Possible to do better than an upper bound for$\int^{\infty}_0 e^{-x}\log(x)\ dx$? I used the series expansion of $e^{-x}$ and the fact that $\log(x)$ was less than $x$ in the $(0, \infty)$ to get an upper bound and so use simple comparison to show this was indeed integrable over $(0,\infty)$. 
Can I do any better? Definitely can't integrate by parts or use any familiar tools but can I write in the form of a summation or something?
According to Wolfram the answer is $\gamma$ the E-M constant.
Edit: Nope that's a horrific mistake: "$\log(x)$ was less than $x$"
Edit: I have realised what I said works in $(1,\infty)$. The problem is $(0,1)$ On $(1,\infty)$ $log(x) < x$ and $e^{x}=1+x+...+\frac{x^3}{6}...\geq \frac{x^3}{6}$ which is integrable.  
so $e^{-x}\log(x) \leq \frac{6}{x^2} $
we integrate that in $(1,\infty)$ to get $6$.
The other region I can't figure out
 A: You should break apart the interval and isolate positive and negative contributions. The correct value is $- \gamma$ not $\gamma$.
On $[0,1]$ we have $e^{-1} \leqslant e^{-x} \leqslant 1$ and $\log x \leqslant \log x e^{-x} \leqslant e^{-1} \log x.$
Hence,
$$\int_0^1 \log x \, dx = -1 \leqslant \int_0^1 \log x e^{-x} \, dx \leqslant - e^{-1}.$$
On $[1 , \infty)$ we have using $\log x = 2 \log \sqrt{x} < 2(\sqrt{x} -1)$
$$0 <
\int_1^\infty \log x e^{-x} \, dx \leqslant \int_1^\infty  2(\sqrt{x}-1) e^{-x}  dx = \sqrt{\pi} \text{erfc}(1) =0.278806.$$
So a first cut (that beats $6$ and at least shows that the value is negative) is
$$-1 \leqslant \int_0^\infty  \log x e^{-x}  dx  \leqslant -0.089073$$
Using the less tight bound $\log x < x -1$ we get $0$ as the upper bound for the integral.
A: First, we write the integral of interest as 
$$\int_0^\infty e^{-t}\log(t)\,dt=\int_0^x e^{-t}\log(t)\,dt+\int_x^\infty e^{-t}\log(t)\,dt \tag 1$$
Note that by integrating by parts with $u=\log(x)$ and $v=-e^{-x}$, we can write the second integral on the right-hand side of $(1)$ as
$$\int_x^\infty e^{-t}\log(t)\,dt=e^{-x}\log(x)+\int_x^\infty \frac{e^{-t}}{t}\,dt \tag 2$$

Now, in THIS ANSWER, I showed that the integral on the right-hand side of $(2)$ satisfies the bounds
$$\frac{1}{2}e^{-x}\ln\left(1+\frac{2}{x}\right)<\int_{x}^{\infty}\frac{e^{-t}}{t}dx<e^{-x}\ln\left(1+\frac{1}{x}\right)\tag 3$$

Combining $(2)$ and $(3)$ reveals
$$\frac12 e^{-x}\log(x(x+2))\le \int_x^\infty e^{-t}\log(t)\,dt\le e^{-x}\log(1+x) \tag 4$$
Letting $x=1$, we find that 
$$\bbox[5px,border:2px solid #C0A000]{\frac12 e^{-1}\log(3)\le \int_1^\infty e^{-t}\log(t)\,dt\le e^{-1}\log(2)} \tag 5$$

Next, we see that the first integral on the right-hand side of $(1)$ can be written
$$\begin{align}
\int_0^1 e^{-t}\log(t)\,dt&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^\infty t^{n} e^{-t}\,dt\\\\
&=-\sum_{n=1}^\infty \frac{(-1)^n}{n\,n!} \tag 6
\end{align}$$
We easily obtain upper and lower bounds for the rapidly convergent alternating series in $(6)$ and find that
$$\bbox[5px,border:2px solid #C0A000]{-0.796599599297056 \le \int_0^1 e^{-t}\log(t)\,dt \le -0.796599599297053} \tag 7$$
Putting together $(5)$ and $(7)$, we obtain
$$-\sum_{n=1}^\infty \frac{(-1)^n}{n\,n!}+\frac12 e^{-1}\log(3) \le \int_0^\infty e^{-t}\log(t)\,dt \le -\sum_{n=1}^\infty \frac{(-1)^n}{n\,n!}+e^{-1}\log(2)$$
Therefore, we find that the integral of interest is bounded by
$$\bbox[5px,border:2px solid #C0A000]{-0.594521161887402 \le \int_0^\infty e^{-t}\log(t)\,dt \le -0.541605001863052}
$$
