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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.

Pedro
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Argon
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3 Answers3

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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.

Mhenni Benghorbal
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Ayman Hourieh
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0

$$\frac{1}{s} (\int_{0}^{\infty} e^{-u}\log u\, du -\log s\int_{0}^{\infty} e^{-u}\, du)=\frac{1}{s}(\log s-\gamma)$$ I must have made a mistake here but cannot find it.

The evaluation of the integral should be positive and equal to one: $$I=\int_0^\infty e^{-u}du=\left |-e^{-u} \right |_0^\infty=1$$ $$\implies \frac {-\log s}{s}\int_0^\infty e^{-u}du=\frac {-\log s}{s}$$ $$\mathcal{L}\{\log x\}=-\frac{1}{s}(\log s+\gamma)$$

user577215664
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  • The rhs in the first formula should be $(-\gamma - \ln s)/s$. You reduced the question to evaluating $\int_{\mathbb R^+} e^{-u} \ln u , du$ by assuming $s > 0$. That integral is equal to $-\gamma$. Then you need to prove that this extends to complex $s$. – Maxim Dec 25 '19 at 08:14
  • Op made a mistake in the last line @Maxim – user577215664 Dec 25 '19 at 08:16
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    Oh, I see. You probably should clarify in some way that you're quoting the question. – Maxim Dec 25 '19 at 08:20
  • in OP 's attempt there is a mistake when he evaluated the integral. It's positive equal to one and not negative as Op evaluated. @Maxim – user577215664 Dec 25 '19 at 08:22
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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

user10316197
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