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Definition of the Laplace Transform

Related calculator: Laplace Transform Calculator

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

F(s)=L(f(t))=∫0∞e−stf(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)=∫0∞e−st⋅1dt=∫0∞e−stdt=lim⁡a→∞(∫0ae−stdt)=lim⁡a→∞(−1se−st∣0a)=lim⁡a→∞(−1se−sa+1se−s⋅0)=lim⁡a→∞(−1se−sa+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}, lim⁡a→∞(−1se−sa+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}, lim⁡a→∞(−1se−sa+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)=∫0∞e−st⋅eatdt=∫0∞e(a−s)tdt=lim⁡b→∞(∫0be(a−s)tdt)=lim⁡b→∞(1a−se(a−s)t∣0b)={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→∞(1a−se(a−s)b−1a−se(a−s)⋅0)=lim⁡b→∞(1a−se(a−s)b−1a−s)=\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{a}-{s}>{0}, we have thatlim⁡b→∞(1a−se(a−s)b−1a−s)=∞\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{a}-{s}<{0}, lim⁡b→∞(1a−se(a−s)b−1a−s)=−1a−s=1s−a\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)=1s−a{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))=∫0∞e−stf′(t)dt=lim⁡a→∞(∫0ae−stf′(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:

lim⁡a→∞(∫0ae−stf′(t)dt)=lim⁡a→∞(∫0ae−stdf(t))=lim⁡a→∞(e−stf(t)∣0a−∫0a(−se−stf(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)}=

=lim⁡a→∞(e−saf(a)−e−s⋅0f(0)+s∫0a(e−stf(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)}=

=lim⁡a→∞(e−saf(a)−f(0))+s∫0∞(e−stf(t))dt=lim⁡a→∞(e−saf(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)}=

=lim⁡a→∞(e−saf(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 t≥B{t}\ge{B}.

This condition implies that lim⁡a→∞(e−saf(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∞(e−st∫0tf(τ)dτ)dt=∫0∞(∫0te−st(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∞(∫0te−st(f(τ)dτ))dt=∫0∞(∫τ∞(e−stf(τ))dt)dτ=∫0∞(f(τ)∫τ∞(e−st)dt)dτ=∫0∞(f(τ)⋅(−1s)e−st∣τ∞)dτ=∫0∞(f(τ)1se−sτ)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.