L = laplace(F) is the Laplace transform of the scalar F with default independent variable t.  If F is not a function of t, then F is  assumed to be a function of the independent variable returned by findsym(F,1).The default return is a function of s.

If F = F(s), then laplace returns a function of t.

By definition,

laplace(F,x) makes L a function of the variable x instead of the default s.

laplace(F,z,x) makes L a function of x instead of the default s. The integration is then with respect to z.

The laplace(M) function computes the element-wise Laplace transform of M.  The result, R, is formed as R[i,j] = laplace(M[i,j]).

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

$\mathrm{laplace}\left(\mathrm{BesselI}\left(0\,2t\right)\right)$

$\frac{1}{\sqrt{s^2-4}}$

$\mathrm{laplace}\left(\mathrm{BesselI}\left(0\,2s\right)\right)$

$\frac{1}{\sqrt{t^2-4}}$

$\mathrm{laplace}\left(\mathrm{BesselI}\left(0\,2t\right)\,x\right)$

$\frac{1}{\sqrt{x^2-4}}$

$\mathrm{laplace}\left(\mathrm{BesselI}\left(0\,z\cdot 2t\right)\,z\,x\right)$

$\frac{1}{\sqrt{-4t^2+x^2}}$

$M≔\mathrm{Matrix}\left(\left[\mathrm{BesselI}\left(0\,2t\right)\,\mathrm{BesselI}\left(0\,z\cdot 2t\right)\right]\right)\:$

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

$\left[\begin{array}{cc}\frac{1}{\sqrt{s^2-4}}& \frac{1}{\sqrt{s^2-4z^2}}\end{array}\right]$