How to approximate the integral $\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt$ Suppose we have the following integral
\begin{equation}
\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt,
\end{equation}
where $b$ is a positive constant. It seems very difficult to derive the exact result. So my question is: is there any approximation result? I'm doubting that the approximation result will be just a simple polynomial of the constant $b$. Note that the term $e^{-e^{-t}}$ converge to 1 very fast due to its double exponential structure. Then how to proceed?
UPDATE:I use MATLAB to plot $f_1(b)=\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt$ and $f_2(b)=\log(b)$ in the region $b\in[0, 10]$. The observation is very interesting: they agree very well except a small constant offset...So I'm strongly doubting that this integral is approximately equal to $\log(b)+constant$!
UPDATE 2: please go to Prove $\int_{-b}^{\infty}\log^{\nu}(t+b)e^{-t}e^{-e^{-t}}dt\xrightarrow{b\rightarrow\infty} \log^{\nu}(b)$
 A: We want to study the integral 

$$I(b)=\int_{-b}^{\infty}\log\left(b+x\right)e^{-x}e^{-e^{-x}}dx$$

for the two cases $b\rightarrow\infty$ and $b\rightarrow 0^+$. 
Let's start with the first one. We write
$$
I(b)=\log(b)\int_{-b}^{\infty}e^{-x}e^{-e^{-x}}+\int_{-b}^{\infty}\log\left(1+\frac{x}{b}\right)e^{-x}e^{-e^{-x}}dx
$$
the first integral is elementary. Furthermore we sub $x/b=y$ in the second. This yields
$$
I(b)=\log(b)(1-e^{-e^{b}})+b\underbrace{\int_{-1}^{\infty}\log\left(1+y\right)e^{-b y}e^{-e^{-b y}}dy}_{J(b)}
$$
Using Taylor expansion we get
$$
J(b)=\sum_{n\geq1}\frac{(-1)^n}{n}\int_{-1}^{\infty} y^n{e^{- b y}} e^{- e^{- b y}}dy
$$
by using the fact that $e^{-e^{-b y}}\le 1$ we may show that the constituents of the above sum  are bounded by terms of $\mathcal{O}(b^{-n-1})$ so to the first term will yield the dominant correction to the $\log(b)$ term.
$$
J(b)\sim\int_{-1}^{\infty} y{e^{- b y}} e^{- e^{- b y}}dy+\mathcal{O}(b^{-3})
$$
now substitue $e^{-by}=\xi$ and use the defintion of the exponential integral to show that ($\gamma$ is the Euler-Marschoni constant)
$$
J(b)\sim\frac{\gamma+b e^{-e^{b}}-\text{Ei}{(-e^b)}}{b^2}+\mathcal{O}(b^{-3})\sim \frac{\gamma}{b^2}+\mathcal{O}(b^{-3})
$$
and therefore 

$$
I(b)\sim\log(b)+\frac{\gamma}{b}+\mathcal{O}(b^{-2})\quad\text{as}\quad b\rightarrow\infty
$$


As $b\rightarrow 0$ we rewrite (because i'm in a hurry this part will be a bit more sketchy)
$$
I(b)=e^{b}\int_0^{\infty}\log(y)e^{-y}e^{-e^{b}e^{-y}}
$$
For small $b$ we might write $e^{b}=1+b+\mathcal{O(b^2)}$ 
$$
I(b)=(1+b)\int_0^{\infty}\log(y)e^{-y}e^{-e^{-y}}(1-be^{-y})+\mathcal{O}(b^2)
$$
or

$$
I(b)=C+b(C-D)+\mathcal{O}(b^2)\quad\text{as}\quad b\rightarrow 0^+
$$

where $C=\int_0^{\infty}\log(y)e^{-y}e^{-e^{-y}}$ and $D=\int_0^{\infty}\log(y)e^{-2y}e^{-e^{-y}}$ are contstants which have to determined numerically $(C\approx -0.155 ,D\approx0.262)$
A: Hint. One may write
$$
\begin{align}
\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt&\stackrel{u=t+b}{=}e^b\int_{0}^{\infty}\log(u)e^{-u}e^{-e^be^{-u}}du
\\\\&=e^b\int_{0}^{\infty}\log(u)e^{-u}\sum_{n=0}^\infty \frac{(-1)^n}{n!}e^{nb}e^{-nu}du
\\\\&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}e^{(n+1)b}\int_{0}^{\infty}\log(u)e^{-(n+1)u}du
\\\\&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}e^{(n+1)b}\left(-\frac{\gamma}{n+1}-\frac{\log(n+1)}{n+1} \right)
\\\\&=-\gamma \left(e^{-e^b}-1\right)+\sum_{n=1}^\infty \frac{(-1)^n}{n!}\cdot \log n \cdot e^{nb} \tag1
\end{align}
$$ where $\gamma$ is the Euler–Mascheroni constant and where the latter infinite sum, being a standard alternating series, satisfies
$$
\left|\sum_{n=1}^\infty\frac{(-1)^n}{n!}\cdot \log n \cdot e^{nb}\,-\,\sum_{n=1}^N\,\frac{(-1)^n}{n!}\cdot \log n \cdot e^{nb}\right|\le \left|\frac{\log N\cdot e^{Nb}}{N!}  \right| \tag2
$$ and is then easy to approximate.
From $(1)$, one may deduce that a closed form can be given in terms of the generalized Bell polynomials:

$$
\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt=-\gamma \left(e^{-e^b}-1\right)+e^{-e^b}\left.\frac{\partial}{\partial n}\text{Bell}(n,-e^b)\right|_{n=0}. \tag3
$$

