# Why does the inverse Fourier transform differs from the Laplace inverse Bromwich integral?

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?

• 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. Feb 27 at 2:28

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 $$e^{-\gamma t}f(t)$$.