# Properties of the Laplace Transform

## Related calculators: Laplace Transform Calculator , Inverse Laplace Transform Calculator

The Laplace transform has a number of interesting properties.

Property 1. Linearity of the Laplace transform: ${L}{\left({a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}={a}{L}{\left({f{{\left({t}\right)}}}\right)}+{L}{\left({g{{\left({t}\right)}}}\right)}$, where ${a}$ and ${b}$ are some constants.

The proof is straightforward through the definition:

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

Now, using linearity, we can write that ${\int_{{0}}^{\infty}}{\left({a}{f{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}+{b}{g{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}\right)}{d}{t}={a}{\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}+{b}\int{{e}}^{{-{s}{t}}}{g{{\left({t}\right)}}}{d}{t}={a}{L}{\left({f{{\left({t}\right)}}}\right)}+{b}{L}{\left({g{{\left({t}\right)}}}\right)}$.

This completes the proof.

Note that there can be more than 2 terms, and the property will still hold.

Example 1 . Calculate ${L}{\left({5}{{e}}^{{{3}{t}}}-{2}{\sin{{\left({7}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}$.

From the table of Laplace transforms, it is known that ${L}{\left({{e}}^{{{3}{t}}}\right)}=\frac{{1}}{{{s}-{3}}}$, ${L}{\left({\sin{{\left({7}{t}\right)}}}\right)}=\frac{{7}}{{{{s}}^{{2}}+{{7}}^{{2}}}}$, and ${L}{\left({\cos{{\left({t}\right)}}}\right)}=\frac{{s}}{{{{s}}^{{2}}+{{1}}^{{2}}}}$; so, using Property 1, we obtain the following:

${L}{\left({5}{{e}}^{{{3}{t}}}-{2}{\sin{{\left({7}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}={5}{L}{\left({{e}}^{{{3}{t}}}\right)}-{2}{L}{\left({\sin{{\left({7}{t}\right)}}}\right)}+{L}{\left({\cos{{\left({t}\right)}}}\right)}=$

$=\frac{{5}}{{{s}-{3}}}-{2}\frac{{7}}{{{{s}}^{{2}}+{49}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}=\frac{{5}}{{{s}-{3}}}-\frac{{14}}{{{{s}}^{{2}}+{49}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}$.

Property 2. Scaling by constant: ${L}{\left({f{{\left({a}{t}\right)}}}\right)}=\frac{{1}}{{s}}{F}{\left(\frac{{s}}{{a}}\right)}$, where ${a}$ is a positive constant.

The proof is straightforward:

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

Using the change of variable technique, we make the substituion ${u}={a}{t}$; then, ${d}{u}={a}{d}{t}$, which yields ${d}{t}=\frac{{{d}{u}}}{{a}}$. Since ${t}$ is changing from ${0}$ to $\infty$, we have that ${u}$ is changing from ${a}\cdot{0}={0}$ to ${a}\cdot\infty=\infty$.

So,

${\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({a}{t}\right)}}}{d}{t}={\int_{{0}}^{\infty}}{{e}}^{{-{s}\frac{{u}}{{a}}}}{f{{\left({u}\right)}}}\frac{{{d}{u}}}{{a}}=\frac{{1}}{{a}}{\int_{{0}}^{\infty}}{{e}}^{{-\frac{{s}}{{a}}{u}}}{f{{\left({u}\right)}}}{d}{u}=\frac{{1}}{{a}}{F}{\left(\frac{{s}}{{a}}\right)}$.

This completes the proof.

Property 3. ${L}{\left({t}{f{{\left({t}\right)}}}\right)}=-\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}$, where ${L}{\left({f{{\left({t}\right)}}}\right)}={F}{\left({s}\right)}$.

Proof.

From the definition, it follows that

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

Differentiating this equality with respect to ${s}$ gives (note that the limits of integration are constant):

$\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}\frac{{d}}{{{d}{s}}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}$,

or

$\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}-{t}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}$,

which finally gives

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

This completes the proof.

Example 2. Calculate ${L}{\left({t}{\sin{{\left({t}\right)}}}\right)}$.

Since ${L}{\left({\sin{{\left({t}\right)}}}\right)}=\frac{{1}}{{{{s}}^{{2}}+{1}}}$, using property 3, we obtain that

${L}{\left({t}{\sin{{\left({t}\right)}}}\right)}=-\frac{{d}}{{{d}{s}}}\frac{{1}}{{{{s}}^{{2}}+{1}}}=\frac{{1}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}\cdot\frac{{{d}{\left({{s}}^{{2}}+{1}\right)}}}{{{d}{s}}}=\frac{{{2}{s}}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}$.

Property 4. ${L}{\left(\frac{{f{{\left({t}\right)}}}}{{t}}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$, where ${F}{\left({s}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}$.

Proof.

This property follows from property 3.

Let ${g{{\left({t}\right)}}}=\frac{{f{{\left({t}\right)}}}}{{t}}$ and ${L}{\left({g{{\left({t}\right)}}}\right)}={G}{\left({s}\right)}$. We need to prove that ${G}{\left({s}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$.

So, ${f{{\left({t}\right)}}}={t}{g{{\left({t}\right)}}}$.

Taking the Laplace transform of both sides gives:

${L}{\left({f{{\left({t}\right)}}}\right)}={L}{\left({t}{g{{\left({t}\right)}}}\right)}$.

Or:

${F}{\left({s}\right)}=-\frac{{d}}{{{d}{s}}}{G}{\left({s}\right)}$ (using property 3).

Integrating this equality with respect to ${s}$ and taking the limits ${s}$ and $\infty$ yields:

${\int_{{s}}^{\infty}}-\frac{{d}}{{{d}{s}}}{G}{\left({s}\right)}{d}{s}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$.

$-{\left(\lim_{{{a}\to\infty}}{G}{\left({a}\right)}-{G}{\left({s}\right)}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$.

Since ${G}{\left({s}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{g{{\left({t}\right)}}}{d}{t}$, we have that $\lim_{{{s}\to\infty}}{G}{\left({s}\right)}={0}$.

So,

$-{\left({0}-{G}{\left({s}\right)}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$.

${G}{\left({s}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}$.

This completes the proof.

Example 3. Calculate ${L}{\left(\frac{{{\sin{{\left({t}\right)}}}}}{{t}}\right)}$.

Since ${L}{\left({\sin{{\left({t}\right)}}}\right)}=\frac{{1}}{{{{s}}^{{2}}+{1}}}$, using property 4, we obtain that

${L}{\left(\frac{{{\sin{{\left({t}\right)}}}}}{{t}}\right)}={\int_{{s}}^{\infty}}\frac{{1}}{{{{p}}^{{2}}+{1}}}{d}{p}=\lim_{{{m}\to\infty}}{\int_{{s}}^{{m}}}\frac{{1}}{{{{p}}^{{2}}+{1}}}{d}{p}=\lim_{{{m}\to\infty}}{\operatorname{atan}{{\left({p}\right)}}}{{\mid}_{{s}}^{{m}}}=$

$=\lim_{{{m}\to\infty}}{\operatorname{atan}{{\left({m}\right)}}}-{\operatorname{atan}{{\left({s}\right)}}}=\frac{\pi}{{2}}-{\operatorname{atan}{{\left({s}\right)}}}={\operatorname{acot}{{\left({s}\right)}}}$.