Definition of the Laplace Transform

The Laplace transform of a function f(t){f{{\left({t}\right)}}}, defined for all t0{t}\ge{0}, is the function F(s){F}{\left({s}\right)}, defined as follows:

F(s)=L(f(t))=0estf(t)dt{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{s} is a complex parameter.

Let's go through a couple of examples.

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

L(1)=0est1dt=0estdt=lima(0aestdt)=lima(1sest0a)=lima(1sesa+1ses0)=lima(1sesa+1s){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{s}<{0}, lima(1sesa+1s)=\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{s}>{0}, lima(1sesa+1s)=1s\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(1)=1s{L}{\left({1}\right)}=\frac{{1}}{{s}}, provided that s>0{s}>{0}.

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

L(eat)=0esteatdt=0e(as)tdt=limb(0be(as)tdt)=limb(1ase(as)t0b)={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)}=

=limb(1ase(as)b1ase(as)0)=limb(1ase(as)b1as)=\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 as>0{a}-{s}>{0}, we have thatlimb(1ase(as)b1as)=\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 as<0{a}-{s}<{0}, limb(1ase(as)b1as)=1as=1sa\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(eat)=1sa{L}{\left({{e}}^{{{a}{t}}}\right)}=\frac{{1}}{{{s}-{a}}}, provided that s>a{s}>{a}.

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

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

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

L(f(t))=0estf(t)dt=lima(0aestf(t)dt){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:

lima(0aestf(t)dt)=lima(0aestdf(t))=lima(estf(t)0a0a(sestf(t))dt)=\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)}=

=lima(esaf(a)es0f(0)+s0a(estf(t))dt)==\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)}=

=lima(esaf(a)f(0))+s0(estf(t))dt=lima(esaf(a))f(0)+sL(f(t))==\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)}=

=lima(esaf(a))+sF(s)f(0)=sF(s)f(0)=\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 AA and BB, such that f(t)<Aeαt{\left|{f{{\left({t}\right)}}}\right|}<{A}{{e}}^{{\alpha{t}}} for all tB{t}\ge{B}.

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

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

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

L(0tf(τ)dτ)=0(est0tf(τ)dτ)dt=0(0test(f(τ)dτ))dt{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{t} is changing from 0{0} to \infty, while τ\tau is changing from 0{0} to t{t}. So, if we change the order of integration, we have that τ\tau is changing from 0 to \infty and t{t} is changing from τ\tau to \infty:

0(0test(f(τ)dτ))dt=0(τ(estf(τ))dt)dτ=0(f(τ)τ(est)dt)dτ=0(f(τ)(1s)estτ)dτ=0(f(τ)1sesτ)dτ=1sL(f(t))=1sF(s){\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(0tf(τ)dτ)=F(s)s{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.