Inverse laplace transform of $\ln$

Question

Can someone help me with the following inverse Laplace transform, have not had trouble with any others thus far but this one is catching me

$\mathcal{L}^{-1}\{ \ln(s^3 + s) \} = ?$

Answer 1

Let $F(s) =\log(s^3+s)$ Then, certainly we can write

$$\begin{align} F(s)&=\log(s^3+s)\\\\ &=3\log(s)+\log\left(1+\frac1{s^2}\right)\\\\ &=3H(s)+G(s) \end{align}$$

where $H(s)=\log(s)$ and $G(s)=\log\left(1+\frac1{s^2}\right)$.

NOTE:

The inverse Laplace Transform, $\displaystyle \mathscr{L}^{-1}\{H\}(t)$, of $H(s)=\log(s)$ does not exist and therefore neither does the inverse Laplace Transform of $F(s)$.

To see this, we note that $\lim_{s\to \infty}\log(s)=\infty$ whereas

$$\lim_{s\to \infty}\int_0^\infty h(t)e^{-st}\,dt=0$$.

We can evaluate, however, the integral given by

$$I(t)=\int_{\sigma -i\infty}^{\sigma +i\infty}\log(s)e^{st}\,dt$$

Choosing the non-positive real axis as the branch cut, we see that

$$\begin{align} I(t)&=\frac{1}{2\pi i}\int_0^{-\infty}e^{tx}(\log(|x|)+i\pi)\,dx+\frac{1}{2\pi i}\int_{-\infty}^0 e^{tx}(\log(|x|)-i\pi)\,dx\\\\ &=\int_0^{-\infty}e^{tx}\,dx\\\\ &=-\frac1t \end{align}$$

But, the integral $\int_0^\infty \frac{e^{-st}}{t}\,dt$ fails to converge due the singularity at $t=0$.

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We can evaluate the inverse Laplace Transform, $\mathscr{L}^{-1}\{G\}(t)$, indirectly by noting that

$$G'(s)=\color{blue}{\frac{2s}{s^2+1}}-\color{red}{\frac2s}$$

Therefore, we find that

$$\mathscr{L}^{-1}\{G'\}(t)=\color{blue}{2\cos(t)}-\color{red}{2}$$

Inasmuch as , $-\mathscr{L}\{t\cdot g\}(s)=-\int_0^\infty t\,g(t)\,e^{-st}\,dt=G'(s)$, we find immediately that

$$\bbox[5px,border:2px solid #C0A000]{G(s)=2\left(\frac{1-\cos(t)}{t}\right)}$$

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We can also calculate the inverse Laplace Transform of $G$ directly. For $t>0$, we have

$$\begin{align} \mathscr{L}^{-1}\{G\}(t)&=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}G(s)e^{st}\,ds\\\\ &=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\left(1+\frac1{s^2}\right)e^{st}\,ds\\\\ &=\text{Res}\left(\log\left(1+\frac1{s^2}\right)e^{st},s=\infty\right)\\\\ &=\text{Res}\left(\frac1{s^2}\log\left(1+s^2\right)e^{(1/s)t},s=0\right)\\\\ \end{align}$$

Expanding $\frac1{s^2}\log\left(1+s^2\right)e^{(1/s)t}$ is a Laurent Series around $s=0$ yields

$$\frac1{s^2}\log\left(1+s^2\right)e^{(1/s)t}=\left(\sum_{n=0}^\infty \frac{(-1)^{n}}{n+1}s^{2n}\right)\left(\sum_{m=0}^\infty \frac{t^m}{m!\,s^m}\right)$$

It is straightforward to show that the coefficient, $a_{-1}$, on the term $s^{-1}$ is given by

$$\begin{align} a_{-1}&=2\sum_{n=0}^\infty \frac{(-1)^{n}t^{2n+1}}{(2n+2)!}\\\\ &=2\left(\frac{1-\cos(t)}{t}\right) \end{align}$$

as was to be shown!

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Putting it all together, we see that

$$\mathscr{L}^{-1}\{F\}=-\frac{1+2\cos(t)}{t}$$

but $\int_0^\infty \left(\frac{1+2\cos(t)}{t}\right)\,e^{-st}\,dt$ does not exist. Therefore, $\log(s^3+s)$ is not a Laplace Transform.