0
$\begingroup$

This might be a repeated question, but I am looking for a more in depth explanation for the relation between inverse Fourier and Laplace transforms. We all know that the inverse Laplace transform is

\begin{equation} f(t) = \frac{1}{2\pi j} \int_{\gamma-j\infty}^{\gamma+j\infty} F(s) e^{st} ds \end{equation}

On the other hand, the inverse Fourier transform is

\begin{equation} f(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(\omega) e^{j\omega t} d\omega \end{equation}

It is easy to substitute $s = j\omega $ into the inverse Laplace transform and obtain a similar inverse Fourier transform like this

\begin{equation} f(t) = \frac{1}{2\pi} \int_{-\infty-j\gamma}^{+\infty-j\gamma} F(\omega) e^{j\omega t} d\omega \end{equation}

However, I have always seen $\gamma=0$, just like the second expression from top. Am I missing something here? Can someone explain why it is not possible to move between the two transforms directly?

$\endgroup$
1
  • $\begingroup$ The LT is a generalized FT. The $\gamma$ term, the real part of $s$, gives us the possibility to shift our line in the complex plane that's parallel to the imaginary axis. $\endgroup$
    – vitamin d
    Feb 27 at 2:28
0
$\begingroup$

In the domain of convergence of $$F(s)=\int_{-\infty}^\infty f(t)e^{-st}dt$$ (for simplicity assuming the integrals converge absolutely, although this is not necessary: convergence in $L^2$ sense, in the sense of distributions..)

then $F(\gamma+i.)$ is the Fourier transform of $e^{-\gamma t}f(t)$ and $$\frac{1}{2i\pi } \int_{\gamma-i\infty}^{\gamma+i\infty} F(s) e^{st} ds=\frac{1}{2i\pi } \int_{-\infty}^{\infty} F(\gamma+i\omega) e^{(\gamma+i\omega)t} di\omega $$ is $e^{\gamma t}$ times the inverse Fourier transform of the Fourier transform of $e^{-\gamma t}f(t)$.

Does it answer your question?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.