I encountered the following integral, which I am unable to solve:
$$\int_0^\infty {x\over a} e^{{x \over a}} e^{{-b \over a}(e^{{x \over b}}-1)} e^{{-c \over d}(e^{{x \over c}}-1)} \mathrm{d}x$$
with $a, b ,c, d> 0$.
I was able to solve it for the case where $b=c$, which has some physical relevance to me:
$$\int_0^\infty {x\over a} e^{{x \over a}} e^{{-c \over a}(e^{{x \over c}}-1)} e^{{-c \over d}(e^{{x \over c}}-1)} \mathrm{d}x$$
using the substitution $z={x \over b}$ I get:
$$\int_0^\infty {b^2 \over a} z e^{z} e^{{-b(d+a) \over da}(e^{z}-1)}\mathrm{d}z$$
which boils down to:
$$\int_0^\infty z e^{z} e^{-\alpha(e^{z}-1)}\mathrm{d}z = e^\alpha E_1(\alpha) /\alpha$$
Where $E_1$ is the exponential integral.
Can anyone suggest a solution or a decent approximation without a special case like $b=c$?