# Inverse Laplace transform of two-sided and one-sided Laplace transform

As I read in Wikipedia, there are two types of Laplace transforms

• One-sided Laplace transform: $$F(s) = \int_{0}^\infty e^{-st} f(t) dt$$

• Two-sided Laplace transform: $$F(s) = \int_{-\infty}^\infty e^{-st} f(t) dt$$

But they give only one formula for Inverse Laplace transform:

$$\hspace{3.0cm} f(t) = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{\gamma - i T}^{\gamma + i T} e^{st} F(s) ds$$

My question is that, does the type of Laplace transform I use affects the Inverse formula ?

### p.s:

I've proved the Inverse Laplace transform above corresponding to Two-sided Laplace transform using Fourier transform. But I've not come up with any idea of proving the correctness of the Inverse Laplace transform corresponding to One-sided Laplace transform.

According to my proof, the Inverse transform above is correct for One-sided transform if $$f$$ satisfies $$f(t) = 0$$ $$\forall t < 0$$. In other words,

$$\hspace{3.0cm} f(t) = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{\gamma - i T}^{\gamma + i T} e^{st} F(s) ds$$, $$\forall t \geq 0$$

The inversion formula is the same for both types of transforms:

$$f(t)=\frac{1}{2\pi j}\int_{\alpha-j\infty}^{\alpha+j\infty}F(s)e^{st}ds\tag{1}$$

The difference is in the choice of the constant $$\alpha$$. The line $$\textrm{Re}\{s\}=\alpha$$ must be inside the region of convergence (ROC). For causal functions (i.e., functions for which $$f(t)=0$$ for $$t<0$$), the ROC is to the right of the pole with the most positive real part, whereas for non-causal functions, the ROC is a vertical strip between two poles.

• Does the choice of $\alpha$ affects the result of the Inverse transform ? Jan 16, 2019 at 13:34
• @HOANGGIANG: Yes, because one function $F(s)$ can have different inverse transforms $f(t)$ depending on the ROC. Jan 16, 2019 at 13:50

I think the easiest way to answer this is just to work through an example. What is the inverse Laplace transform of, say, $$F(s) = \frac{2s + 1}{(s+2)(s-1)} = \frac{1}{s+2} + \frac{1}{s-1}?$$

It's a bit of trick question; there is no function $$f(t)$$ whose Laplace transform converges to $$F(s)$$ over the whole complex plane. The region of convergence (ROC) of a Laplace transform is always a vertical strip in the complex plane of finite or infinite width, with any strip boundaries containing singularities. In order to specify the Laplace transform enough to invert it, we need to specify which strip is the region of convergence. Then we use the usual Mellin's inverse formula $$f(t) = \frac{1}{2 \pi i}\int_{\alpha - i \infty}^{\alpha + i \infty} e^{st} F(s) ds,$$ with the integration contour contained within the specified strip containing the real part $$\alpha$$.

In this example, $$F(s)$$ has poles at $$s = -2$$ and $$s = 1$$, so there are three possible ROCs:

1. $$\mathrm{Re}(s) < -2$$,
2. $$-2 < \mathrm{Re}(s) < 1$$, or
3. $$\mathrm{Re}(s) > 1$$.

We will need the following unilateral Laplace transforms and ROCs: $$\mathcal{L}\left[ e^{a t} u(t) \right] = \frac{1}{s - a},\qquad \mathrm{Re}(s) > a,$$ $$\mathcal{L}\left[ e^{a t} u(-t) \right] = -\frac{1}{s - a},\qquad \mathrm{Re}(s) < a,$$ where $$u(t)$$ denotes the unit step function.

1. If the ROC is specified to be $$\mathrm{Re}(s) < -2$$, then we need to use the "left-unilateral" version of the Laplace transform for both terms in $$F(s)$$, and we get the inverse $$f(t) = \begin{cases} -e^{-2 t} - e^t, &t < 0\\ 0, &t > 0. \end{cases}$$

2. If the ROC is specified to be $$-2 < \mathrm{Re}(s) < 1$$, then we need to use the "right-unilateral" version of the Laplace transform for the first term in $$F(s)$$ but the "left-unilateral" version for the second term, and we get the inverse $$f(t) = \begin{cases} -e^t, &t < 0\\ e^{-2 t}, &t > 0. \end{cases}$$ Note that only in this case is the Fourier transform $$\tilde{f}(\omega)$$ of $$f(t)$$ the restriction $$\tilde{f}(\omega) = F(i \omega)$$ of the Laplace transform, because the ROC of the Laplace transform contains the imaginary axis that corresponds to the Fourier transform.

3. If the ROC is specified to be $$\mathrm{Re}(s) > 1$$, then we need to use the "right-unilateral" version of the Laplace transform for both terms in $$F(s)$$, and we get the inverse $$f(t) = \begin{cases} 0, &t < 0\\ e^{-2 t} + e^t, &t > 0. \end{cases}$$ This case corresponds to the standard right-unilateral Laplace transform.

This is not to be an answer but following the accepted answer. For non-causal functions, the ROC is a vertical strip shown below $$\sigma_1<\textrm{Re}\{s\}=\alpha<\sigma_2$$ between two poles.