# 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,

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

$$f(t)=\frac{1}{2\pi j}\int_{\alpha-j\infty}^{\alpha+j\infty}F(s)e^{st}ds\tag{1}$$
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 ? – HOANG GIANG Jan 16 '19 at 13:34
• @HOANGGIANG: Yes, because one function $F(s)$ can have different inverse transforms $f(t)$ depending on the ROC. – Matt L. Jan 16 '19 at 13:50