I was looking at a table of common Laplace transforms of functions when I came across the transform for $\log x$. Apparently, the transform is as follows:

$$\mathcal{L} \left\{ \log x\right\}=-\frac{1}{s}\left(\log s + \gamma\right)$$

where $\gamma$ is Euler's Constant.

Clearly, because $\gamma$ is present, the integral

$$\mathcal{L} \left\{ \log x\right\}=\int_{0}^{\infty} e^{-st}\log t dt$$

must be turned into a sum at some point. This integral as well looks very similar to

$$\Gamma'(1)=\int_{0}^{\infty} e^{-t}\log t dt=-\gamma$$

How should $\mathcal{L} \left\{ \log x\right\}$ be solved?

Here is my attempt:

Letting $u=st \Rightarrow t =\frac{1}{s}u \Rightarrow dt = \frac{du}{s}$ so we have

$$ \mathcal{L} \left\{ \log x\right\}=\int_{0}^{\infty} e^{-st}\log t \, dt= \frac{1}{s} \int_{0}^{\infty} e^{-u}\log (\frac{1}{s}u)du = \frac{1}{s} (\int_{0}^{\infty} e^{-u}\log u\, du -\log s\int_{0}^{\infty} e^{-u}\, du)=\frac{1}{s}(\log s-\gamma) $$

I must have made a mistake here but cannot find it.