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| Meta Title | Laplace Transform | Brilliant Math & Science Wiki |
| Meta Description | 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. ... |
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| Boilerpipe Text | Aareyan Manzoor
,
Sare Gamapa
,
Tapas Mazumdar
,
and
contributed
Contents
Definition
Properties
Laplace Transform of some Popular Functions
Calculating Laplace Transform
Inverse Laplace Transform
The Convolution Theorem
Solving Differential Equation
Evaluating Improper Integrals
The Laplace transform maps a function of \(t\) to a function of \(s.\) We define
\[\mathcal{L}\left\{f\right\}\left(s\right) := \int\limits_{0}^{\infty} f(t)e^{-st}\,\text{dt}.\]
\(\mathcal{L}\left\{f+g\right\}=\mathcal{L}\left\{f\right\}+\mathcal{L}\left\{g\right\}.\)
\(\mathcal{L}\left\{cf\right\}=c\mathcal{L}\left\{f\right\}\), where \(c\) is a constant.
\(\mathcal{L}\left\{f^{(n)}\right\}=s^n\mathcal{L}\left\{f\right\}-\displaystyle\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0).\) (This is proved later in the wiki.)
Note: Here, we are transforming \(f(t),\) a function of \(t,\) into \(F(s),\) a function of \(s.\)
\(f(t)\) \(\hspace{15mm}\)
\(F(s)=\mathcal{L}\{f(t)\}\)
\(t^n\)
\(\dfrac{\Gamma(n+1)}{s^{n+1}},\) where \(\Gamma\) is the
gamma function.
\(e^{at}\)
\(\dfrac{1}{s-a}\)
\(\sin(at)\)
\(\dfrac{a}{s^2+a^2}\)
\(\cos(at)\)
\(\dfrac{s}{s^2+a^2}\)
\(e^{at}f(t)\)
\(F(s-a)\)
\(t^nf(t)\)
\((-1)^n F^{(n)}(s)\)
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
\[\begin{align}
\mathcal{L}\{\cos(at)\}
&=\mathcal{L}\left\{\dfrac{1}{2}(e^{ait}+e^{-ait})\right\}\\
&=\dfrac{1}{2}\left(\mathcal{L}\{e^{ait}\}+\mathcal{L}\{e^{-ait}\}\right)\\
&=\dfrac{1}{2}\left( \dfrac{1}{s-ai}+\dfrac{1}{s+ai}\right)\\
&=\dfrac{s}{s^2+a^2}.\ _\square
\end{align}\]
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
\[\dfrac{5}{2}e^{3t}-\dfrac{5}{2}e^t.\ _\square\]
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. \(_\square\) |
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# Laplace Transform
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**Aareyan Manzoor**, **Sare Gamapa**, **Tapas Mazumdar**, and
2 others
- **Jenna Nieminen**
- **Jimin Khim**
contributed
The ​**Laplace transform**​ is an important tool in [differential equations](https://brilliant.org/wiki/differential-equations-formulate-a-statement/ "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](https://brilliant.org/wiki/dirichlet-integral/ "Dirichlet integral").
#### Contents
- [Definition](https://brilliant.org/wiki/laplace-transform/#definition)
- [Properties](https://brilliant.org/wiki/laplace-transform/#properties)
- [Laplace Transform of some Popular Functions](https://brilliant.org/wiki/laplace-transform/#laplace-transform-of-some-popular-functions)
- [Calculating Laplace Transform](https://brilliant.org/wiki/laplace-transform/#calculating-laplace-transform)
- [Inverse Laplace Transform](https://brilliant.org/wiki/laplace-transform/#inverse-laplace-transform)
- [The Convolution Theorem](https://brilliant.org/wiki/laplace-transform/#the-convolution-theorem)
- [Solving Differential Equation](https://brilliant.org/wiki/laplace-transform/#solving-differential-equation)
- [Evaluating Improper Integrals](https://brilliant.org/wiki/laplace-transform/#evaluating-improper-integrals)
## Definition
> 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}.\\\]
## Properties
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.)
## Laplace Transform of some Popular Functions
Note: Here, we are transforming \\(f(t),\\) a function of \\(t,\\) into \\(F(s),\\) a function of \\(s.\\)
| | |
|---|---|
| \\(f(t)\\) \\(\\hspace{15mm}\\) | \\(F(s)=\\mathcal{L}\\{f(t)\\}\\) |
| \\(t^n\\) | \\(\\dfrac{\\Gamma(n+1)}{s^{n+1}},\\) where \\(\\Gamma\\) is the [gamma function.](https://brilliant.org/wiki/gamma-function/ "gamma function.") |
| \\(e^{at}\\) | \\(\\dfrac{1}{s-a}\\) |
| \\(\\sin(at)\\) | \\(\\dfrac{a}{s^2+a^2}\\) |
| \\(\\cos(at)\\) | \\(\\dfrac{s}{s^2+a^2}\\) |
| \\(e^{at}f(t)\\) | \\(F(s-a)\\) |
| \\(t^nf(t)\\) | \\((-1)^n F^{(n)}(s)\\) |
Proofs:
> 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}\\\]
> ***
> 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\\\]
## Calculating Laplace Transform
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\\\]
## Inverse Laplace Transform
> 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\\\]
> 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
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](https://brilliant.org/wiki/proving-trigonometric-identities/ "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\\\]
## Solving Differential Equation
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](https://brilliant.org/wiki/integration-by-parts/ "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](https://brilliant.org/wiki/induction/ "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](https://brilliant.org/wiki/laplace-transform/ "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\\\]
## Evaluating Improper Integrals
> \\\[\\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](https://brilliant.org/wiki/tabular-integration/ "tabular integration") is solved in less than 5 minutes with Laplace transform. \\(\_\\square\\)
**Cite as:** Laplace Transform. *Brilliant.org*. Retrieved from <https://brilliant.org/wiki/laplace-transform/>
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| Readable Markdown | **Aareyan Manzoor**, **Sare Gamapa**, **Tapas Mazumdar**, and
contributed
#### Contents
- [Definition](https://brilliant.org/wiki/laplace-transform/#definition)
- [Properties](https://brilliant.org/wiki/laplace-transform/#properties)
- [Laplace Transform of some Popular Functions](https://brilliant.org/wiki/laplace-transform/#laplace-transform-of-some-popular-functions)
- [Calculating Laplace Transform](https://brilliant.org/wiki/laplace-transform/#calculating-laplace-transform)
- [Inverse Laplace Transform](https://brilliant.org/wiki/laplace-transform/#inverse-laplace-transform)
- [The Convolution Theorem](https://brilliant.org/wiki/laplace-transform/#the-convolution-theorem)
- [Solving Differential Equation](https://brilliant.org/wiki/laplace-transform/#solving-differential-equation)
- [Evaluating Improper Integrals](https://brilliant.org/wiki/laplace-transform/#evaluating-improper-integrals)
> 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.\\)
| | |
|---|---|
| \\(f(t)\\) \\(\\hspace{15mm}\\) | \\(F(s)=\\mathcal{L}\\{f(t)\\}\\) |
| \\(t^n\\) | \\(\\dfrac{\\Gamma(n+1)}{s^{n+1}},\\) where \\(\\Gamma\\) is the [gamma function.](https://brilliant.org/wiki/gamma-function/ "gamma function.") |
| \\(e^{at}\\) | \\(\\dfrac{1}{s-a}\\) |
| \\(\\sin(at)\\) | \\(\\dfrac{a}{s^2+a^2}\\) |
| \\(\\cos(at)\\) | \\(\\dfrac{s}{s^2+a^2}\\) |
| \\(e^{at}f(t)\\) | \\(F(s-a)\\) |
| \\(t^nf(t)\\) | \\((-1)^n F^{(n)}(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}\\\]
> ***
> 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\\\]
> 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](https://brilliant.org/wiki/proving-trigonometric-identities/ "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](https://brilliant.org/wiki/integration-by-parts/ "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](https://brilliant.org/wiki/induction/ "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](https://brilliant.org/wiki/laplace-transform/ "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](https://brilliant.org/wiki/tabular-integration/ "tabular integration") is solved in less than 5 minutes with Laplace transform. \\(\_\\square\\) |
| Shard | 152 (laksa) |
| Root Hash | 15741013377293584752 |
| Unparsed URL | org,brilliant!/wiki/laplace-transform/ s443 |