Let $X_1,X_2,\dots X_n$ be a random sample from the density $$f(x;\theta)=e^{-(x-\theta)} e^{-e^{-(x-\theta)}}, \quad -\infty<x<\infty ,\quad -\infty<\theta<\infty$$ Find the method of moments estimator of $\theta$.
What I attempted:-
Here $E(X)=\int_{-\infty}^{\infty} xe^{-(x-\theta)} e^{-{e^{-(x-\theta)}}}dx=\int_{-\infty}^{\infty} (x-\theta)e^{-(x-\theta)} e^{-{e^{-(x-\theta)}}}dx+\theta$.
The integral of the final expression can be written as \begin{equation}
\begin{aligned}
\int_{-\infty}^{\infty} (x-\theta)e^{-(x-\theta)} e^{-{e^{-(x-\theta)}}}dx&=
\int_{-\infty}^{\infty}t e^{-t} e^{-{e^{-t}}}dt \quad \mbox{where $t=x-\theta$}\\
&=\int_{0}^{\infty} \log y \hspace{1mm} e^{-y}dy \quad \mbox{where $y=e^{-t}$}\\
&=-\log y \hspace{1mm} e^{-y}|_0^{\infty}+\int_{0}^{\infty}\frac{e^{-y}}{y}dy
\end{aligned}
\end{equation}
I was not able to proceed beyond that. I don't think that the integral exist.
However, it can be shown that $e^{-X}$ is an exponential variate with mean $e^{-\theta}$. Is this information of any help?
Don't use integration by parts on $\int_0^\infty \log (y)e^{-y}dy$. Look closely and you'll find that integration by parts leads to the indeterminate form $-\infty +\infty$. In fact the integral $$\int_0^\infty \log( y) e^{-y}dy$$ converges; its value is well known to be the opposite of the Euler-Mascheroni constant.