Tauberian Theorem: If f and its derivative are a piecwise continuous functions on [0, ∞) of some exponential order, then its Laplace transform fL satisfies
$$\label{EqLaplace.4} \lim_{\lambda \to \infty} \,\lambda\,f^L (\lambda ) = \lim_{\lambda\to \infty} \,\lambda \int_0^{\infty} f(t)\,e^{-\lambda\,t}\,{\text d}t = \lim_{t\to 0} f(t) = f(0^{+}) .$$
$$\label{EqLaplace.5} \lim_{\lambda \to 0} \,\lambda\,f^L (\lambda ) = \lim_{t\to \infty} f(t) .$$

Table of Laplace transforms

It is useful to have a “library” of Laplace transforms at hand; some common ones are listed below.
f(t)           fL(λ)
$$\displaystyle e^{at}\,f(t)$$ $$\displaystyle f^L (\lambda - a)$$
$$\displaystyle a \,e^{-ab\,t}\, f\left( a\,t \right)$$ $$\displaystyle f^L \left( \frac{\lambda}{a} + b \right)$$
$$\displaystyle e^{at}$$ 1/(λ - 𝑎)
sin ωt $$\displaystyle \frac{\omega}{\lambda^2 + \omega^2}$$
$$\displaystyle t\,\sin \omega t$$ $$\displaystyle \frac{2\omega\lambda}{\left( \lambda^2 + \omega^2 \right)^2}$$
$$\displaystyle e^{kt}\,\sin \omega t$$   $$\displaystyle \frac{\omega}{(\lambda -k )^2 + \omega^2}$$
sinh ωt   $$\displaystyle \frac{\omega}{\lambda^2 - \omega^2}$$
t sinh ωt   $$\displaystyle \frac{2\,\omega\lambda}{\left( \lambda^2 - \omega^2 \right)^2}$$
$$\displaystyle \cos \omega t - \omega t\,\sin \omega t$$ $$\displaystyle \frac{\lambda \left( \lambda^2 - \omega^2 \right)}{\left( \lambda^2 + \omega^2 \right)^2}$$
$$\displaystyle \cos \omega t + \omega t\,\sin \omega t$$ $$\displaystyle \frac{\lambda \left( \lambda^2 + 3\,\omega^2 \right)}{\left( \lambda^2 + \omega^2 \right)^2}$$
tp  $$\displaystyle \frac{\Gamma (p+1)}{\lambda^{p+1}}$$
$$\displaystyle \left( \pi \,t \right)^{-1/2}$$ $$\displaystyle \lambda^{-1/2}$$
$$\displaystyle \frac{\sin \omega t}{t}$$ $$\displaystyle \arctan \frac{\omega}{\lambda}$$
$$\displaystyle \frac{2}{t} \left( 1 - \cos \omega t \right)$$ $$\displaystyle \ln \left( 1 + \frac{\omega^2}{\lambda^2} \right)$$

f(t)         fL(λ)
$$\displaystyle f\ast g(t) = \int_0^t f(\tau )\,g(t-\tau )\,{\text d}\tau = g\ast f(t)$$ fLgL
H(t)  1/λ
δ(t)
$$\displaystyle t^p\, e^{kt}$$   $$\displaystyle \frac{\Gamma (p+1)}{(\lambda -k)^{p+1}}$$
cos ωt $$\displaystyle \frac{\lambda}{\lambda^{2} + \omega^2}$$
t cos ωt $$\displaystyle \frac{\lambda^2 - \omega^2}{\left( \lambda^{2} + \omega^2 \right)^2}$$
$$e^{kt}\cos \omega t$$ $$\displaystyle \frac{\lambda -k}{(\lambda - k)^2 + \omega^2}$$
cosh ωt $$\frac{\lambda}{\lambda^{2} - \omega^2}$$
t cosh ωt $$\frac{\lambda^2 + \omega^2}{\left( \lambda^{2} - \omega^2 \right)^2}$$
$$\displaystyle \sin \omega t - \omega t\,\cos \omega t$$ $$\displaystyle \frac{2\,\omega^3}{\left( \lambda^2 + \omega^2 \right)^2}$$
$$\displaystyle \sin \omega t + \omega t\,\cos \omega t$$ $$\displaystyle \frac{2\,\omega \lambda^2}{\left( \lambda^2 + \omega^2 \right)^2}$$
$$\displaystyle \frac{1}{t} \left( 1 - e^{-t} \right)$$ $$\displaystyle \ln \left( 1 + \frac{1}{\lambda} \right)$$
$$\displaystyle \frac{2}{t}\,\sinh at$$ $$\displaystyle \ln \frac{\lambda + a}{\lambda - a} = 2\,\mbox{arctanh} \frac{a}{\lambda}$$
$$\displaystyle \frac{2}{t} \left( 1 - \cosh at \right)$$ $$\displaystyle \ln \left( 1 - \frac{a^2}{\lambda^2} \right)$$
Here fL denotes the Laplace transform of function f(t), Γ(ν) is the gamma function of Euler, and H(t) is the Heaviside function, that is,
$f^L = {\cal L} \left[ f(t) \right] (\lambda ) = \int_0^{\infty} e^{-\lambda\,t}f(t)\,{\text d}t , \qquad \Gamma (\nu ) = \int_0^{\infty} t^{\nu -1} e^{-t} {\text d}t , \qquad H(t) = \begin{cases} 1 , & \ \mbox{ if } t > 0, \\ 1/2 , & \ \mbox{ if } t = 0, \\ 0, & \ \mbox{ if } t < 0. \end{cases}$

The error function erf is defined by

$\mbox{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} {\text d}t .$
Its Laplace transform is
${\cal L}_{t\to\lambda} \left[ \mbox{erf}(\sqrt{t}) \right] = \frac{1}{\lambda \sqrt{1+\lambda}} .$

Expressions with Bessel and Modified Bessel Functions

Original function f(t) Laplace transform fL
$$\displaystyle J_0 \left( at \right)$$ $$\displaystyle \frac{1}{\sqrt{\lambda^2 + a^2}}$$
$$\displaystyle I_0 \left( at \right)$$ $$\displaystyle \frac{1}{\sqrt{\lambda^2 - a^2}}$$
$$\displaystyle J_0 \left( 2\sqrt{at} \right)$$ $$\displaystyle \frac{1}{\lambda}\, e^{-a/\lambda}$$
$$\displaystyle I_0 \left( 2\sqrt{at} \right)$$ $$\displaystyle \frac{1}{\lambda}\, e^{a/\lambda}$$
$$\displaystyle J_{\nu} \left( at \right) , \qquad \nu > 1$$ $$\displaystyle \frac{a^{\nu}}{\sqrt{\lambda^2 + a^2} \left( \lambda + \sqrt{\lambda^2 + a^2} \right)^{\nu}}$$
$$\displaystyle I_{\nu} \left( at \right) , \qquad \nu > 1$$ $$\displaystyle \in \frac{a^{\nu}}{\sqrt{\lambda^2 - a^2} \left( \lambda + \sqrt{\lambda^2 - a^2} \right)^{\nu}}$$
$$\displaystyle t^{\nu} J_{\nu} \left( at \right) , \qquad \nu > -\frac{1}{2}$$ $$\displaystyle 2^{\nu} \pi^{-1/2} \Gamma \left( \nu + \frac{1}{2} \right) a^{\nu} \left( \lambda^2 + a^2 \right)^{-\nu -1/2}$$
$$\displaystyle t^{\nu} I_{\nu} \left( at \right)$$ $$\displaystyle 2^{\nu} \pi^{-1/2} \Gamma \left( \nu + \frac{1}{2} \right) a^{\nu} \left( \nu^2 - a^2 \right)^{-\nu -1/2}$$

Elementary Properties of the Laplace Transforms

1. Linearity: $${\cal L} \left[ \alpha\,f(t) + \beta\, g(t) \right] = \alpha\, {\cal L} \left[ f \right] + \beta\,{\cal L} \left[ g \right] = \alpha\, f^L + \beta \, g^L .$$
For arbitrary cosnatnts α and β, we have
\begin{align*} {\cal L} \left[ \alpha f(t) + \beta g(t) \right] &= \int_0^{\infty} e^{-\lambda t} \left[ \alpha f(t) + \beta g(t) \right] {\text d}t \\ &= \int_0^{\infty}\left[ \alpha f(t)\, e^{-\lambda t} + \beta g(t)\, e^{-\lambda t} \right] {\text d}t \\ &= \int_0^{\infty}\alpha f(t)\, e^{-\lambda t} {\text d}t + \int_0^{\infty} \beta g(t) \, e^{-\lambda t} {\text d}t \\ &= \alpha \int_0^{\infty} f(t)\, e^{-\lambda t} {\text d}t + \beta \int_0^{\infty} g(t) \, e^{-\lambda t} {\text d}t \\ &= \alpha {\cal L} \left[ f(t) \right] + \beta {\cal L} \left[ g(t) \right] = \alpha\,f^L + \beta \,g^L . \end{align*}
2. The derivative rule:
$$\label{EqTable.2} {\cal L} \left[ f^{(n)} (t) \right] = \lambda^n - \sum_{k=1}^n \lambda^{n-k} f^{(k-1)} (+0) .$$
In particular,
$$\label{EqTable.3} {\cal L} \left[ f' (t) \right] = {\cal L} \left[ \frac{{\text d}f (t)}{{\text d}t} \right] = \lambda\,f^L (\lambda ) - f(+0) .$$
$$\label{EqTable.4} {\cal L} \left[ f'' (t) \right] = {\cal L} \left[ \frac{{\text d}^2 f (t)}{{\text d}t^2} \right] = \lambda^2 f^L (\lambda ) - f'(+0) - \lambda\,f(+0) .$$
By mathematical induction, utilizing integration by parts
${\cal L} \left[ f' (t) \right] = \int_0^{\infty} e^{-\lambda t} f' (t)\, {\text d}t = \left. e^{-\lambda t} f (t) \right\vert_{t=0}^{t=\infty} - \lambda \int_0^{\infty} e^{-\lambda t} f (t)\, {\text d}t = -f(+0) + \lambda\,f^L (\lambda ) .$ Assuming the property holds for n = k ∈ ℤ+, we have
${\cal L} \left[ f^{(k)} (t) \right] = \lambda^k f^L - \lambda^{k-1} f(+0) - \lambda^{k-2} f' (+0) - \cdots - \lambda\,f^{(k-2)} (+0) - f^{(k-1)} (+0) .$
We now prove for n = k + 1.
\begin{align*} {\cal L} \left[ f^{(k+1)}(t) \right] &= \int_0^{\infty} e^{-\lambda t} f^{(k+1)}(t) \,{\text d}t = \lim_{A\to \infty} \int_0^A e^{-\lambda t} f^{(k+1)}(t) \,{\text d}t \\ &= \lim_{A\to \infty} \left[ \left. e^{-\lambda t} f^{(k)}(t) \right\vert_{t=0}^{t=A} + \lambda \int_0^A f^{(k)}(t) e^{-\lambda t} {\text d}t \right] \\ &= \lambda\,{\cal L} \left[ f^{(k)}(t) \right] - f^{(k)}(+0) \end{align*}
and the statement follows.
3. Convolution rule: $${\cal L} \left[ f \ast g \right] = f^L \,g^L .$$
Recall that the convolution of two functions f and g is
$$\label{EqTable.1} \left( f \ast g \right) (t) = \int_0^t f(t-\tau )\,g(\tau ) \,{\text d} \tau = \int_0^t g(t-\tau )\,f(\tau ) \,{\text d} \tau = \left( g \ast f\right) (t) .$$
Application of the Laplace transformation to the convolution integral \eqref{EqTable.1} yields
${\cal L} \left[ f*g \right] (\lambda ) = \int_0^{\infty} e^{-\lambda t} {\text d} t \int_0^{t} f(\tau )\, g(t-\tau )\, {\text d}\tau .$
Reversing the order of integration gives
${\cal L} \left[ f*g \right] (\lambda ) = \int_0^{\infty} {\text d}\tau .f(\tau ) \, e^{-\lambda \tau} \int_{\tau}^{\infty} {\text d} t\,g(t-\tau )\, e^{-\lambda (t- \tau )}$
Substitution u = −τ shows that
${\cal L} \left[ f*g \right] (\lambda ) = \int_0^{\infty} {\text d}\tau .f(\tau ) \, e^{-\lambda \tau} \int_{0}^{\infty} {\text d} u\, g(u) \, e^{-\lambda u} = f^L \cdot g^L .$
To finish the proof, we need to show that the convolution is a function-original.
4. Shift rule: $${\cal L} \left[ f(t-a)\,H(t-a) \right] = e^{-a\lambda} \,f^L (\lambda ) .$$
\begin{align*} {\cal L} \left[ H(t-a)\,f(t-a) \right] &= \int_0^{\infty} e^{-\lambda t} H(t-a)\,f(t-a)\, {\text d} t = \int_a^{\infty} e^{-\lambda t}\,f(t-a)\, {\text d} t \\ &= \int_0^{\infty} e^{-\lambda \left( \tau + a \right)} f\left( \tau \right) {\text d} \tau \end{align*}
because of substitution t = τ + 𝑎. Then
${\cal L} \left[ H(t-a)\,f(t-a) \right] = \int_0^{\infty} e^{-\lambda \left( \tau + a \right)} f\left( \tau \right) {\text d} \tau = e^{-\lambda a} f^L (\lambda ) .$
5. Similarity rule: $${\cal L} \left[ f(kt) \right] = \frac{1}{k}\, f^L \left( \frac{\lambda}{k} \right) .$$
\begin{align*} {\cal L} \left[ f(kt) \right] &= \int_0^{\infty} e^{-\lambda t} f(kt)\,{\text d}t = \frac{1}{k} \int_0^{\infty} e^{-\lambda u/k} f(u)\,{\text d}u \\ &= \frac{1}{k} \, f^L \left( \frac{\lambda}{k} \right) . \end{align*}
6. Attenuation rule: $${\cal L} \left[ e^{-at} \, f(t) \right] = f^L \left( \lambda +a \right) .$$
${\cal L} \left[ e^{-at} \, f(t) \right] = \int_0^{\infty} e^{-at} e^{-\lambda t} f(t)\, {\text d}t = \int_0^{\infty} e^{-at - \lambda t} f(t)\, {\text d}t = f^L (\lambda + a ) .$
7. Differentiation rule: $$\frac{{\text d}}{{\text d} \lambda} \, f^L (\lambda ) = - {\cal L} \left[ t\, f(t) \right] .$$ In deneral, for a positive integer n,
${\cal L} \left[ t^n f(t) \right] = \left( -1 \right)^n \frac{{\text d}^n}{{\text d}\lambda^n} \,{\cal L} \left[ f(t) \right] = \left( -1 \right)^n \frac{{\text d}^n}{{\text d}\lambda^n} \,f^L (\lambda ) .$
Using mathematical induction,
${\cal L} \left[ t\, f(t) \right] = \int_0^{\infty} t\,f(t)\,e^{-\lambda t} {\text d}t = - \int_0^{\infty} f(t)\,\frac{\text d}{{\text d}\lambda}\,e^{-\lambda t} {\text d}t = -\frac{\text d}{{\text d}\lambda} \int_0^{\infty} f(t)\,e^{-\lambda t} {\text d}t = - \frac{\text d}{{\text d}\lambda} \,f^L (\lambda ) .$
Assuming that the property holds for n = k ∈ ℤ+, then
${\cal L} \left[ t^k f(t) \right] = \left( -1 \right)^k \frac{{\text d}^k}{{\text d}\lambda^k} \,{\cal L} \left[ f(t) \right] = \left( -1 \right)^k \frac{{\text d}^n}{{\text d}\lambda^k} \,f^L (\lambda ) .$
We now prove that the property holds for n = k+1,
\begin{align*} \left( -1 \right)^{k+1} \frac{{\text d}^{k+1}}{{\text d}\lambda^{k+1}}\, f^L (\lambda ) &= \left( -1 \right) \left( -1 \right)^k \frac{{\text d}}{{\text d}\lambda} \left[ \frac{{\text d}^{k}}{{\text d}\lambda^{k}}\, f^L (\lambda ) \right] \\ &\quad \\ &= \left( -1 \right) \frac{{\text d}}{{\text d}\lambda} \left[ \left( -1 \right)^k \frac{{\text d}^{k}}{{\text d}\lambda^{k}}\, f^L (\lambda ) \right] = \left( -1 \right) \frac{{\text d}}{{\text d}\lambda} \,{\cal L} \left[ t^k f(t) \right] \\ &\quad \\ &= \left( -1 \right) \frac{{\text d}}{{\text d}\lambda} \int_0^{\infty} \left[ t^k f(t) \, e^{-\lambda t} \right] {\text d} t \\ &\quad \\ &= \int_0^{\infty} \frac{\partial}{\partial \lambda} \left[ t^k f(t)\, e^{-\lambda t} \right] {\text d} t \\ &\quad \\ &= \int_0^{\infty} e^{-\lambda t} t^{k+1} f(t) \, {\text d} t = {\cal L} \left[ t^{k+1} f(t) \right] . \end{align*}
8. Integration rule: $${\cal L} \left[ t^n \ast f(t) \right] = \frac{n!}{\lambda^{n+1}} \, f^L (\lambda ) .$$
In particular, for n = 0,
${\cal L} \left[ \int_0^t f(s)\,{\text d}s \right] = \frac{1}{\lambda}\, f^L (\lambda ) .$
The property follows from the convolution rule.
9. The Laplace transform of periodic functions.
If $$f(t) = f(t+ \omega ) ,$$ then $$\displaystyle f^L (\lambda ) = \int_0^{\infty} f(t )\,e^{-\lambda\,t} \,{\text d} t = \frac{1}{1- e^{-\omega\lambda}} \, \int_0^{\omega} \,f(t ) \,e^{-\lambda \, t} \,{\text d} t .$$
If f(t) is periodic with period ω and its Laplace transform exists, then
\begin{align*} f^L (\lambda ) &= {\cal L} \left[ f(t) \right] = \int_0^{\infty} e^{-\lambda t} f(t)\, {\text d}t \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + \int_{\omega}^{\infty} e^{-\lambda t} f(t)\, {\text d}t \qquad \mbox{make substitution at the latter} \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + \int_0^{\infty} e^{-\lambda (u+\omega )} f(u + \omega )\, {\text d}u \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + e^{-\lambda\omega} \int_0^{\infty} e^{-\lambda u} f(u)\, {\text d}u \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + e^{-\lambda\omega} f^L . \end{align*}
The required identity follows from the equation
$f^L = \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + e^{-\lambda\omega} f^L .$
10. The Laplace transform of anti-periodic functions.
If $$f(t) = -f(t+ \omega ) ,$$ then $$\displaystyle f^L (\lambda ) = \int_0^{\infty} f(t )\,e^{-\lambda\,t} \,{\text d} t = \frac{1}{1+ e^{-\omega\lambda}} \, \int_0^{\omega} \,f(t ) \,e^{-\lambda \, t} \,{\text d} t .$$
Proof is similar to the previous one.
11. Division by t:
$$\displaystyle {\cal L} \left[ \frac{f(t)}{t} \right] (\lambda ) = \int_{\lambda}^{\infty} f^L (u)\,{\text d}u .$$
\begin{align*} \int_{\lambda}^{\infty} f^L (s)\,{\text d}s &= \int_{\lambda}^{\infty} {\text d}s \int_0^{\infty} e^{-st} f(t) \,{\text d}t \\ &= \int_0^{\infty} {\text d}t \int_{\lambda}^{\infty} e^{-st} f (s)\,{\text d}s \\ &= \int_0^{\infty} {\text d}t\left[ \lim_{k\to\infty} \int_{\lambda}^{k} e^{-st} f (s)\,{\text d}s \right] \\ &= \int_0^{\infty} {\text d}t\left[ \lim_{k\to\infty} \left. f(t) \,\frac{e^{-st}}{-t} \right\vert_{s}^k \right] = \int_0^{\infty} {\text d}t \lim_{k\to\infty} \left[ e^{-st} - e^{-kt} \right] \frac{f(t)}{t} \\ &= {\cal L} \left[ \frac{1}{t}\, f(t) \right] . \end{align*}

Table of Laplace transform properties in Mathematica:

nvlpltList = {1, E^(a t), t^2, t^p, Sqrt[t], t^(n - 1/2), Sin[a t], Cos[a t], t Sin[a t], t Cos[a t](*#10*), Sin[a t] - a t Cos[a t], Sin[a t] + a t Cos[a t], Cos[a t] - a t Sin[a t], Cos[a t] + a t Sin[a t], Sin[a t + b], Cos[a t + b], Sinh[a t], Cosh[a t], E^(a t) Sin[b t], E^(a t) Cos[b t](*#20*), E^(a t) Sinh[b t], E^(a t) Cosh[b t], t^n E^(a t), f[c t], HeavisideTheta[t], DiracDelta[t - 1], HeavisideTheta[t] f[t - c], HeavisideTheta[t] g[t], E^(c t) f[t], t^n f[t](*#30*), 1/t f[t], \!$$\*SubsuperscriptBox[\($Integral]$$, $$0$$, $$t$$]$$f[ v] \[DifferentialD]v$$\), \!$$\*SubsuperscriptBox[\(\[Integral]$$, $$0$$, $$t$$]$$f[ t - \[Tau]]\ g[\[Tau]] \[DifferentialD]\[Tau]$$\), f[t + T], f'[t], f''[t], (f^(n))[t]}; lpltList = LaplaceTransform[#, t, s] & /@ invlpltList // FullSimplify; TableForm[ Table[{i, invlpltList[[i]], lpltList[[i]]}, {i, 1, Length[lpltList]}]] Definition: The full-wave rectifier of a function f(t), defined on a finite interval 0≤tT, is a periodic function with period T that is equal to f(t) on the interval [0,T]. The half-wave rectifier of a function f(t), defined on a finite interval 0≤tT, is a periodic function with period 2T that coincides with f(t) on the interval [0,T] and is identically zero on the interval [T,2T]. half = Plot[f[t], {t, 0, 4*Pi}, PlotStyle -> Thickness[0.015], AspectRatio -> 1, Axes -> False]; txt = Graphics[Text[Style["f(t)", FontSize -> 14, Red], {1.5, 0.3}]]; t0 = Graphics[Text[Style["0", FontSize -> 14, Black], {-0.1, -0.5}]]; t1 = Graphics[Text[Style["T", FontSize -> 14, Black], {3.1, -0.5}]]; t2 = Graphics[Text[Style["2T", FontSize -> 14, Black], {6.3, -0.5}]]; t3 = Graphics[Text[Style["3T", FontSize -> 14, Black], {9.35, -0.5}]]; txt2 = Graphics[Text[Style["f(t)", FontSize -> 14, Red], {7.8, 0.3}]]; Show[txt, half, t0, t1, t2, t3, txt2]  Full-wave rectifier. Half-wave rectifier. Example 1: Consider a linear function f(t) = t on the interval [0, ∞). For a positive T, its full-wave rectifier is \[ g(t) = t \left[ H(t) - H(t-T) \right] + \left( t-T \right) \left[ H(t-T) - H(t-2T) \right] + \left( t-2T \right) \left[ H(t-2T) - H(t-3T) \right] + \cdots .$
 We plot full wave rectifier of function f(t) = t: f[t_] = Piecewise[{{t, 0 < t < 1}, {t-1, 1 < t < 2}, {t - 2, 2 < t < 3}, {t-3, 3 < t < 4}, {t - 4, 4 < t < 5}}]; plot = Plot[f[t], {t, 0, 3.5}, PlotStyle -> Thickness[0.01], AspectRatio -> 1/3, Axes -> False, PlotRange -> {{-0.4, 3.8}, {-0.6, 1.5}}]; ar = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{-0.2, 0}, {3.8, 0}}]}]; ar2 = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{0, -0.2}, {0, 1.2}}]}]; tt = Graphics[{Black, Text[Style["t", 18], {3.6, 0.3}]}]; t1 = Graphics[{Black, Text[Style["T", 18], {1, -0.3}]}]; t2 = Graphics[{Black, Text[Style["2T", 18], {2, -0.3}]}]; t3 = Graphics[{Black, Text[Style["3T", 18], {3, -0.3}]}]; Show[plot, ar, ar2, tt, t1, t2, t3] Full-wave rectifier of function f = t. Mathematica code

Since the full wave rectifier is a periodic function with period T, its Laplace transform is
${\cal L} \left[ g(t) \right] = \frac{1}{1 - e^{-\lambda T}} \, \int_0^T t\,e^{-\lambda t} {\text d}t = \frac{1}{\lambda^2} - \frac{T}{1 - e^{-\lambda T}} \,\frac{1}{\lambda} \,e^{-\lambda T} .$
Assuming[ ss > 0, Integrate[t*Exp[-ss*t], {t, 0, TT}]]
(1 - E^(-ss TT) (1 + ss TT))/ss^2

Now we consider the half-wave rectifier that corresponds to the linear function. It is a 2T periodic function:
$h(t) = t \left[ H(t) - H(t-T) \right] + \left( t-2T \right) \left[ H(t-2T) - H(t-3T) \right] + \left( t-4T \right) \left[ H(t-4T) - H(t-5T) \right] + \cdots .$

 We plot full wave rectifier of function f(t) = t: f[t_] = Piecewise[{{t, 0 < t < 1}, {0, 1 < t < 2}, {t - 2, 2 < t < 3}, {0, 3 < t < 4}, {t - 4, 4 < t < 5}}]; plot = Plot[f[t], {t, 0, 5.3}, PlotStyle -> Thickness[0.01], AspectRatio -> 1/3, Axes -> False, PlotRange -> {{-0.4, 5.5}, {-0.6, 1.5}}]; ar = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{-0.2, 0}, {5.5, 0}}]}]; ar2 = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{0, -0.2}, {0, 1.2}}]}]; tt = Graphics[{Black, Text[Style["t", 18], {5.2, 0.3}]}]; t1 = Graphics[{Black, Text[Style["T", 18], {1, -0.3}]}]; t2 = Graphics[{Black, Text[Style["2T", 18], {2, -0.3}]}]; t3 = Graphics[{Black, Text[Style["3T", 18], {3, -0.3}]}]; t4 = Graphics[{Black, Text[Style["4T", 18], {4, -0.3}]}]; Show[plot, ar, ar2, tt, t1, t2, t3, t4] Half-wave rectifier of function f = t. Mathematica code

Again, using rule 9, we obtain the Laplace transform of the half-wave rectifier:
${\cal L} \left[ h(t) \right] = \frac{1}{1 - e^{-2\lambda T}} \, \int_0^T t\,e^{-\lambda t} {\text d}t = \frac{1}{1 - e^{-2\lambda T}} \,\frac{1}{\lambda^2} \left( 1 - e^{-\lambda T} \right) \left( 1 + \lambda T \right) = \frac{1}{1 + e^{-\lambda T}} \,\frac{1}{\lambda^2} \left( 1 + \lambda T \right) .$
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Example 2: The sine integral function is defined by
$\mbox{Si}(x) = \int_0^x \frac{\sin t}{t}\,{\text d}t .$
We know that the Laplace transform of the sine function is $${\cal L} \left[ \sin t \right] = (\lambda^2 + 1)^{-1} .$$ Integrating, we obtain according to rule 11 that
$\int_0^{\infty} e^{-\lambda t} \frac{\sin t}{t}\,{\text d}t = \int_{\lambda}^{\infty} {\cal L} \left[ \sin t \right] (p)\,{\text d}p = \left[ \arctan \lambda \right]_{p = \lambda}^{\lambda = \infty} = \frac{\pi}{2} - \arctan \lambda$
because $$\int \frac{{\text d}t}{t^2 +1} = \arctan t .$$ Now we have
\begin{align*} {\cal L} \left[ \mbox{Si}(t) \right] &= \int_0^{\infty} e^{-\lambda t} {\text d}t \left( \int_0^t \frac{\sin x}{x}\,{\text d}x \right) = \left[ \frac{e^{-\lambda t}}{-\lambda} \,\int_0^t \frac{\sin x}{x}\,{\text d}x\right]_{t=0}^{\infty} + \frac{1}{\lambda} \int_0^t e^{-\lambda t} \frac{\sin t}{t}\,{\text d}t \\ &= \frac{1}{\lambda} \left( \frac{\pi}{2} - \arctan \lambda \right) = \frac{1}{\lambda} \,\mbox{arccot}(\lambda ) , \qquad \Re\lambda > 1. \end{align*}
We check with Mathematica
Simplify[LaplaceTransform[SinIntegral[t], t, s]]
Unfortunately, Mathematica does not know how to present the answer in an accurate way.    ■
Example 3: Consider the scaled complementary error function
$f(t) = \mbox{erfc}\left( \frac{a}{\sqrt{t}} \right) = \frac{2}{\sqrt{\pi}} \,\int_{a/\sqrt{t}}^{\infty} e^{-x^2} {\text d}x = 1 - \mbox{erf} \left( \frac{a}{\sqrt{t}} \right) = 1 - \frac{2}{\sqrt{\pi}} \,\int_{0}^{a/\sqrt{t}} e^{-x^2} {\text d}x ,$
where 𝑎 is a positive number. We denote its Laplace transform by F(p), so
\begin{align*} F(p) &= {\cal L}_{t\to p} \left[ f(t) \right] = \frac{2}{\sqrt{\pi}} \int_0^{\infty} e^{-pt} {\text d}t \left( \int_{a/\sqrt{t}}^{\infty} e^{-x^2} {\text d}x \right) \\ &= \frac{2}{\sqrt{\pi}} \left( \frac{e^{-pt}}{-p} \left[ \int_{a/\sqrt{t}}^{\infty} e^{-x^2} {\text d}x \right]_{t=0}^{\infty} + \frac{1}{p} \,\int_0^{\infty} {\text d}t \,e^{-pt} \left( - e^{- a^2 /t} \right) \left( - \frac{a}{2}\, \frac{{\text d}t}{t^{/2}} \right) \right) \\ &= \frac{a}{p} \,\frac{1}{\sqrt{\pi}} \int_0^{\infty} e^{-pt-a^2 /t} \frac{{\text d}t}{t^{3/2}} \\ &= \frac{2a}{p\sqrt{\pi}} \int_0^{\infty} e^{-p/x^2 - (ax)^2} {\text d}x \qquad\mbox{where} \quad \frac{1}{\sqrt{t}} = x. \end{align*}
To evaluate this integral, let us denote p = b². We introduce an auxiliary integral:
$I(a,b) = \int_0^{\infty} e^{-b^2/x^2 - (ax)^2} {\text d}x ,$
so $$I(a, 0) = \sqrt{\pi} /(2a) .$$ Then
$F(p) = F \left( b^2 \right) = \frac{2a}{p\sqrt{\pi}} \, I(a,b) .$
Substitution 𝑎x = u yields
$I(a,b) = \frac{1}{b} \int_0^{\infty} e^{-u^2 - a^2 b^2 /u^2} {\text d}u .$
Also
$\frac{\text d}{{\text d}b} \,I(a,b) = -2b \int_0^{\infty} e^{-u^2 - a^2 b^2 /u^2} \frac{{\text d}u}{u^2} = -2 \int_0^{\infty} e^{-z^2 -a^2 b^2 /z^2} {\text d}z, \qquad z= \frac{b}{x} .$
Thus,
$\frac{\text d}{{\text d}b} \,I(a,b) = -2a\,I (a,b) \qquad \Longrightarrow \qquad I(a,b) = \frac{\sqrt{\pi}}{2a} \,e^{-2ab} .$
Therefore,
$F(p) = {\cal L}_{t\to p} \left[ \mbox{erfc}\left( \frac{a}{\sqrt{t}} \right) \right] = \frac{1}{p} \, e^{-2a\sqrt{p}} , \qquad \Re p > 0.$
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Example 4: The cosine integral function is
$\mbox{Ci}(t) = \int_{\infty}^t \frac{\cos x}{x} \,{\text d}x = - \int_t^{\infty} \frac{\cos x}{x} \,{\text d}x .$
Let f(t) = Ci(t). Since its derivative is
$\mbox{Ci}'(t) = f' (t) =\frac{\cos t}{t} ,$
we find its Laplace transform to be
${\cal L}_{t\to p} \left[ t\,f'(t) \right] = {\cal L}_{t\to p} \left[ \cos t\right] = \frac{p}{p^2 +1} .$
Using the derivative rule, we get
$\frac{\text d}{{\text d}p} \int_0^{\infty} e^{-pt} f' (t) \,{\text d}t = \frac{p}{p^2 +1} .$
This gives
$\frac{\text d}{{\text d}p} \left[ -f(0) + p\,F(p) \right] = \frac{p}{p^2 +1} .$
Integration yields
$p\,F(p) = p\,{\cal L}_{t\to p} \left[ f(t) \right] = p\,{\cal L}_{t\to p} \left[ \mbox{Ci}(t) \right] = \frac{1}{2}\,\ln \left( 1 + p^2 \right) + c ,$
where c is a constant of integration, which we determine using the Tauberian theorem
$\lim_{p\to 0} p\,F(p) = \lim_{t\to \infty} f(t) = c = 0.$
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