Laplace Transform

The ​Laplace transform​ is an important tool in differential equations, most often used for its handling of non-homogeneous differential equations. It can also be used to solve certain improper integrals like the Dirichlet integral.

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

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

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We have

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

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We have

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

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We have

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

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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. $_\square$

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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

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

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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

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

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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

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

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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

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

Find

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

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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

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

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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

$$(\sin*\cos)(t).$$

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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

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

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

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

Here make a substitution:

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

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

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

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 $f'(0)=0.$

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First take the Laplace transform of both sides:

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

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

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

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This famous integral can be proved in one line:

$$\int_0^\infty\frac{ \color{#3D99F6}{\sin(x)} }{x}\, dx=\int_{0}^{\infty}\mathcal{L}\{ {\color{#3D99F6}{\sin(x)}} \}(s)\; ds=\int_{0}^{\infty}\frac{1}{s^{2}+1}\, ds=\arctan s\bigg|_{0}^{\infty}=\dfrac{\pi}{2}.\ _\square$$

Even without this, we can solve some integrals like:

Find

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

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First,

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

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