How to figure of the Laplace transform for $\log x$? 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.
The Laplace transform of the power function is:

$$
\int_0^\infty e^{-st} t^a dt = \frac{\Gamma(a+1)}{s^{a+1}}
$$

Differentiate with respect to $a$ using differentiation under the integral sign:

$$
\int_0^\infty e^{-st} t^a \log{t} dt = \frac{\Gamma'(a+1) s^{a+1} - \Gamma(a+1) s^{a+1} \log{s}}{(s^{a+1})^2}
= \frac{\Gamma'(a+1) - \Gamma(a+1) \log{s}}{s^{a+1}}
$$

Now plug in $a = 0$ to get what you want.
The only mistake you have done is in the last step.You have written log(s) instead of -log(s).
Solve the last integral carefully , You are missing the negative sign and hence the variation in the answer