The ifourier function applies the inverse Fourier transform to M using the definition

The ifourier(M) calling sequence computes the element-wise inverse Fourier transform of M.  The result, R, is formed as R[i,j] = ifourier(M[i,j], v, u).

ifourier(F) is the inverse Fourier transform of the scalar F with default independent variable w.  If F is not a function of w, then F is  assumed to be a function of the independent variable returned by findsym(F,1). By default, the return value is a function of x.

If F = F(x), then ifourier returns a function of t. The integration above proceeds with respect to w.

ifourier(F,u) makes F a function of the variable u instead of the default x. The integration above proceeds with respect to w.

ifourier(F,v,u) takes F to be a function of v instead of the default w. The integration is then with respect to v.

$\mathrm{with}\left(\mathrm{MTM}\right)\:$

$\mathrm{ifourier}\left(\frac{3}{1+w^2}\right)$

$\frac{3\mathrm{Heaviside}\left(x\right){\ⅇ}^{-x}}{2}+\frac{3{\ⅇ}^x\mathrm{Heaviside}\left(-x\right)}{2}$

$\mathrm{ifourier}\left(\frac{3}{1+x^2}\right)$

$\frac{3\mathrm{Heaviside}\left(t\right){\ⅇ}^{-t}}{2}+\frac{3{\ⅇ}^t\mathrm{Heaviside}\left(-t\right)}{2}$

$\mathrm{ifourier}\left(\frac{3}{1+w^2}\,s\right)$

$\frac{3\mathrm{Heaviside}\left(s\right){\ⅇ}^{-s}}{2}+\frac{3{\ⅇ}^s\mathrm{Heaviside}\left(-s\right)}{2}$

$\mathrm{ifourier}\left(\frac{z\cdot 3}{1+w^2}\,z\,t\right)$

$\frac{-3\mathrm{I}\mathrm{Dirac}\left(1\,t\right)}{w^2+1}$

$M≔\mathrm{Matrix}\left(\left[\frac{3}{1+w^2}\,\frac{z\cdot 3}{1+w^2}\right]\right)\:$

$\mathrm{ifourier}\left(M\right)$

$\left[\begin{array}{cc}\frac{3\mathrm{Heaviside}\left(x\right){\ⅇ}^{-x}}{2}+\frac{3{\ⅇ}^x\mathrm{Heaviside}\left(-x\right)}{2}& \frac{3z\left(\mathrm{Heaviside}\left(x\right){\ⅇ}^{-x}+{\ⅇ}^x\mathrm{Heaviside}\left(-x\right)\right)}{2}\end{array}\right]$