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.
Best Answer
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.