# How to pass from Laplace transform hat(Psi)_z to Psi_z?

How to pass from Laplace transform ${\stackrel{^}{\mathrm{\Psi }}}_{z}$ to ${\mathrm{\Psi }}_{z}$?
Let's define the Laplace Transform
${\stackrel{^}{\mathrm{\Psi }}}_{z}={\int }_{\mathbb{R}}{e}^{-izt}{\mathrm{\Psi }}_{t}\phantom{\rule{thinmathspace}{0ex}}dt$
and the Laplace Transform
${\stackrel{^}{\mathrm{\Phi }}}_{z}={\int }_{\mathbb{R}}{e}^{-izt}{\mathrm{\Phi }}_{z}\phantom{\rule{thinmathspace}{0ex}}dt$
If I know that
${\stackrel{^}{\mathrm{\Psi }}}_{z}={\int }_{\mathbb{R}}{e}^{-izt}{\mathrm{\Psi }}_{t}\phantom{\rule{thinmathspace}{0ex}}dt={\stackrel{^}{\mathrm{\Phi }}}_{z}\left(I-{\stackrel{^}{\mathrm{\Phi }}}_{z}{\right)}^{-1}$
where I is the identity matrix - how can I get ${\mathrm{\Psi }}_{z}?$?
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You're actually using Fourier transforms, not Laplace transforms. You're not allowed to just define the Laplace transform however you want. The inverse Fourier transform is actually easier to write down than the inverse Laplace Transform. For your definition of the Fourier transform (the non-unitary angular frequency version), its inverse is defined by
${\mathrm{\Psi }}_{z}=\frac{1}{2\pi }{\int }_{\mathbb{R}}{e}^{izt}{\stackrel{^}{\mathrm{\Psi }}}_{z}\phantom{\rule{thinmathspace}{0ex}}dz,$
making
${\mathrm{\Psi }}_{z}=\frac{1}{2\pi }{\int }_{\mathbb{R}}{e}^{izt}{\stackrel{^}{\mathrm{\Phi }}}_{z}{\left(I-{\stackrel{^}{\mathrm{\Phi }}}_{z}\right)}^{\phantom{\rule{negativethinmathspace}{0ex}}-1}\phantom{\rule{thinmathspace}{0ex}}dz.$
Further progress is not possible without knowing more.