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 $(C\approx -0.155 ,D\approx0.262)$
