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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,

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

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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.

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

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