1

Let $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ be a continous function such that $\exists \alpha \in \mathbb{R}$: $t \mapsto e^{\alpha t}f(t) \in L^1(\mathbb{R}_+)$. Let $\hat{f}(\lambda) := \int_0^\infty{e^{-\lambda t}f(t) dt}$ be its Laplace transform defined for $Re(\lambda) \geq \alpha$.

My question is: what natural conditions should we add on $f$ such that the inverse Laplace formula holds, that is: $$ \forall t \geq 0, \forall \sigma > \alpha, f(t) = \lim_{T \rightarrow \infty} \frac{1}{2\pi i} \int_{\sigma - i T}^{\sigma + i T}{e^{zt} \hat{f}(z) dz}$$

What are the standard references on this question?

Quentin
  • 592

1 Answers1

2

If $\int_{0}^{\infty}e^{-\alpha t}|f(t)|dt < \infty$ for some real $\alpha$, then the Laplace transform $\mathscr{L}\{f\}(\lambda)=\int_{0}^{\infty}e^{-\lambda t}f(t)dt$ is defined and continuous for $\Re\lambda \ge \alpha$, and it is holomorphic in the open region as well. The proposed inversion integral is the limit of the following as $R\uparrow\infty$, assuming $\sigma \ge \alpha$: \begin{align} & \frac{1}{2\pi i}\int_{\sigma-iR}^{\sigma+iR}e^{\lambda x}\int_{0}^{\infty}e^{-\lambda t}f(t)dt d\lambda \\ & = \int_{0}^{\infty}\frac{1}{2\pi i}\int_{\sigma-iR}^{\sigma+iR}e^{-\lambda (t-x)}d\lambda f(t)dt \\ & = \int_{0}^{\infty}\frac{1}{2\pi i}(e^{-(\sigma+iR)(t-x)}-e^{-(\sigma-iR)(t-x)})f(t)dt \\ & = \int_{0}^{\infty}\frac{1}{\pi}e^{-\sigma(t-x)}\frac{\sin(R(t-x))}{t-x}f(t)dt \\ & = e^{\sigma x}\frac{1}{\pi}\int_{0}^{\infty}\frac{\sin(R(t-x))}{t-x}e^{-\sigma t}f(t)dt. \end{align} This is a well-known integral from Fourier Analysis. Because $e^{-\alpha t}f(t)$ is absolutely integrable, so is $e^{-\sigma t}f(t)$ because $\sigma \ge \alpha$ by assumption. If $f$ is piecewise smooth, or satisfies other similar Fourier conditions, then the above converges as $R\rightarrow\infty$ to $$ e^{\sigma x} \left[e^{-\sigma x}\frac{f(x-0)+f(x+0)}{2}\right]. $$ And that's what you want. And the result does not depend on the choice of $\sigma$, provided $\sigma \ge \alpha$.

Disintegrating By Parts
  • 92,626
  • 6
  • 78
  • 173
  • Thank you for your help, that is very useful. What do you mean by $f$ being piecewise smooth? For instance, is $f$ continuous enough? If I understood correctly, the problem can be reduce to the following: if $g \in L^1(\mathbb{R})$ and $\eta_T(x) = T sinc(Tx)$, which regularity on $g$ ensure that: $\lim_{T \rightarrow \infty} (g * \eta_T) (x) = g(x) \forall x \in \mathbb{R}$, where $*$ denotes the convolution product, right? – Quentin Apr 25 '18 at 15:16
  • I have question regarding your solution or in general of the inverse Laplace formula. As you sad we have convergence of the last integral towards $\frac{f(x+0)+f(x-0)}2$, that means for $x=0$ the Laplace inversion formula converges towards $f(0)/2$? That means, for a continuous differentiable function of exponential order, the Laplace inversion formula holds everywhere except at $t=0$ ? – Salfalur Feb 03 '23 at 17:08
  • @Salfalur : $f(0+0)/2$, which is half of the right hand limit of $f$ at $0$. – Disintegrating By Parts Feb 03 '23 at 23:50