frac{dy}{\ln y}$$ $$\Rightarrow I = -\frac{1}{n \cdot a} \bigg(\int_{0}^{1} \frac{dy}{\ln y} - \int_{0}^{e^{-n}} \frac{dy}{\ln y}\bigg)$$

The above integral does not converge since $$\int_{0}^{1} \frac{dy}{\ln y} = \infty$$ That is, the first integral diverges.

Side notes:

We have $$\int_{0}^{x} \frac{dy}{\ln y} = li(x)$$

where $li(x)$ is the Logarithmic Integral Function.

Hence the above integral can be rewritten for the limits as: $$I = \boxed{-\frac{1}{na}\cdot \big(li(1) - li(e^{-n})\big)}$$

However, note that $li(1) = \infty$ and hence the above integral diverges.

Answer 2

Concerning the problem of the antiderivative $$I = \int e^{-ne^{-at}}dt$$ change variable $$ne^{-at}=u \implies t=-\frac{\log \left(\frac{u}{n}\right)}{a}\implies dt=-\frac{du}{a u}$$ This makes $$I=-\frac 1a\int\frac{e^{-u}}{ u}\,du=-\frac{\text{Ei}(-u)}{a}$$ where appears the exponential integral function.

Back to $t$, this will give $$I=-\frac{\text{Ei}\left(-n\,e^{-a t} \right)}{a}$$ If there is no problem for $t \to 0$, as user1952500 already answered, there is a major one for $t \to \infty$.

If you consider the case $n=a=1$, you would notice that $I$ almost behaves as $t$