Definition of the Laplace Transform

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 $$${s}$$$.

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