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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.

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make the change of variables $y=st$ and that is exactly what you get. – mike Apr 29 '12 at 15:37
up vote 11 down vote accepted

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.

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(+1) nice answer. – Mhenni Benghorbal Jul 5 '13 at 19:28
@MhenniBenghorbal Thanks for the correction! – Ayman Hourieh Jul 5 '13 at 19:33

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