# Definition of the Laplace Transform

The Laplace transform of a function ${f{{\left({t}\right)}}}$, defined for all ${t}\ge{0}$, is the function ${F}{\left({s}\right)}$, defined as follows:

${F}{\left({s}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}$, where ${s}$ is a complex parameter.

Let's go through a couple of examples.

Example 1. Calculate ${L}{\left({1}\right)}$.

${L}{\left({1}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}\cdot{1}{d}{t}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{d}{t}=\lim_{{{a}\to\infty}}{\left({\int_{{0}}^{{a}}}{{e}}^{{-{s}{t}}}{d}{t}\right)}=\lim_{{{a}\to\infty}}{\left(-\frac{{1}}{{s}}{{e}}^{{-{s}{t}}}{{\mid}_{{0}}^{{a}}}\right)}=\lim_{{{a}\to\infty}}{\left(-\frac{{1}}{{s}}{{e}}^{{-{s}{a}}}+\frac{{1}}{{s}}{{e}}^{{-{s}\cdot{0}}}\right)}=\lim_{{{a}\to\infty}}{\left(-\frac{{1}}{{s}}{{e}}^{{-{s}{a}}}+\frac{{1}}{{s}}\right)}$

Note that if ${s}<{0}$, $\lim_{{{a}\to\infty}}{\left(-\frac{{1}}{{s}}{{e}}^{{-{s}{a}}}+\frac{{1}}{{s}}\right)}=\infty$ and the integral is divergent.

In case ${s}>{0}$, $\lim_{{{a}\to\infty}}{\left(-\frac{{1}}{{s}}{{e}}^{{-{s}{a}}}+\frac{{1}}{{s}}\right)}=\frac{{1}}{{s}}$ and the integral is convergent.

So, ${L}{\left({1}\right)}=\frac{{1}}{{s}}$, provided that ${s}>{0}$.

Example 2. Calculate ${L}{\left({{e}}^{{{a}{t}}}\right)}$.

${L}{\left({{e}}^{{{a}{t}}}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}\cdot{{e}}^{{{a}{t}}}{d}{t}={\int_{{0}}^{\infty}}{{e}}^{{{\left({a}-{s}\right)}{t}}}{d}{t}=\lim_{{{b}\to\infty}}{\left({\int_{{0}}^{{b}}}{{e}}^{{{\left({a}-{s}\right)}{t}}}{d}{t}\right)}=\lim_{{{b}\to\infty}}{\left(\frac{{1}}{{{a}-{s}}}{{e}}^{{{\left({a}-{s}\right)}{t}}}{{\mid}_{{0}}^{{b}}}\right)}=$

$=\lim_{{{b}\to\infty}}{\left(\frac{{1}}{{{a}-{s}}}{{e}}^{{{\left({a}-{s}\right)}{b}}}-\frac{{1}}{{{a}-{s}}}{{e}}^{{{\left({a}-{s}\right)}\cdot{0}}}\right)}=\lim_{{{b}\to\infty}}{\left(\frac{{1}}{{{a}-{s}}}{{e}}^{{{\left({a}-{s}\right)}{b}}}-\frac{{1}}{{{a}-{s}}}\right)}$

Note that if ${a}-{s}>{0}$, we have that$\lim_{{{b}\to\infty}}{\left(\frac{{1}}{{{a}-{s}}}{{e}}^{{{\left({a}-{s}\right)}{b}}}-\frac{{1}}{{{a}-{s}}}\right)}=\infty$ and the integral is divergent.

In case that ${a}-{s}<{0}$, $\lim_{{{b}\to\infty}}{\left(\frac{{1}}{{{a}-{s}}}{{e}}^{{{\left({a}-{s}\right)}{b}}}-\frac{{1}}{{{a}-{s}}}\right)}=-\frac{{1}}{{{a}-{s}}}=\frac{{1}}{{{s}-{a}}}$ and the integral is convergent.

So, ${L}{\left({{e}}^{{{a}{t}}}\right)}=\frac{{1}}{{{s}-{a}}}$, provided that ${s}>{a}$.

Note that we put a restriction on ${s}$ in order to compute the Laplace transform. In general, all Laplace transforms have restrictions on ${s}$.

Let's go through the last two 'non-standard' examples.

Example 3. Calculate the Laplace transform of the derivative: ${L}{\left({f{'}}{\left({t}\right)}\right)}$.

${L}{\left({f{'}}{\left({t}\right)}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{'}}{\left({t}\right)}{d}{t}=\lim_{{{a}\to\infty}}{\left({\int_{{0}}^{{a}}}{{e}}^{{-{s}{t}}}{f{'}}{\left({t}\right)}{d}{t}\right)}$

Using integration by parts:

$\lim_{{{a}\to\infty}}{\left({\int_{{0}}^{{a}}}{{e}}^{{-{s}{t}}}{f{'}}{\left({t}\right)}{d}{t}\right)}=\lim_{{{a}\to\infty}}{\left({\int_{{0}}^{{a}}}{{e}}^{{-{s}{t}}}{d}{f{{\left({t}\right)}}}\right)}=\lim_{{{a}\to\infty}}{\left({{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{{\mid}_{{0}}^{{a}}}-{\int_{{0}}^{{a}}}{\left(-{s}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}\right)}{d}{t}\right)}=$

$=\lim_{{{a}\to\infty}}{\left({{e}}^{{-{s}{a}}}{f{{\left({a}\right)}}}-{{e}}^{{-{s}\cdot{0}}}{f{{\left({0}\right)}}}+{s}{\int_{{0}}^{{a}}}{\left({{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}\right)}{d}{t}\right)}=$

$=\lim_{{{a}\to\infty}}{\left({{e}}^{{-{s}{a}}}{f{{\left({a}\right)}}}-{f{{\left({0}\right)}}}\right)}+{s}{\int_{{0}}^{\infty}}{\left({{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}\right)}{d}{t}=\lim_{{{a}\to\infty}}{\left({{e}}^{{-{s}{a}}}{f{{\left({a}\right)}}}\right)}-{f{{\left({0}\right)}}}+{s}{L}{\left({f{{\left({t}\right)}}}\right)}=$

$=\lim_{{{a}\to\infty}}{\left({{e}}^{{-{s}{a}}}{f{{\left({a}\right)}}}\right)}+{s}{F}{\left({s}\right)}-{f{{\left({0}\right)}}}={s}{F}{\left({s}\right)}-{f{{\left({0}\right)}}}$

On this stage, we need to add a restriction: assume that there are constants $A$ and $B$, such that ${\left|{f{{\left({t}\right)}}}\right|}<{A}{{e}}^{{\alpha{t}}}$ for all ${t}\ge{B}$.

This condition implies that $\lim_{{{a}\to\infty}}{\left({{e}}^{{-{s}{a}}}{f{{\left({a}\right)}}}\right)}={0}$, provided that ${s}>\alpha$.

So, ${L}{\left({f{'}}{\left({t}\right)}\right)}={s}{L}{\left({f{{\left({t}\right)}}}\right)}-{f{{\left({0}\right)}}}$, provided that ${s}>\alpha$.

Example 4. Calculate the Laplace transform of the integral: ${L}{\left({\int_{{0}}^{{t}}}{f{{\left(\tau\right)}}}{d}\tau\right)}$.

${L}{\left({\int_{{0}}^{{t}}}{f{{\left(\tau\right)}}}{d}\tau\right)}={\int_{{0}}^{\infty}}{\left({{e}}^{{-{s}{t}}}{\int_{{0}}^{{t}}}{f{{\left(\tau\right)}}}{d}\tau\right)}{d}{t}={\int_{{0}}^{\infty}}{\left({\int_{{0}}^{{t}}}{{e}}^{{-{s}{t}}}{\left({f{{\left(\tau\right)}}}{d}\tau\right)}\right)}{d}{t}$

On this stage, we need to change the order of integration. ${t}$ is changing from ${0}$ to $\infty$, while $\tau$ is changing from ${0}$ to ${t}$. So, if we change the order of integration, we have that $\tau$ is changing from 0 to $\infty$ and ${t}$ is changing from $\tau$ to $\infty$:

${\int_{{0}}^{\infty}}{\left({\int_{{0}}^{{t}}}{{e}}^{{-{s}{t}}}{\left({f{{\left(\tau\right)}}}{d}\tau\right)}\right)}{d}{t}={\int_{{0}}^{\infty}}{\left({\int_{\tau}^{\infty}}{\left({{e}}^{{-{s}{t}}}{f{{\left(\tau\right)}}}\right)}{d}{t}\right)}{d}\tau={\int_{{0}}^{\infty}}{\left({f{{\left(\tau\right)}}}{\int_{\tau}^{\infty}}{\left({{e}}^{{-{s}{t}}}\right)}{d}{t}\right)}{d}\tau={\int_{{0}}^{\infty}}{\left({f{{\left(\tau\right)}}}\cdot{\left(-\frac{{1}}{{s}}\right)}{{e}}^{{-{s}{t}}}{{\mid}_{\tau}^{\infty}}\right)}{d}\tau={\int_{{0}}^{\infty}}{\left({f{{\left(\tau\right)}}}\frac{{1}}{{s}}{{e}}^{{-{s}\tau}}\right)}{d}\tau=\frac{{1}}{{s}}{L}{\left({f{{\left({t}\right)}}}\right)}=\frac{{1}}{{s}}{F}{\left({s}\right)}$

So, ${L}{\left({\int_{{0}}^{{t}}}{f{{\left(\tau\right)}}}{d}\tau\right)}=\frac{{{F}{\left({s}\right)}}}{{s}}$.

For a list of common Laplace transforms, see the table of Laplace transforms.