Can the integral
$$ \int_0^\infty \frac{\log(x)-\log(a)}{x-a}e^{-x} \mathrm{d}x $$
be expressed in terms of some simple special function? I have searched through integral tables but couldn't find anything.
I am mainly interested in values $a \geq 1$, but the integral should be well-defined for all $a>0$.
This integral may be interpreted as a Laplace transform. To see this, let $u=x/a$. Then, $$ \int_0^\infty \frac{\log{x}-\log{a}}{x-a} \mathrm{e}^{-x}\,dx = \int_0^\infty \frac{\log{u}}{u-1} \mathrm{e}^{-au}\,du.$$
In other words, if you knew the Laplace transform of $\frac{\log{x}}{x-1}$, this integral would be solved. Mathematica can do this, but the result is not helpful:
$$ e^{-a} \left(-G_{2,3}^{3,0}\left(-a\left| \begin{array}{c} 1,1 \\ 0,0,0 \\ \end{array} \right.\right)+(\log (a)+\gamma +2 i \pi ) (-\Gamma (0,-a))\right) $$
Edit: Summarizing the comments below, the branch cut that Mathematica chose apparently introduced an imaginary component. Let's fix this. First, we re-rewrite the integral as $$ \mathrm{e}^{-a}\left[\int_{-1}^0 + \int_0^\infty\right]\left\{\frac{\log(x+1)}{x}\mathrm{e}^{-ax}\,dx\right\}.$$
The second integral in that can be taken exactly by Mathematica to be
The first integral, however, let us write that as $I(a)$. Clearly, $$\frac{d^n I}{da^n} = -\int_0^1 x^{n-1}\log(1-x)\mathrm{e}^{ax}\,dx$$
We can (mostly) evaluate these when $a=0$; so let us Taylor expand $I(a)$ about $a=0$. We will obtain $$ I(a) = \frac{\pi^2}{6} + \sum_{n=1}^\infty \frac{H_n}{(n+1)!} a^n $$ where $H_n$ denotes the $n$ Harmonic number. (Partial sums of the harmonic series.)
I don't think we can improve much on this; to summarize, the integral, as originally asked, should be $$ \mathrm{e}^{-a}G_{2,3}^{3,1}\left(a\left| \begin{array}{c} 0,1 \\ 0,0,0 \\ \end{array} \right.\right) + \mathrm{e}^{-a}\left[\frac{\pi^2}{6} + \sum_{n=1}^\infty \frac{H_n}{(n+1)!} a^n\right]. $$