2
$\begingroup$

I was looking for a way to express the Weiestrass transform of $f(t)$, $$\mathcal{W}\{f(t)\}(s)=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}f(t)\exp\left(-\frac{(s-t)^2}{4}\right)dt$$ in terms of the unilateral Laplace transform of $f(t)$, $$\mathcal{L}\{f(t)\}(s)=\int_{0}^{\infty}f(t)\exp(-st)dt$$ where $s = \sigma + i\omega$.


I have thought of proceeding in two steps:

STEP 1: Expressing $\mathcal{W}\{f(t)\}(s)$ in terms of the bilateral Laplace transform of $f(t)$, $$\mathcal{B}\{f(t)\}(s)=\int_{-\infty}^{\infty}f(t)\exp(-st)dt.$$

I have tried using the substitution $s=2w$ to get $$\begin{align}\mathcal{W}\{f(t)\}(2w)&=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}f(t)\exp\left(-\frac{4w^2-4wt+t^2}{4}\right)dt \\&=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}f(t)\exp(-w^2)\exp(wt)\exp\left(-\frac{t^2}{4}\right)dt\end{align}$$ And then substituting $z=-w$ as in step two, to get $$\begin{align} \mathcal{W}\{f(t)\}(-2z)&=\frac{\exp(-z^2)}{\sqrt{4\pi}}\int_{-\infty}^{\infty}f(t)\exp(-zt)\exp\left(-\frac{t^2}{4}\right)dt\\ &=\frac{\exp(-z^2)}{\sqrt{4\pi}}\cdot\mathcal{B}\left\{f(t)\exp\left(\frac{t^2}{4}\right)\right\}(z)\end{align}$$ $$\mathcal{W}\{f(t)\}(s)=\frac{1}{\sqrt{4\pi}}\exp\left(-\frac{s^2}{4}\right)\cdot\mathcal{B}\left\{f(t)\exp\left(\frac{t^2}{4}\right)\right\}\left(-\frac{s}{2}\right) \tag{1}$$

STEP 2: Expressing the bilateral Laplace transform in terms of the unilateral Laplace transform.

Starting from the fact that $$\int_{-\infty}^{\infty}f(t)\exp(-st)dt=\int_{0}^{\infty}f(t)\exp(-st)dt+\int_{-\infty}^{0}f(t)\exp(-st)dt$$ and, by substituting $u=-t$, $$\int_{-\infty}^{0}f(t)\exp(-st)dt=-\int_{\infty}^{0}f(u)\exp(-su)du=\int_{0}^{\infty}f(-t)\exp(st)dt=\mathcal{L}\{f(-t)\}(-s)$$ we end up with $$\mathcal{B}\{f(t)\}(s)=\mathcal{L}\{f(t)\}(s)+\mathcal{L}\{f(-t)\}(-s) \tag{2} $$


Now, here are my questions:

  • Are there any dumb mistakes anywhere?
  • How do I get the RHS of the equation $(1)$ to be some function of $\mathcal{B}\{f(t)\}(s)$?
  • Similarly, how can I express $\mathcal{L}\{f(-t)\}(-s)$ in equation $(2)$ as some function of $\mathcal{L}\{f(t)\}(s)$?

Thanks in advance!

$\endgroup$

1 Answer 1

3
$\begingroup$

The Weierstrass and Laplace transformations are related by this well known theorem:

Theorem. $f(t)$ is a Weierstrass-transformable generalized function with the strip of definition $\{s: \sigma_1 < \Re s < \sigma_2\}$ for $(\mathcal W\{f(t)\}) (s)$ if and only if $\mathrm e^{-t^2/4}f(t)$ is a Laplace-transformable generalized function with the strip of definition $\{z: -\sigma_2/2 < \Re z < -\sigma_1/2\}$ for $\big(\mathcal B\{ \mathrm e^{-t^2/4}f(t)\}\big)(z)$. In this case, $$ (\mathcal W\{f(t)\}) (s)=\frac{\mathrm e^{-s^2/4}}{\sqrt{4\pi}}\big(\mathcal B\{ \mathrm e^{-t^2/4}f(t)\}\big)\left(-\tfrac{s}{2}\right)\qquad \sigma_1 < \Re s < \sigma_2 $$ where $(\mathcal Bf(t))(s)= (\mathcal Lf(t))(s) + (\mathcal Lf(-t))(-s) $.

$\endgroup$
1
  • $\begingroup$ Any reference for this? $\endgroup$ Commented Aug 10, 2021 at 1:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .