Calculating the integral of exponential of exponential

Question

I've been trying to integrate the following expression: $$\int_{0}^{\infty} (1- \exp(-a e^{-bx})) dx$$ I started to solve this integral by first using the substitution method. With $t = e^{-bx}$, then the integral is: $$\frac{1}{b} \int_{0}^{1} \frac{1}{t}(1-e^{-at})dt$$ The only thing that comes to my mind is solving this using integration by parts, but it ends up complicating the integration way too much. For example, let $u = \frac{1}{t}$ and $dv = 1 - e^{-at}$. Is there any easier way to solve this integral?

Answer 1

We begin with the integral $I$ given by

$$I=\int_0^1 \frac{1-e^{-at}}{t}\,dt\tag1$$

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Enforcing the substitution $t\mapsto t/a$ in $(1)$ reveals

$$I=\int_0^a \frac{1-e^{-t}}{t}\,dt \tag2$$

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Integrating by parts the integral in $(2)$ with $u=1-e^{-t}$ and $v=\log(t)$, we find that

$$\begin{align} I&=(1-e^{-a})\log(a)-\int_0^a \log(t)e^{-t}\,dt\\\\ &=(1-e^{-a})\log(a)+\gamma+\int_a^\infty \log(t)e^{-t}\,dt\tag3 \end{align}$$

where $\gamma =-\int_0^\infty \log(t) e^{-t}\,dt$ is the Euler-Mascheroni constant.

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A subsequent integration by parts in $(3)$ with $u=\log(t)$ and $v=-e^{-t}$ yields

$$\bbox[5px,border:2px solid #C0A000]{I=\log(a)+\gamma+\Gamma(0,a)}\tag 4$$

where $\Gamma(0,a)=\int_a^\infty \frac{e^{-t}}{t}\,dt$ is the upper Incomplete Gamma function.

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Note that we could have alternatively expressed $(4)$ as

$$\bbox[5px,border:2px solid #C0A000]{I=\log(a)+\gamma-\text{Ei}(-a)}$$

in terms of the Exponential integral $\text{Ei}(-a)=-\int_a^\infty \frac{e^{-t}}{t}\,dt$.

Answer 2

The integral of $\dfrac {e^{a x} } x$ does not work out easily.

Here is how it is done on ProofWiki:

https://proofwiki.org/wiki/Primitive_of_Exponential_of_a_x_over_x

$$\int \frac {e^{a x} \, \mathrm d x} x = \ln |x| + \sum_{k \mathop \ge 1} \frac {(a x)^k} {k \times k!} + C$$

This will allow you to actually get the answer to your substituted integral, and it should be straightforward to get the answer out from there. Just that it ain't very pretty.