Laplace transform ($\mathcal{L}$)

The Laplace transform is $\hat{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ and converts differentiation into multiplication by $p$ plus initial data.

For bounded locally integrable $f$ on the nonnegative real line, suppose its Laplace transform $F$ extends holomorphically to an open neighborhood of the closed right half-plane. Then the ordinary improper integral converges to $F(0)$. Contour damping for bounded Laplace transforms bounds the difference from the finite-interval transform by $2\Vert f{\Vert }_{\infty }/R$ in the limit for every radius $R$. Extension merely to the open right half-plane is insufficient, as the constant function and transform $1/z$ show.

The displayed kernel has residue one at zero and vanishes at the endpoints of a radius-$R$ imaginary-axis segment. On either radius-$R$ semicircle, $|1+z^2/R^2|=2|\mathrm{Re}z|/R$, cancelling the inverse real-part bound on a bounded function's Laplace transform tail. Use the full transform only on a right semicircle and a thin analytic leftward path; deform the finite-interval entire transform to the large left semicircle. This proves the Newman Tauberian theorem without assuming analytic continuation across an entire left half-disk.

For functions whose Laplace transforms converge on a common right half-plane, the causal convolution $(f\ast g)(t)={\int }_0^{t}f(s)g(t-s)ds$ satisfies $\mathcal{L}(f\ast g)=\mathcal{L}(f)\mathcal{L}(g)$. This converts a transformed boundary forcing into a known time-domain integral representation.

Under suitable regularity and growth conditions, a function with Laplace transform $F$ is recovered by the inverse contour integralwhere the vertical line lies to the right of every singularity of $F$.

A Laplace transform $G_T(p)={\int }_0^T{e}^{-ps}g(s)ds$ can grow exponentially as $\mathrm{Re}p\to -\infty$. At an evaluation time $t<T$, multiplying by $e^{pt}$ does not automatically permit closing the inverse contour to the left. Either retain the growth-bearing contour pieces, use a causal time convolution, or justify a deformation with the time-truncated transform. Resolvent-pole information alone is insufficient.

Terms singular at $s=0$ in a Laplace transform encode persistent late-time behaviour. In particular,contributes $At+B$ to the inverse transform; singularities farther into the left half-plane contribute decaying terms under the usual analyticity and growth assumptions.

Integration by parts giveswhenever the boundary term at infinity vanishes.

For $a\ge 0$, $\mathcal{L}\{f(t-a)H(t-a)\}(s)=e^{-as}\mathcal{L}\{f\}(s)$.

The delayed unit step has

The Heaviside function is zero before its switching time and one after it, allowing delayed signals to be written algebraically.

If $g$ has period $T$, then

The Laplace transform of a matrix-valued function is obtained by transforming each entry. A constant matrix may be taken outside on the corresponding side of the integral; in particular, $\mathcal{L}\{AB\}=A\mathcal{L}\{B\}$ when $A$ is constant.

Forthe derivative rule givesand hence $Y(s)=(sI-A{)}^{-1}(y_0+G(s))$ wherever the transforms converge and $sI-A$ is invertible.