Dgdrxrt

$int_K^infty1-exp(-exp(-x))dx$

By change of variable we have $int_0^exp(-K)frac1-exp(-u)udu$ but I can't spit the two because integral of $frac1u$ from 0 to K diverges, and therefore there will be an $infty-infty$ term when calculating the integral.

Wolframalpha gives but how?

I'll use $G$ for the lower limit because reasons. Assume that all parameters are positive. The inner integral is elementary:
$$I(gamma) =
int_G^infty (1 - exp(-N v e^-gamma/c)) ,e^-lambda v ,dv =
e^-G lambda left( frac 1 lambda - frac
exp left(-G N e^-gamma/c right)
N e^-gamma/c + lambda right).$$
The antiderivative of the result has a closed form in terms of the incomplete gamma function:
$$int I(-c ln u) ,d(-c ln u) =
frac c e^-G lambda lambda left(
int frac e^-G N u u du -
int frac N e^-G N u N u + lambda du -
ln u right) =\
frac c e^-G lambda lambda
(e^G lambda Gamma(0, G N u + G lambda) - Gamma(0, G N u) - ln u).$$
$Gamma(0, u)$ is continuous on $(0, infty)$ and is $-ln u - gamma_e + O(u)$ at $0^+$, $gamma_e$ being Euler's constant. Therefore
$$int_1^0 I(-c ln u) ,d(-c ln u) =
frac c lambda
(Gamma(0, G lambda) - Gamma(0, G N + G lambda) +
e^-G lambda (Gamma(0, G N) + ln (G N) + gamma_e)).$$

The function
$$
vlongmapstoleft[1-expleft(-Nve^-gamma/cright)right]exp(-lambda v)
$$
has elementary primitive
$$
vlongmapsto
frace^-Nv e^-gamma/c + gamma/c - lambda v
lambda e^gamma/c + N
- frace^-lambda v
lambda.
$$
and
$$
int_0^inftyleft[1-expleft(-Nve^-gamma/cright)right]exp(-lambda v),dv = fracNlambda(lambda e^gamma/c + N),
$$
also with elementary primitive... (to be continued)