Laplace Transform | Brilliant Math & Science Wiki (2024)

The Laplace transform maps a function of \(t\) to a function of \(s.\) We define

\[\mathcal{L}\left\{f\right\}\left(s\right) := \int\limits_{0}^{\infty} f(t)e^{-st}\,\text{dt}.\]

1. \(\mathcal{L}\left\{f+g\right\}=\mathcal{L}\left\{f\right\}+\mathcal{L}\left\{g\right\}.\)

2. \(\mathcal{L}\left\{cf\right\}=c\mathcal{L}\left\{f\right\}\), where \(c\) is a constant.

3. \(\mathcal{L}\left\{f^{(n)}\right\}=s^n\mathcal{L}\left\{f\right\}-\displaystyle\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0).\) (This is proved later in the wiki.)

Note: Here, we are transforming \(f(t),\) a function of \(t,\) into \(F(s),\) a function of \(s.\)

We have

\[\begin{align}\mathcal{L}\{t^n\}&=\int_0^\infty t^n e^{-st} \text{dt}\\&=\dfrac{1}{s^{n+1}}\int_0^\infty x^n e^{-x} \text{dx} \qquad (\text{since } x=st\implies \text{dx}=s\text{dt})\\ &=\dfrac{\Gamma(n+1)}{s^{n+1}}.\ _\square\end{align}\]

We have

\[\begin{align}\mathcal{L}\{e^{at}\}&=\int_0^\infty e^{at}e^{-st} \text{dt}\\&=\int_0^\infty e^{at-st} \text{dt}\\&=\left. \dfrac{e^{at-st}}{a-s}\right|_0^\infty \\&=\dfrac{1}{s-a}.\ _\square\end{align}\]

We have

\[\begin{align}\mathcal{L}\{\sin(at)\}&=\mathcal{L}\left\{\dfrac{-i}{2}\big(e^{ait}-e^{-ait}\big)\right\}\\&=\dfrac{-i}{2}\left(\mathcal{L}\{e^{ait}\}-\mathcal{L}\{e^{-ait}\}\right)\\&=\dfrac{-i}{2}\left( \dfrac{1}{s-ai}-\dfrac{1}{s+ai}\right)\\&=\dfrac{a}{s^2+a^2}.\ _\square\end{align}\]

We have

\[\begin{align}\mathcal{L}\{\cos(at)\}&=\mathcal{L}\left\{\dfrac{1}{2}(e^{ait}+e^{-ait})\right\}\\&=\dfrac{1}{2}\left(\mathcal{L}\{e^{ait}\}+\mathcal{L}\{e^{-ait}\}\right)\\&=\dfrac{1}{2}\left( \dfrac{1}{s-ai}+\dfrac{1}{s+ai}\right)\\&=\dfrac{s}{s^2+a^2}.\ _\square\end{align}\]

See Also

Differential Equations - Laplace Transforms7.1: Introduction to the Laplace TransformLaplace transform - MATLAB laplaceDifferential Equations - Laplace Transforms

We have

\[\mathcal{L}\left\{e^{at}f(t)\right\}=\int_0^\infty e^{at}f(t)e^{-st}\text{dt}=\int_0^\infty f(t)e^{-(s-a)t}\text{dt}=F(s-a).\]

What we have in the integral is \((s-a)\) instead of \(s\), so the function gets shifted. \(_\square\)

Consider

\[F(s)=\int_0^\infty f(t)e^{-st}\text{dt}.\]

Differentiate with respect to \(s\) \(n\) times to get

\[F^{(n)}(s)=\int_0^\infty (-t)^n f(t)e^{-st}\text{dt}\implies \mathcal{L}\{t^nf(t)\}=(-1)^n F^{(n)}(s).\ _\square\]

Some examples are shown here, which demonstrate how to calculate the Laplace transform of some given functions.

Find

\[\mathcal{L}\big\{5e^{6t}\sin(5t)+6e^{5t}\cos(7t)\big\}.\]

First, split it as two Laplace transforms:

\[5\mathcal{L}\big\{e^{6t}\sin(5t)\big\}+6\mathcal{L}\big\{e^{5t}\cos(7t)\big\}.\]

Now, we know \(\mathcal{L}\{\sin(5t)\}=\frac{5}{s^2+25}\) and \(\mathcal{L}\{\cos(7t)\}=\frac{s}{s^2+49}.\) Since the exponential function shifts the Laplace transform,

\[5\mathcal{L}\big\{e^{6t}\sin(5t)\big\}+6\mathcal{L}\big\{e^{5t}\cos(7t)\big\}=5\dfrac{5}{(s-6)^2+25}+6\dfrac{s-5}{(s-5)^2+49},\]

which is the answer. \(_\square\)

Find

\[\mathcal{L}\left\{\dfrac{\sin(t)}{t}\right\}.\]

First,

\[\mathcal{L}\left\{t^1\dfrac{\sin(t)}{t}\right\}=-\mathcal{L}\left\{\dfrac{\sin(t)}{t}\right\}'(s).\]

So,

\[\mathcal{L}\left\{\dfrac{\sin(t)}{t}\right\}=-\int \dfrac{1}{s^2+1}\text{ds}=-\arctan(s)+C.\]

A neat trick to find the \(C\) is puttin \(s=\infty\); then the \(e^{-st}\) becomes zero and the integral also becomes zero, so we have

\[\lim_{s\to\infty}-\arctan(s)+C=0\implies C=\dfrac{\pi}{2}.\]

Therefore,

\[\mathcal{L}\left\{\dfrac{\sin(t)}{t}\right\}=\dfrac{\pi}{2}-\arctan(s).\ _\square\]

If \(\mathcal{L}\{f(t)\}=F(s),\) then the inverse Laplace transform of \(F(s)\) is \(\mathcal{L}^{-1}\{F(s)\}=f(t)\).

We see some examples of how to calculate Laplace inverse.

Find

\[\mathcal{L}^{-1}\left\{\dfrac{5}{s^2-4s+3}\right\}.\]

We can ignore the constant (in this case 5) as it doesn't affect our Laplace transform much. We can factor the denominator into easy linear factors, so let's try that

\[ 5\mathcal{L}^{-1}\left\{\dfrac{1}{(s-1)(s-3)}\right\}=\dfrac{5}{2}\mathcal{L}^{-1}\left\{\dfrac{1}{s-3}-\dfrac{1}{s-1}\right\}.\]

We use the fact \(\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}\)(proved earlier), so this becomes

\[\dfrac{5}{2}e^{3t}-\dfrac{5}{2}e^t.\ _\square\]

See Also

Laplace Transform (Definition, Formula, Properties and Examples) | Laplace Transform Differential Equations

Find

\[\mathcal{L}^{-1}\left\{\dfrac{2}{(s-9)^2+1}\right\}.\]

We first note that

\[\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+1}\right\}=2\sin(t).\]

Since we are shifting the LHS by 9, we multiply by \(e^{9t}\) to get

\[\mathcal{L}^{-1}\left\{\dfrac{2}{(s-9)^2+1}\right\}=2e^{9t}\sin(t).\ _\square\]

Find

\[\mathcal{L}^{-1}\left\{\dfrac{2(s-1729)}{(s-2)^2+1}\right\}.\]

We can write this as

\[2\mathcal{L}^{-1}\left\{\dfrac{s-2}{(s-2)^2+1}\right\}-3454\mathcal{L}^{-1}\left\{\dfrac{1}{(s-2)^2+1}\right\}.\]

By doing what we did in the last problem, we have this to be equal to

\[2e^{2t}\cos(t)-3454e^{2t}\sin(t).\ _\square\]

The convolution theorem for Laplace transform is a useful tool for solving certain Laplace transforms. First, we must define convolution.

The convolution of two functions is given by

\[(f*g)(t)=\int_0^t f(t-\tau) g(\tau)\, \text{d}\tau.\]

Here is an example of convolution:

Find the convolution

\[(\sin*\cos)(t). \]

We use the definition

\[I=(\sin*\cos)(t) =\int_0^t \sin(t-\tau) \cos(\tau)\, \text{d}\tau.\]

We can use trigonometric identities to write this as

\[I=\int_{0}^{t} \big(\sin(t)\cos(\tau)-\cos(t)\cos(\tau)\big) \cos(\tau)\, \text{d}\tau=\sin(t)\int_{0}^{t} \cos^2(\tau)\, d\tau-\cos(t)\int_{0}^{t} \cos(\tau)\sin(\tau)\, d\tau.\]

Both integrals can be solved using double- or half-angle identities to get

\[\begin{align}I&=\dfrac{\sin(t)}{2}\int_{0}^{t} 1+\cos(2\tau)\, d\tau-\dfrac{\cos(t)}{2} \int_{0}^t \sin(2\tau)\, d\tau\\&=\dfrac{t\sin(t)}{2} +\dfrac{\sin(t)\sin(2t)}{4}-\dfrac{\cos(t)\big(1-\cos(2t)\big)}{4}.\end{align}\]

Plugging in double-angle identities, we will get the last two terms to cancel out, which gives us

\[I=\dfrac{t\sin(t)}{2}.\ _\square\]

This will be useful in Laplace transforms because of the convolution theorem:

The convolution theorem states that

\[\mathcal{L}(f*g)=\mathcal{L}(f)\mathcal{L}(g).\]

Start with

\[\begin{align}\mathcal{L}(f)\mathcal{L}(g)&=\int_0^\infty e^{-sx} f(x)\, dx \int_0^\infty e^{-sy} g(y)\, dy\\&= \int_0^\infty\int_0^\infty e^{-s(x+y)} f(x) g(y)\, dx\, dy. \end{align}\]

Here make a substitution:

\[\begin{align}1t=x+y \to y&=t-x\\\\dx\, dy &= dx\, dt\\t &= \Big|_0^\infty, x=\Big|_0^t. \end{align}\]

Then the integral turns into

\[ \mathcal{L}(f)\mathcal{L}(g)=\int_0^\infty\int_0^t e^{-st} f(x) g(t-x)\, dx\, dt = \mathcal{L}(f*g).\ _\square\]

We first start with the following theorem:

\[\mathcal{L}\big\{f^{(n)}(t)\big\}=s^n\mathcal{L}\{f\}-\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0)\]

We see the base case at \(n=1\) is true by using integration by parts. So assume this is true for \(n=k,\) then

\[\mathcal{L}\big\{f^{(k)}(t)\big\}=s^{k}\mathcal{L}\{f\}-\sum_{i=1}^{k}s^{n-i}f^{(i-1)}(0).\]

Consider

\[\mathcal{L}\big\{f^{(k+1)}(t)\big\}=\int_0^\infty f^{(k+1)}(t)e^{-st}\text{dt}.\]

Now consider \(\begin{cases} v'=f^{(k+1)}(t)\implies v= f^{(k)}(t)\\ u=e^{-st}\implies u'=-se^{-st}\end{cases}\), then

\[\begin{align}\int_0^\infty f^{(k+1)}(t)e^{-st}\text{dt}&=\left. e^{-st}f^{(k)}(t)\right|_0^\infty+s\int_0^\infty f^{(k)}(t)e^{-st}\text{dt}\\\\\mathcal{L}\big\{f^{(k+1)}(t)\big\}&=f^{(k)}(0)+s\mathcal{L}\big\{f^{(k)}(t)\big\}\\&=s^{k+1}\mathcal{L}\{f\}-\sum_{i=1}^{k+1}s^{n-i}f^{(i-1)}(0)\end{align}\]

by induction. Hence proved. \(_\square\)

How do we use this? FIrst note that in a simple form \(\mathcal{L}\{f'(t)\}=s\mathcal{L}\{f(t)\}-f(0)\) and \(\mathcal{L}\{f''(t)\}=s^2\mathcal{L}\{f(t)\}-sf(0)-f'(0)\) as our examples are mainly \(2^\text{nd}\) ODEs.

Solve the \(2^\text{nd}\) ODE

\[f''+2f'+2f=\sin(t)\]

with \(f(0)=0\) and \(f'(0)=0.\)

First take the Laplace transform of both sides:

\[\begin{align}\mathcal{L}\{f''(t)\}+2\mathcal{L}\{f'(t)\}+2\mathcal{L}\{f(t)\}&=\dfrac{1}{s^2+1}\\(s^2+2s+2)\mathcal{L}\{f(t)\}-(2s+1)f(0)-f'(0)&=\dfrac{1}{s^2+1}\\\\\mathcal{L}\{f(t)\} &=\dfrac{1}{(s^2+1)(s^2+2s+2)}\\&=\dfrac{1}{5(s^2+1)}-\dfrac{2s}{5(s^2+1)}+\dfrac{2s}{5\big((s+1)^2+1\big)}+\dfrac{3}{5\big((s+1)^2+1\big)} \\&=\dfrac{1}{5(s^2+1)}-\dfrac{2s}{5(s^2+1)}+\dfrac{2(s+1)}{5\big((s+1)^2+1\big)}-\dfrac{2}{5\big((s+1)^2+1\big)}+\dfrac{3}{5\big((s+1)^2+1\big)} \\&=\dfrac{1}{5(s^2+1)}-\dfrac{2s}{5(s^2+1)}+\dfrac{2(s+1)}{5\big((s+1)^2+1\big)}+\dfrac{1}{5\big((s+1)^2+1\big)}.\end{align}\]

Taking Laplace inverse,

\[f(t)=\dfrac{\sin(t)-2\cos(t)+2e^{-t}\cos(t)+e^{-t}\sin(t)}{5}.\ _\square\]

\[\int_0^\infty \dfrac{f(t)}{t}e^{-at}\text{dt}=\int_a^\infty \mathcal{L}\{f(t)\}(s)\text{ds} \implies (\text{if } a=0)\int_0^\infty \dfrac{f(t)}{t}\text{dt}=\int_0^\infty \mathcal{L}\{f(t)\}(s)\text{ds}\]

Consider the right integral:

\[\int_{0}^{\infty} \mathcal{L} \{f(t)\}(s) ds = \int_{s=0}^{\infty} \int_{t=0}^\infty f(t) e^{-st} dt.\]

Changing the order of integration and performing inner integration on variable \(s,\) we get the result

\[ \int_{t=0}^{\infty} f(t) \int_{s=0}^\infty e^{-st} dt= \int_0^\infty \frac{f(t)}{t} dt.\ _\square \]

So let's see how to apply this:

Prove the Dirichlet integral

\[\int_0^\infty \dfrac{\sin(t)}{t}\text{dt}=\dfrac{\pi}{2}.\]

This famous integral can be proved in one line:

\[\int_0^\infty\frac{ \color{blue}{\sin(x)} }{x}\, dx=\int_{0}^{\infty}\mathcal{L}\{ {\color{blue}{\sin(x)}} \}(s)\; ds=\int_{0}^{\infty}\frac{1}{s^{2}+1}\, ds=\arctan s\bigg|_{0}^{\infty}=\dfrac{\pi}{2}.\ _\square\]

Even without this, we can solve some integrals like:

Find

\[\int_0^\infty t^{1729}\sin(t)e^{-t}\text{dt}.\]

First,

\[\begin{align}\int_0^\infty t^{1729}\sin(t)e^{-t}\text{dt}&=\mathcal{L}\big\{t^{1729}\sin(t)\big\}(1)\\&=\left.\dfrac{(-1)^{1729}}{2i} \dfrac{d^{1729}}{ds^{1729}} \dfrac{1}{s+i}-\dfrac{1}{s-i}\right|_{s=1}\\&=\left. \dfrac{-1}{2i} \left( \dfrac{-1729!}{(s-i)^{1730}}-\dfrac{-1729!}{(s+i)^{1730}}\right)\right|_{s=1}\\&=\dfrac{1729!}{2^{865}}.\end{align}\]

So, a seemingly difficult integral that would have taken forever with tabular integration is solved in less than 5 minutes with Laplace transform. \(_\square\)