How to approximate the integral $\int_{-b}^{\infty}(t+b)^{\nu}e^{-t}e^{-e^{-t}}dt$ Suppose that we have the following integral
\begin{equation}
\int_{-b}^{\infty}(t+b)^{\nu}e^{-t}e^{-e^{-t}}dt,
\end{equation}
where $b$ is a positive constant and $\nu$ is any number (real or complex). How can we get an approximated result? I'm doubting that the form $b^{c_1\nu+c_2\nu^2}$ can approximate it very well.
 A: Not easy.
Since you want an approximation, we can do as follows. 
The first thing to analyze is the trivial case in which $b = 0$. This will help us later. In this case you have
$$\int_0^{+\infty} t^{\nu} e^{-t} e^{-e^{-t}} \ \text{d}t$$
Since the range of integration is $\mathbb{R}^+$ you are allowed to use Taylor Series for $e^{-e^{-t}}$, and we get:
$$\sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!} \int_0^{+\infty} t^{\nu} e^{-t} e^{-kt}\ \text{d}t = \sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!} \int_0^{+\infty} t^{\nu} e^{-t(1+k)}\ \text{d}t$$
This is a standard integration which gives you
$$\int_0^{+\infty} t^{\nu} e^{-t(1+k)}\ \text{d}t = (1+k)^{-1-\nu}\ \Gamma(1+\nu)$$
Where $\Gamma(a)$ represent the well known Gamma Function.
Hence
$$ \sum_{k = 0}^{+\infty}\frac{(-1)^k}{k!}  (1+k)^{-1-\nu}\ \Gamma(1+\nu) = \Gamma(1-\nu) \sum_{k = 0}^{+\infty}\frac{(-1)^k}{(k+1)!}  (1+k)^{-\nu}$$
Unfortunately, there is no sufficient data to be able to tell if the series does converge and where, because we know nothing about $\nu$. If it's real, complex, imaginary, greater than $2$, between $0$ and $1$ and so on.
I can plot you this series, for $k$ from $0$ to $100$, and $\nu$ from $-8$ to $+8$, to show you the partial behavior, if it helps:

Now we come to a more general case, if it is possible.
When $b\neq 0$ we have your integral.
$$\int_{-b}^{+\infty} (t+b)^{\nu} e^{-t} e^{-e^{-t}}\ \text{d}t$$
General Integral
For this integral, which is a beast, it's conveniente to split it into two parts. 
The first one:
$$\int_0^{+\infty} (t+b)^{\nu} e^{-t} e^{-e^{-t}}\ \text{d}t$$
"can be evaluated" by a giant use of Taylor series for both $e^{-t}$ and $e^{-e^{-t}}$. And it's not over. So, using the series we get (easy to verify):
$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \frac{(-1)^k (-1)^j}{k!\ j!} \int_0^{+\infty} (t+b)^{\nu} t^k\ e^{-tj}\ \text{d}t$$
Again we use Taylor series for the last term $e^{-tj}$, getting
$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \sum_{n = 0}^{+\infty} \frac{(-1)^k (-1)^j (-1)^n}{k!\ j!\ n!} j^n\int_0^{+\infty} (t+b)^{\nu} t^k\ t^n\ \text{d}t$$
The integral now is
$$\int_0^{+\infty} (t+b)^{\nu} t^{k+n}\ \text{d}t  = b^{1 + \nu + n + k} \frac{\Gamma(-1-\nu - j - n)\Gamma(1 + j + n)}{\Gamma(-\nu)}$$
And so you start seeing how difficult is to get an approximate form, since this result only holds for $\Re(\nu + j + n) >-1$. 
But it's something, right?
So this part will give you in the end
$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \sum_{n = 0}^{+\infty} \frac{(-1)^k (-1)^j (-1)^n}{k!\ j!\ n!} j^n b^{1 + \nu + n + k} \frac{\Gamma(-1-\nu - j - n)\Gamma(1 + j + n)}{\Gamma(-\nu)}$$
Good luck in summing that. Even if you may try with the first $3-4$ terms each series, that would be simple, actually. I cannot say anything about the convergence of those series, though. The parameter $\nu$ is too general.
For what concerns the second integral, we have
$$\int_{-b}^0 (t+b)^{\nu} e^{-t} e^{-e^{-t}}\ \text{d}t$$
What we can do here are two passages: the first is to use the Binomial theorem for $(t+b)^{\nu}$ term, and the second is again a Taylor series for the second exponential, allowed because 
$$e^{-e^{-t}}$$
has two parts: the "higher" $e^{-t}$, which becomes great for $t < 0$, and the second part $e^{-e^{-t}}$ which becomes very small for $e^{-t}$ large.
Hence
$$(t+b)^{\nu} = \sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} t^m$$
$$e^{-e^{-t}} = \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} e^{-pt}$$
Putting all together you get
$$ \sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} \sum_{p = 0}^{+\infty} \frac{(-1)^p}{p!} \int_{-b}^0 t^m e^{-t} e^{-tp}\ \text{d}t$$
The very last integral is trivial
$$ \int_{-b}^0 t^m e^{-t} e^{-tp}\ \text{d}t = \frac{(-1)^m }{(-p-1)^{m+1}} (\Gamma (m+1)-\Gamma (m+1,-b (p+1)))$$
Where the last function is the Incomplete Gamma Function. You can manipulate a bit the minus signs, and in the end with the previous sums you have:
$$ \sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} \sum_{p = 0}^{+\infty} \frac{(-1)^{p+1}}{(p+1)!} (\Gamma (m+1)-\Gamma (m+1,-b (p+1)))$$
The final Union
Summing this part to the previous and we get the (actually not so approximate, but rather quite exact for the conditions we established about the parameters) solution:

$$\sum_{k =0}^{+\infty} \sum_{j = 0}^{+\infty} \sum_{n = 0}^{+\infty} \frac{(-1)^k (-1)^j (-1)^n}{k!\ j!\ n!} j^n b^{1 + \nu + n + k} \frac{\Gamma(-1-\nu - j - n)\Gamma(1 + j + n)}{\Gamma(-\nu)}  + \\\\ +\sum_{m = 0}^{+\nu} \binom{\nu}{m} b^{\nu - m} \sum_{p = 0}^{+\infty} \frac{(-1)^{p+1}}{(p+1)!} (\Gamma (m+1)-\Gamma (m+1,-b (p+1)))$$
