I am trying to calculate the Laplace Transformation of the following function: $$f(x) = \theta(t+1)-\theta(t-1)$$

where $\theta(t)$ is the Heaviside step function defined as: $${\displaystyle \theta(x):={\begin{cases}1,&x>0\\0,&x\leq 0\end{cases}}}$$

My code to calculate this is:

f2[t_] :=  Piecewise[{{1, t < 1 && t > -1}},0]
Plot[f2[x],{x, -2, 2},Evaluated -> True]
lapl[s_] := LaplaceTransform[f2[t], t, s] 

But the output I get from Mathematica is $$\frac{1-cosh(s)+sinh(s)}{s}$$ When running the following transformation integral in wolfram alpha $$\int_{-\infty}^{\infty}(\theta(t+1)-\theta(t-1))*e^{-st}dt$$ the result I get is: $\frac{2sinh(s)}{s}$ for $\Re(s)>0$ which seems to be correct.

My question is: Why are the results different? I am using Wolfram Engine Version 13.1 in Wolfram Cloud Version 1.64.


1 Answer 1


Your integral is wrong. Laplace transform is defined from $0$ to $\infty$ not from $-\infty$ to $\infty$. The 2-sided Laplace transform is defined from $-\infty$ to $\infty$ but that is not the Mathematica command uses.

enter image description here

Mathematica result matches the integral directly:

f2[t_] := Piecewise[{{1, t < 1 && t > -1}}, 0]
LaplaceTransform[f2[t], t, s]

Mathematica graphics

Integrate[f2[t]*Exp[-s*t], {t, 0, Infinity}];
ExpToTrig[%] // FullSimplify

Mathematica graphics

  • 2
    $\begingroup$ I see I was not aware that the bilateral laplace transform was another function, thanks. $\endgroup$ Nov 29, 2022 at 3:57

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