Calculation of the Inverse Laplace Transform of $\frac{1}{p}$ by contour integration.

Question

I am always told in my lessons of control engineering that the inverse Laplace Transform of $\frac{1}{p}$ is the Heaviside step function $\theta(t)$. But I have a problem when I calculate the inverse transform by its definition, which is (again according to my lessons)

$$\mathcal{L}^{-1}(F(p))=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}F(p)e^{pt}\;\mathrm{d}p $$

Here is how I would begin to handle the problem but I can't get to the end.

$$\begin{align} \mathcal{L}^{-1}\left(\frac{1}{p}\right)&=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\;\frac{1}{p}e^{pt}\;\mathrm{d}p \end{align}$$

In order to calculate this integral, I calculate the contour integration by defining the contour $C_R$ as being the half circle of radius $R$ and of diameter $[\sigma-iR, \sigma+iR]$ with $\sigma>0$. This half circle contains $p=0$ and the contour integral

$$\begin{alignat}{2} &&\frac{1}{2\pi i}\oint_{C_R}\frac{e^{pt}}{p}\;\mathrm{d}p &= \frac{1}{2\pi i}\int_{T_R} \frac{e^{pt}}{p}\;\mathrm{d}p+\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\;\frac{1}{p}e^{pt}\;\mathrm{d}p \\ &\Leftrightarrow\quad&\text{Res}_{p=0}\frac{e^{pt}}{p}&= \frac{1}{2\pi i}\int_{T_R} \frac{e^{pt}}{p}\;\mathrm{d}p+\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\;\frac{1}{p}e^{pt}\;\mathrm{d}p\\ &\Leftrightarrow \quad&1-\frac{1}{2\pi i}\int_{T_R} \frac{e^{pt}}{p}\;\mathrm{d}p&= \mathcal{L}^{-1}\left(\frac{1}{p}\right) \end{alignat}$$

where $T_R$ is the arc of the half circle but I don't really know how to show that the integral on $T_R$ is nil... I tried by noting that on $T_R$, we had $p=\sigma+Re^{i\theta} = \sigma+R\cos\theta+iR\sin\theta $ that

$$\begin{align} \lim_{R\to\infty}\int_{T_R} \frac{e^{pt}}{p}\;\mathrm{d}p &= \lim_{R\to\infty}\int_{\pi/2}^{-\pi/2} \frac{e^{[(\sigma+R\cos\theta)+iR\sin\theta]t}}{\sigma+Re^{i\theta}}iRe^{i\theta}\;\mathrm{d}\theta \\ &=\lim_{R\to\infty}\int_{\pi/2}^{-\pi/2} \frac{e^{[(\sigma+R\cos\theta)+iR\sin\theta]t}}{\sigma+Re^{i\theta}}iRe^{i\theta}\;\mathrm{d}\theta \\ &\leq \lim_{R\to\infty}\int_{\pi/2}^{-\pi/2} \frac{e^{[(\sigma+R\cos\theta)]t}}{|\sigma+Re^{i\theta}|}R\;\mathrm{d}\theta \\ \end{align}$$

and here is where I'm stuck... I don't know how to handle it. Furthermore, I don't know how the heaviside step function $\theta(t)$ appears.

Answer 1

The appearance of the Heaviside $\theta$ is due to imposing a condition of convergence on the circular arc section of the contour, $T_R$ as you call it. When $t \gt 0$, substitute $p = R e^{i \phi}$ and we have

$$\int_{T_R} dp \frac{e^{p t}}{p} = i R \int_{\pi/2}^{3 \pi/2} d\phi\, e^{i \phi} \frac{e^{R t \cos{\phi}} e^{i R t \sin{\phi}}}{i R e^{i \phi}}$$

Note that $\cos{\phi} \lt 0$ when $\phi\in (\pi/2, 3 \pi/2)$, so that the integral vanishes in the limit of $R \to \infty$ by Jordan's lemma only when $t \gt 0$. In this region, there is a pole at $p=0$ and by the residue theorem, $f(t) = 1$ here.

When $t \lt 0$, however, the integral about $T_R$ does not vanish, but blows up. Thus, we may not use this contour. Rather, we go the other way, using a new segment $S_R$ which goes to the right of $\Re{z} = \sigma$. Here, $\cos{\phi} \gt 0$ and so Jordan's lemma applies when $t \lt 0$ (and not when $t \gt 0$). In this region, there are no poles (this is why we always choose $\sigma$ to be to the right of all of the poles of $\hat{f}(p)$), so the integral over the closed contour is zero.

Therefore, the ILT of $\hat{f}(p) = 1/p$ is $f(t) = \theta(t)$.