nyuuzyou/wikis huggingface dataset:
Laplace Transform Determination/Use of Tables

== Solution Technique for Laplace Transform ==
To find the Laplace transform of a function $f$, one can evaluate it using the following technique:
:$(1): \quad$ Inspect a table of established Laplace transforms for something similar
:$(2): \quad$ Use whatever miscellaneous methods may be useful.
== Sources ==
* : Chapter $1$: The Laplace Transform: Methods of Finding Laplace Transforms: $6$. Use of Tables
Category:Laplace Transform Determination
Laplace Transform Exists if Function Piecewise Continuous and of Exponential Order

== Theorem ==
Let $f$ be a real function which is:
:piecewise continuous in every closed interval $\closedint 0 N$
:of exponential order $\gamma$ for $t > N$.
Then the Laplace transform $\map F s$ of $\map f t$ exists for all $s > \gamma$.
== Proof ==
For all $N \in \Z_{>0}$:
:$\ds \int_0^\infty e^{-s t} \map f t \rd t = \int_0^N e^{-s t} \map f t \rd t + \int_N^\infty e^{-s t} \map f t \rd t$
We have that $f$ is piecewise continuous in every closed interval $\closedint 0 N$.
Hence the first of the integrals on the exists.
Also, as $\map f t$ is of exponential order $\gamma$ for $t > N$, so does the second integral on the .
Indeed:
Thus the Laplace transform exists for $s > \gamma$.
== Also see ==
* Laplace Transform of Reciprocal of Square Root for a real function which is not of exponential order but which does have a Laplace transform
== Sources ==
* : Chapter $1$: The Laplace Transform: Sufficient Conditions for Existence of Laplace Transforms: Theorem $1 \text{-} 1$
* : Chapter $1$: The Laplace Transform: Solved Problems: Miscellaneous Problems: $47$
Category:Laplace Transforms
Laplace Transform of 1

== Theorem ==
Let $f: \R \to \R$ be the function defined as:
:$\forall t \in \R: \map f t = 1$
Then the Laplace transform of $\map f t$ is given by:
:$\laptrans {\map f t} = \dfrac 1 s$
for $\map \Re s > 0$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of some Elementary Functions: $1$.
* : $\S 32$: Table of Special Laplace Transforms: $32.25$
Category:Laplace Transforms
Category:Laplace Transform of 1
Laplace Transform of 1/Proof 2

== Theorem ==
== Proof ==
From Laplace Transform of Derivative:
:$(1): \quad \laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
under suitable conditions.
Then:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transform of Derivative: $15 \ \text{(a)}$
Category:Laplace Transform of 1
Laplace Transform of Bessel Function of the First Kind

== Theorem ==
Let $J_n$ denote the Bessel function of the first kind of order $n$.
Then the Laplace transform of $J_n$ is given as:
:$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $3$
Category:Laplace Transforms of Bessel Functions
Category:Laplace Transform of Bessel Function of the First Kind
Laplace Transform of Bessel Function of the First Kind of Order One

== Theorem ==
Let $J_1$ denote the Bessel function of the first kind of order $1$.
Then the Laplace transform of $J_1$ is given as:
:$\laptrans {\map {J_1} t} = \dfrac {\sqrt {s^2 + 1} - s} {\sqrt {s^2 + 1} }$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Bessel Functions: $35$
Category:Laplace Transforms of Bessel Functions
Laplace Transform of Bessel Function of the First Kind of Order Zero

== Theorem ==
Let $J_0$ denote the Bessel function of the first kind of order $0$.
Then the Laplace transform of $J_0$ is given as:
:$\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Laplace Transforms of Bessel Functions
Category:Laplace Transform of Bessel Function of the First Kind of Order Zero
Laplace Transform of Bessel Function of the First Kind of Order Zero/Corollary

== Corollary to Laplace Transform of Bessel Function of the First Kind of Order Zero ==
Let $J_0$ denote the Bessel function of the first kind of order $0$.
Then the Laplace transform of $\map {J_0} {a t}$ is given as:
:$\laptrans {\map {J_0} {a t} } = \dfrac 1 {\sqrt {s^2 + a^2} }$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $2$
* : Chapter $1$: The Laplace Transform: Solved Problems: Bessel Functions: $34 \ \text{(b)}$
Category:Laplace Transform of Bessel Function of the First Kind of Order Zero
Laplace Transform of Bessel Function of the First Kind of Order Zero/Proof 2

== Theorem ==
== Proof ==
From Bessel Function of the First Kind of Order Zero:
Hence:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Bessel Functions: $34 \ \text{(a)}$
Category:Laplace Transform of Bessel Function of the First Kind of Order Zero
Laplace Transform of Bessel Function of the First Kind of Order Zero/Proof 3

== Theorem ==
== Proof ==
By definition of Bessel function of the first kind, $\map {J_0} t$ satisfies Bessel's equation:
From Laplace Transform of Derivative:
:$\laptrans {\map {J_0'} t} = s \laptrans {\map {J_0} t} - \map {J_0} 0$
and from Laplace Transform of Second Derivative:
:$\laptrans {\map {J_0} t} = s^2 \laptrans {\map {J_0} t} - s \, \map {J_0} 0 - \map {J_0'} 0$
From Bessel Function of the First Kind of Order $0$:
:$\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$
from which it follows immediately that:
:$\map {J_0} 0 = 1$
From Derivative of Bessel Function of the First Kind of Order $0$ we have:
from which it follows that:
:$\map {J'_0} 0 = 0$
Then from Derivative of Laplace Transform:
and:
:$\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$
Thus by taking the Laplace transform of $(1)$ we have:
Let $y = \laptrans {\map {J_0} t}$.
Now we have that:
and:
:$\ds \lim_{t \mathop \to 0} \map {J_0} t = 1$
Thus by the Initial Value Theorem of Laplace Transform:
:$c = 1$
and so:
:$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Bessel Functions: $34 \ \text{(a)}$
Category:Laplace Transform of Bessel Function of the First Kind of Order Zero
Laplace Transform of Complex Power

== Theorem ==
Let $q$ be a constant complex number with $\map \Re q > -1$.
Let $t^q$ be the the principal branch of the $q$-th complex power whose domain contains the half-plane $\map \Re s > 0$.
Then $t^q$ has a Laplace transform given by:
:$\laptrans {t^q} = \dfrac {\map \Gamma {q + 1} } {s^{q + 1} }$
where $\Gamma$ denotes the gamma function.
== Proof ==
By definition of Laplace transform for a function not continuous at zero,
:$\ds \laptrans {t^q} = \lim_{\varepsilon \mathop \to 0^+} \lim_{L \mathop \to +\infty} \map I {\varepsilon, L}$
where:
:$\ds \map I {\varepsilon, L} = \int_\varepsilon^L t^q e^{-s t} \rd t$
Let $n \in \Z_{>0}$.
From Laplace Transform of Positive Integer Power:
:$\laptrans {t^n} = \dfrac {n!} { s^{n + 1} }$
From Gamma Function Extends Factorial, a reasonable Ansatz is:
:$\laptrans {t^q} = \dfrac {\map \Gamma {q + 1} } { s^{q + 1} }$
From Poles of Gamma Function, $\map \Gamma {-1}$ is undefined.
This suggests that having $q > -1$ would be a good idea for $q$ wholly real.
Reasons for insisting $\map \Re q > -1$ for complex $q$ will become apparent during the course of the proof.
With the aim of expressing $\map I {\varepsilon, L}$ in a form similar to the integral defining $\Gamma$, we use Complex Riemann Integral is Contour Integral to express $\map I {\varepsilon, L}$ as a contour integral.
Write $s = \sigma + i \omega$ for $\sigma, \omega \in \mathbb R$ and $\sigma > 0$.
From Complex Power by Complex Exponential is Analytic, the integrand for $\map I {\varepsilon, L}$ is analytic.
Thus the conditions for Contour Integration by Substitution are satisfied for $\map \Re s > 0$.
Substitute:
Denote:
:$I_C := \ds \int_C u^q e^{-u} \rd u$
Consider the simple closed contour:
:$C = C_1 + C_2 - C_3 + C_4$
defined as a concatenation of four contours, where:
:$C_1$ is the line segment connecting $\sigma \varepsilon$ to $L \sigma$
:$C_2$ is the line segment connecting $L\sigma$ to $L\sigma + i L \omega$
:$C_3$ is the line segment connecting $i \varepsilon$ to $L\sigma + i L \omega$
:$C_4$ is the circular arc connecting $i \varepsilon$ to $\sigma \varepsilon$ whose center is at the origin.
Note that $C_1$, $C_2$, and $C_4$ are positively oriented with respect to the interior of $C$, while $C_3$ is negatively oriented with respect to the interior of $C$.
The contours are illustrated in the following graph:
:
By the Cauchy-Goursat Theorem:
:$I_C = I_{C_1} + I_{C_2} - I_{C_3} + I_{C_4} = 0$
Consider $I_{C_2}$ as $L \to +\infty$:
Now consider $I_{C_4}$ for $\varepsilon \to 0^+$:
Thus $I_{C_4} \to 0$ as $\varepsilon \to 0^+$ and $I_{C_2} \to 0$ as $L \to +\infty$.
Thus, taking limits on $I_C = 0$:
:$I_C = I_{C_1} - I_{C_3} = 0$
That is, $I_{C_1} = I_{C_3}$ in the limit:
Thus:
and the Ansatz is proved correct.
Category:Laplace Transforms
Laplace Transform of Constant Mapping

== Theorem ==
Let $a \in \R$ be a real constant.
Let $f_a: \R \to \R$ or $\C$ be the constant mapping, defined as:
:$\forall t \in \R: \map {f_a} t = a$
Let $\laptrans {f_a}$ be the Laplace transform of $f_a$.
Then:
:$\laptrans {\map {f_a} t} = \dfrac a s$
for $\map \Re s > a$.
== Proof 1 ==
== Proof 2 ==
== Examples ==
Category:Laplace Transforms
Category:Constant Mappings
Category:Laplace Transform of Constant Mapping
Laplace Transform of Constant Mapping/Examples/Example 1

== Example of Laplace Transform of Constant Mapping ==
Let $\map f t$ be the real function defined as:
:$\forall t \in \R: \map f t = \begin {cases} 0 & : t < 0 \\ 5 & : 0 \le t < 3 \\ 0 & : t \ge 3 \end {cases}$
Then the Laplace transform of $f$ is given by:
:$\laptrans {\map f t} = \dfrac {5 \paren {1 - e^{-3 s} } } s$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transforms of some Elementary Functions: $4$
Category:Examples of Laplace Transforms
Category:Laplace Transform of Constant Mapping
Laplace Transform of Constant Multiple

== Theorem ==
Let $f$ be a function such that $\LL f$ exists.
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.
Let $\map F s$ denote $\laptrans {\map f t}$.
Let $a \in \C$ or $\R$ be constant.
Then:
:$a \laptrans {\map f {a t} } = \map F {\dfrac s a}$
== Proof ==
== Also presented as ==
This result can also be given as:
:$\laptrans {\map f {a t} } = \dfrac 1 a \map F {\dfrac s a}$
== Also known as ==
This property of the Laplace transform operator is sometimes seen referred to as the change of scale property.
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $4$. Change of scale property: Theorem $1 \text{-} 5$
* : Chapter $1$: The Laplace Transform: Solved Problems: Translation and Change of Scale Properties: $11$
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.4$
Category:Laplace Transforms
Laplace Transform of Constant Multiple/Examples/Example 1

== Examples of Use of Laplace Transform of Constant Multiple ==
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
:$\laptrans {\sin 3 t} = \dfrac 3 {s^2 + 9}$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $4$. Change of scale property
Category:Examples of Laplace Transforms
Laplace Transform of Cosine

== Theorem ==
Let $\cos$ be the real cosine function.
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Then:
:$\laptrans {\cos a t} = \dfrac s {s^2 + a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
* Laplace Transform of Sine
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of some Elementary Functions: $6$.
* : $\S 32$: Table of Special Laplace Transforms: $32.33$
Category:Laplace Transforms
Category:Cosine Function
Category:Laplace Transform of Cosine
Laplace Transform of Cosine/Proof 4

== Theorem ==
== Proof ==
By definition of the Laplace Transform:
:$\ds \laptrans {\cos at} = \int_0^{\to +\infty} e^{-s t} \cos at \rd t$
From Integration by Parts:
:$\ds \int f g' \rd t = f g - \int f'g \rd t$
Here:
So:
Consider:
:$\ds \int e^{-s t} \sin a t \rd t$
Again, using Integration by Parts:
:$\ds \int h j \,' \rd t = h j - \int h'j \rd t$
Here:
So:
Substituting this into $(1)$:
Evaluating at $t = 0$ and $t \to +\infty$:
Category:Laplace Transform of Cosine
Laplace Transform of Cosine Integral Function

== Theorem ==
:$\laptrans {\map \Ci t} = \dfrac {\map \ln {s^2 + 1} } {2 s}$
where:
:$\laptrans f$ denotes the Laplace transform of the function $f$
:$\Ci$ denotes the cosine integral function.
== Proof 1 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $9$
Category:Laplace Transform of Cosine Integral Function
Category:Laplace Transforms
Category:Cosine Integral Function
Laplace Transform of Cosine Integral Function/Proof 1

== Theorem ==
== Proof ==
Let $\map f t := \map \Ci t = \ds \int_t^\infty \dfrac {\cos u} u \rd u$.
Then:
By the Initial Value Theorem of Laplace Transform:
:$\ds \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$
which leads to:
:$c = 0$
Thus:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: The Sine, Cosine and Exponential Integrals: $37$
Category:Laplace Transform of Cosine Integral Function
Laplace Transform of Cosine of Root over Root

== Theorem ==
:$\laptrans {\dfrac {\cos \sqrt t} {\sqrt t} } = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }$
where $\laptrans f$ denotes the Laplace transform of the function $f$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $5$
Category:Laplace Transforms
Category:Cosine Function
Category:Laplace Transform of Cosine of Root over Root
Laplace Transform of Derivative

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any interval of the form $0 \le t \le A$.
Let $f$ be of exponential order $a$.
Let $\laptrans f$ denote the Laplace transform of $f$.
Let $f'$ be piecewise continuous with one-sided limits on said intervals.
Then $\laptrans f$ exists for $\map \Re s > a$, and:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
=== Discontinuity at $t = 0$ ===
=== Discontinuity at $t = a$ ===
=== Finite Discontinuities at $t = a_i$ for $i = 1, 2, \ldots, n$ ===
== Proof ==
Consider:
:$\ds \int_0^A e^{-s t} \map {f'} t \rd t$
By hypothesis, $f'$ is piecewise continuous with one-sided limits.
So by Piecewise Continuous Function with One-Sided Limits is Darboux Integrable, this integral exists.
This means that integration by parts can be invoked:
:$\ds \int h j\,' \rd t = h j - \int h' j \rd t$
Here:
So:
Now, take the limit as $t = A \to +\infty$:
Recall that $f$ is of exponential order $a$:
This implies, from Complex Exponential Tends to Zero and the Squeeze Theorem for Functions:
:$\ds \lim_{A \mathop \to +\infty} e^{-s A} \map f A = 0$
which produces:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
== Examples ==
== Also see ==
* Derivative of Laplace Transform
* Laplace Transform of Second Derivative
* Laplace Transform of Higher Order Derivatives
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $5$. Laplace transform of derivatives: Theorem $1 \text{-} 6$
* : Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transform of Derivative: $13$
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.7$
* : $\S 6.2$
* 
Category:Laplace Transforms
Category:Derivatives
Category:Laplace Transforms of Derivatives
Laplace Transform of Derivative/Discontinuity at t = 0

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any interval of the form $0 < t \le A$.
Let $f$ be of exponential order $a$.
Let $f'$ be piecewise continuous with one-sided limits on said intervals.
Let $\laptrans f$ denote the Laplace transform of $f$.
Let $f$ fail to be continuous at $t = 0$, but let:
:$\ds \lim_{t \mathop \to 0} \map f t = \map f {0^+}$
exist.
Then $\laptrans f$ exists for $\map \Re s > a$, and:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f {0^+}$
== Proof ==
See Laplace Transform of Derivative/Discontinuity at t = a and use $a = 0$ and $\map f {0^-} = 0$.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $5$. Laplace transform of derivatives: Theorem $1 \text{-} 7$
Category:Laplace Transforms of Derivatives
Laplace Transform of Derivative/Discontinuity at t = a

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any interval of the form $0 < t \le A$.
Let $f$ be of exponential order $a$.
Let $f'$ be piecewise continuous with one-sided limits on said intervals.
Let $\laptrans f$ denote the Laplace transform of $f$.
Let $f$ have a jump discontinuity at $t = a$.
Then:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0 - e^{-a s} \paren {\map f {a^+} - \map f {a^-} }$
== Proof ==
See Laplace Transform of Derivative with Finite Discontinuities and use $n = 1$ and $a_1 = a$.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $5$. Laplace transform of derivatives: Theorem $1 \text{-} 8$
Category:Laplace Transforms of Derivatives
Laplace Transform of Derivative/Examples/Example 1

== Examples of Use of Laplace Transform of Derivative ==
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
:$\laptrans {-3 \sin 3 t} = \dfrac {-9} {s^2 + 9}$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $5$. Laplace transform of derivatives
Category:Examples of Laplace Transforms
Laplace Transform of Derivative with Finite Discontinuities

== Theorem ==
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any interval of the form $0 < t \le A$.
Let $f$ be of exponential order $a$.
Let $f'$ be piecewise continuous with one-sided limits on said intervals.
Let $\laptrans f$ denote the Laplace transform of $f$.
Let $f$ have a finite number of jump discontinuities at $t = a_i$ for $i = 1, 2, \ldots, n$.
Then:
:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0 - \ds \sum_{i \mathop = 1}^n e^{-a_i s} \paren {\map f {a_i^+} - \map f {a_i^-} }$
== Proof ==
The proof is found similarly to the proof in Laplace Transform of Derivative but requires breaking the integral into $n$ integrals of finite range between the discontinuities and one improper integral from $a_n^+$ to $+\infty$. Integration by parts yields the summation when evaluating $e^{-s t}x(t)$ and the remaining integrals can all be recombined into one improper integral that is identically equal to $s \laptrans {\map f t}$.
Specifically, as in the proof in Laplace Transform of Derivative,
Breaking the integral into $n+1$ integrals over continuous intervals gives
:$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = \int_0^{a_1^-} e^{-s t} \map {f'} t \rd t + \int_{a_1^+}^{a_2^-} e^{-s t} \map {f'} t \rd t + \cdots + \int_{a_{n-1}^+}^{a_n^-} e^{-s t} \map {f'} t \rd t + \lim_{A \mathop \to +\infty} \int_{a_n^+}^A e^{-s t} \map {f'} t \rd t$
Integration by parts on all integrals yields
:$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = a_1^-} + s \int_0^{a_1^-} e^{-s t} \map {f} t \rd t + \bigintlimits {e^{-s t} \map f t} {t \mathop = a_1^+} {t \mathop = a_2^-} + s \int_{a_1^+}^{a_2^-} e^{-s t} \map {f} t \rd t + \cdots + \bigintlimits {e^{-s t} \map f t} {t \mathop = a_{n-1}^+} {t \mathop = a_n^-} + s \int_{a_{n-1}^+}^{a_n^-} e^{-s t} \map {f} t \rd t + \lim_{A \mathop \to +\infty} \bigintlimits {e^{-s t} \map f t} {t \mathop = a_n^+} {t \mathop = A} + \lim_{A \mathop \to +\infty} s \int_{a_n^+}^A e^{-s t} \map {f} t \rd t$
These terms combine to form
:$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = \sum_{i \mathop = 1}^n e^{-a_i s} \paren {\map f {a_i^+} - \map f {a_i^-} } + \lim_{A \mathop \to +\infty} \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = A} + \lim_{A \mathop \to +\infty} s \int_0^A e^{-s t} \map {f} t \rd t$
where the proof in Laplace Transform of Derivative simplifies the terms other than the summation and gives the final result
:$\ds \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t = s \laptrans {\map f t} - \map f 0 - \ds \sum_{i \mathop = 1}^n e^{-a_i s} \paren {\map f {a_i^+} - \map f {a_i^-} }$
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $5$. Laplace transform of derivatives: Theorem $1 \text{-} 8$
Category:Laplace Transforms of Derivatives
Laplace Transform of Dirac Delta Function

== Theorem ==
Let $\map \delta t$ denote the Dirac delta function.
The Laplace transform of $\map \delta t$ is given by:
:$\laptrans {\map \delta t} = 1$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Warning ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $12$
Category:Laplace Transforms
Category:Dirac Delta Function
Category:Laplace Transform of Dirac Delta Function
Laplace Transform of Dirac Delta Function/Lemma

== Lemma for Laplace Transform of Dirac Delta Function ==
Let $F_\epsilon: \R \to \R$ be the real function defined as:
:$\map {F_\epsilon} t = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le t \le \epsilon \\ 0 & : t > \epsilon \end{cases}$
Then:
:$\laptrans {\map {F_\epsilon} t} = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $41$
Category:Laplace Transform of Dirac Delta Function
Laplace Transform of Dirac Delta Function/Warning

== Warning on Laplace Transform of Dirac Delta Function ==
Mathematically speaking, $\ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} t$ does not actually exist.
Hence $\ds \laptrans {\lim_{\epsilon \mathop \to 0} \map {F_\epsilon} t}$ is not actually defined.
However, it is useful to consider $\map \delta t = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} t$ to be such that $\laptrans {\map \delta t} = 1$.
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $42 \ \text{(b)}$
Category:Laplace Transform of Dirac Delta Function
Laplace Transform of Dirac Delta Function by Function

== Theorem ==
Let $\map f t: \R \to \R$ or $\R \to \C$ be a function.
Let $\map \delta t$ denote the Dirac delta function.
Let $c$ be a positive constant real number.
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.
Then:
:$\laptrans {\map \delta {t - c} \map f t} = e^{- s c} \map f c$
== Proof ==
Category:Laplace Transforms
Category:Dirac Delta Function
Laplace Transform of Error Function

== Theorem ==
:$\laptrans {\map \erf t} = \dfrac 1 s \map \exp {\dfrac {s^2} 4} \map \erfc {\dfrac s 2}$
where:
:$\laptrans f$ denotes the Laplace transform of the function $f$
:$\erf$ denotes the error function
:$\erfc$ denotes the complementary error function
:$\exp$ denotes the exponential function.
== Proof ==
By Derivative of Error Function, we have: 
:$\ds \map {\frac \d {\d t} } {\map \erf t} = \frac 2 {\sqrt \pi} e^{-t^2}$
By Primitive of Exponential Function, we have: 
:$\ds \int e^{-s t} \rd t = -\frac {e^{-s t} } s$
So: 
We have: 
We also have: 
Therefore: 
== Sources ==
* : Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $6$
Category:Laplace Transforms
Category:Error Function