Suppose $X$ is non-central exponentially distributed with location $k$ and rate $\lambda$. Then, what is $E(\log(X))$.
I know that for $k=0$, the answer is $-\log(\lambda) - \gamma$ where $\gamma$ is the Euler-Mascheroni constant. What about when $k > 0$?
The desired integral can be wrestled into submission by brute-force manipulations; here, we instead try to give an alternative derivation with a slightly more probabilistic flavor.
Let $X \sim \mathrm{Exp}(k,\lambda)$ be a noncentral exponential random variable with location parameter $k > 0$ and rate parameter $\lambda$. Then $X = Z + k$ where $Z \sim \mathrm{Exp}(\lambda)$.
Note that $\log(X/k) \geq 0$ and so, using a standard fact for computing the expectation of nonnegative random variables, $$ \newcommand{\e}{\mathbb E}\newcommand{\rd}{\mathrm d}\renewcommand{\Pr}{\mathbb P} \e \log(X/k) = \int_0^\infty \Pr(\log(X/k) > z)\,\rd z = \int_0^\infty \Pr(Z > k(e^z - 1)) \,\rd z \>. $$ But, $\Pr(Z > k(e^z -1)) = \exp(-\lambda k(e^z - 1))$ on $z \geq 0$ since $Z \sim \mathrm{Exp}(\lambda)$ and so $$ \e \log(X/k) = e^{\lambda k} \int_0^\infty \exp(-\lambda k e^z) \, \rd z = e^{\lambda k} \int_{\lambda k}^\infty t^{-1} e^{-t} \,\rd t \>, $$ where the last equality follows from the substitution $t = \lambda k e^z$, noting that $\rd z = \rd t / t$.
The integral on the right-hand size of the last display is just $\Gamma(0,\lambda k)$ by definition and so $$ \e \log X = e^{\lambda k} \Gamma(0,\lambda k) + \log k \>, $$ as confirmed by @Procrastinator's Mathematica computation in the comments to the question.
NB: The equivalent notation $\mathrm E_1(x)$ is also often used in place of $\Gamma(0,x)$.
