What is the Laplace Inverse Transform of $\ln(s)/(s(s+a))$?

Question

I've come across a term $\frac{\ln(s)}{s(s+a)}$ in a transformed differential equation.

I am trying to take the inverse transform of it, but I have no idea how to approach it.

If anyone knows how to solve this, that would be greatly appreciated.

Answer 1

Let $$F(s)=\ln s\\G(s)=\dfrac{1}{s(s+a)}$$therefore we're gonna find the Inverse Laplace Transform (ILT) of $F(s)G(s)$, also if we denote the ILTs of $F(s)$ and $G(s)$ with $f(t)$ and $g(t)$ respectively the $ILT$ of $F(s)G(s)$ is $f(t)*g(t)$ where $*$ denotes the convolution operator. Also we have$$f(t)=\dfrac{u(t)}{t}\\g(t)=\dfrac{1}{a}(1-e^{-at})u(t)$$where $u(t)$ denotes unit step function. Therefore$$f(t)*g(t)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau=\dfrac{1}{a}\int_{0}^{t}\dfrac{1-e^{-a\tau}}{t-\tau}d\tau$$which has no closed form so:$$\dfrac{1}{a}\int_{0}^{t}\dfrac{1-e^{-a\tau}}{t-\tau}d\tau=L^{-1}\left(\dfrac{\ln s}{s(s+a)}\right)$$

Answer 2

It have a closed form:

$$\mathcal{L}_s^{-1}\left[\frac{\log (s)}{s (s+a)}\right](t)=-\frac{\gamma }{a}-\frac{e^{-a t} E_1(-a t)}{a}-\frac{\log (t)}{a}+\frac{e^{-a t} \log (t)}{a}-\frac{e^{-a t} \log (-a t)}{a}$$ for: $a>0$

where: $ E_1$ is exponential integral function and $\gamma$ is Euler's constant.

Solution:

$\color{blue}{\mathcal{L}_s^{-1}\left[\frac{\log (s)}{s (s+a)}\right](t)=}\\\mathcal{L}_s^{-1}\left[\underset{n\to 0}{\text{lim}}\frac{\partial }{\partial n}\frac{s^n}{s (s+a)}\right](t)=\\\underset{n\to 0}{\text{lim}}\frac{\partial }{\partial n}\left(\mathcal{L}_s^{-1}\left[\frac{s^n}{s (s+a)}\right](t)\right)=\\\underset{n\to 0}{\text{lim}}\frac{\partial }{\partial n}\frac{e^{-a t} t^{-n} \left(-a t E_n(-a t)+n (-a t)^n \Gamma (-n)\right)}{a \Gamma (1-n)}=\\\underset{n\to 0}{\text{lim}}\frac{e^{-a t} t^{-n} \left(a t E_n(-a t) \left(\gamma -H_{-n}+\log (t)\right)+(-a t)^n \Gamma (1-n) (\log (t)-\log (-a t))-G_{2,3}^{3,0}\left(-a t\left| \begin{array}{c}1+n,1+n \\1,n,n \\\end{array}\right.\right)\right)}{a \Gamma (1-n)}=\color{blue}{\\-\frac{\gamma }{a}-\frac{e^{-a t} E_1(-a t)}{a}-\frac{\log (t)}{a}+\frac{e^{-a t} \log (t)}{a}-\frac{e^{-a t} \log (-a t)}{a}}$

EDITED:

With the same method:

$\mathcal{L}_s^{-1}\left[\frac{\log (s)}{s (s-a)}\right](t)=\frac{\gamma }{a}+\frac{e^{a t} \Gamma (0,a t)}{a}-\frac{\log (a)}{a}+\frac{e^{a t} \log (a)}{a}+\frac{\log (a t)}{a}$

and:

$\mathcal{L}_s^{-1}\left[\frac{\log (s)}{s^2}\right](t)=t-\gamma t-t \log (t)$ for: $a=0$

Answer 3

We assume that $a>0$.

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The inverse Laplace Transform of $F(s)=\frac{\log(s)}{s(s+a)}$ is given by the Bromwich integral

$$\begin{align} f(t)&\equiv\mathscr{L^{-1}}\{F\}(t)\\\\ &=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}\,ds\\\\ &=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\log(s)}{s(s+a)}\,e^{st}\,ds\\\\ &=\underbrace{\frac1{2\pi ia}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\log(s)}{s}\,e^{st}\,ds}_{=g(t)}-\underbrace{\frac1{2\pi ia}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\log(s)}{s+a}\,e^{st}\,ds}_{h(t)}\tag1 \end{align}$$

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We proceed by evaluating the first integral on the right-hand side of $(1)$ by deforming the Bromwich contour around the branch cut along the negative real axis and the pole at $s=0$. Letting $\epsilon>0$, we find

$$\begin{align} g(t)&=\frac1{2\pi i a}\left(\int_{-\epsilon}^{-\infty}\frac{\log(|s|)+i\pi}{s}\,e^{st}\,ds-\int_{-\epsilon}^{-\infty}\frac{\log(|s|)-i\pi}{s}\,e^{st}\,ds\right)\\\\ &+\frac1{2\pi i a}\int_{-\pi}^\pi \frac{\log(\epsilon e^{i\phi})}{\epsilon e^{i\phi}}\,e^{\epsilon e^{i\phi} t}\,i\epsilon e^{i\phi}\,d\phi\\\\ &=\frac1a \int_{\epsilon t}^\infty \frac{e^{-s}}{s}\,ds+\frac1a \log(\epsilon)\\\\ &=-\frac1a(\gamma+\log(t))+o(1) \end{align}$$

where $o(1)\to 0$ as $\epsilon\to 0$. Letting $\epsilon\to 0^+$ yields

$$\bbox[5px,border:2px solid #C0A000]{g(t)=-\frac1a(\gamma+\log(t))}\tag 2$$

It is straightforward to show that $\mathscr{L}\{g\}(s)=\frac{\log(s)}{as}$ .

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Next, we evaluate the second integral on the right-hand side of $(1)$ by deforming the Bromwich contour around the branch cut along the negative real axis and the pole at $s=-a$. Letting $\epsilon>0$, we find

$$\begin{align} h(t)&=\frac1{2\pi i a}\left(\int_0^{-a+\epsilon}\frac{\log(|s|)+i\pi}{s+a}\,e^{st}\,ds-\int_0^{-a+\epsilon}\frac{\log(|s|)-i\pi}{s+a}\,e^{st}\,ds\right)\\\\ &+\frac1{2\pi i a}\left(\int_{-a-\epsilon}^{-\infty}\frac{\log(|s|)+i\pi}{s+a}\,e^{st}\,ds-\int_{-a-\epsilon}^{-\infty}\frac{\log(|s|)-i\pi}{s+a}\,e^{st}\,ds\right)\\\\ &+\frac1{2\pi i a}\int_0^\pi \frac{\log(a)+i\pi +\log\left(1-\frac{\epsilon e^{i\phi}}{a}\right)}{\epsilon e^{i\phi}}\,e^{-at}e^{\epsilon e^{i\phi} t}\,i\epsilon e^{i\phi}\,ds\\\\ &+\frac1{2\pi i a}\int_{-\pi}^0 \frac{\log(a)-i\pi +\log\left(1-\frac{\epsilon e^{i\phi}}{a}\right)}{\epsilon e^{i\phi}}\,e^{-at}e^{\epsilon e^{i\phi}t}\,i\epsilon e^{i\phi}\,ds\\\\ &=-\frac{e^{-at}}a \left(\int_0^{at}\frac{e^s-1}{s}\,ds+\log(a)-\log(\epsilon)\right)\\\\ &-\frac{e^{-at}}a \left(\gamma+\log(t)+\log(\epsilon)\right)\\\\ &+\frac{e^{-at}}a \log(a)+o(1) \end{align}$$

where $o(1)\to 0$ as $\epsilon\to 0$. Letting $\epsilon\to 0^+$ yields

$$\bbox[5px,border:2px solid #C0A000]{h(t)=-\frac{e^{-at}}a\left(\gamma+\log(t)+\int_0^{at}\frac{e^s-1}{s}\,ds\right)}\tag 3$$

Using Frullani's integral, it is straightforward to show that $\mathscr{L}\{h\}(s)=\frac{\log(s)}{a(s+a)}$

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Substituting $(2)$ and $(3)$ into $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{f(t)=-\frac1a(\gamma+\log(t))+\frac{e^{-at}}a\left(\gamma+\log(t)+\int_0^{at}\frac{e^s-1}{s}\,ds\right)}$$

Answer 4

The derivation from the properties of the convolution would proceed as follows. Let $1/t$ be the distribution defined as $$\left (\frac 1 t, \phi \right) = \int_0^1 \frac {\phi(t) - \phi(0)} t dt + \int_1^\infty \frac {\phi(t)} t dt.$$ Also, let $$f(t) = \frac {1 - e^{-at}} a H(t).$$ Then $$\mathcal L^{-1}[\ln p] = -\frac 1 t - \gamma \delta(t), \\ \mathcal L^{-1}\left[ \frac 1 {p(p+a)} \right] = f(t), \\ \mathcal L^{-1}\left[ \frac {\ln p} {p(p+a)} \right] = -\frac 1 t * f(t) - \gamma \delta(t) * f(t) = \\ -\left( \frac 1 \tau, f(t - \tau) \right) - \gamma(\delta(\tau), f(t - \tau)) = \\ -\int_0^t \frac {f(t - \tau) - f(t)} \tau d\tau - \int_1^t \frac {f(t)} \tau d\tau - \gamma f(t).$$ And carrying out the integration indeed yields $$-\frac {e^{-a t} E_1(-a t) + e^{-a t} \ln(-a) + \ln t +\gamma} a,$$ valid for all real $a$, with a removable singularity at zero.