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AFewNotes On The Dirac Delta Function And
The Laplace Transform
David Meyer
dmm613@gmail.com
Last update: October 28, 2022
1 Introduction
These notes began life as some thoughts on the Dirac Delta Function and evolved into notes on several
related topics including Laplace Transforms. The Dirac Delta function has all kinds of crazy and interesting
properties. More TBD.
2 The Dirac Delta Function
The Dirac Delta Function is defined as shown in Figure 1. In the limit (ǫ → 0) the Dirac Delta function
is written δ (t) or sometimes δ(t − a). As we will see in a moment, the δ (t) form of the delta function
a a,ǫ
is useful when we want to use the Mean Value Theorem for Integrals [2] to evaluate integrals involving the
delta function.
δa,ǫ(t)
(1 a≤t≤a+ǫ
δa,ǫ(t) = ǫ
0 otherwise
1
ǫ
a a+ǫ t
Figure 1: The Dirac Delta Function δa,ǫ(t)
So δa,ǫ(t) is defined to be
(1 a≤t≤a+ǫ
δa,ǫ(t) = ǫ
0 otherwise
1
and has the constraint that
Z ∞δa,ǫ(t) = 1
0
That is, δa,ǫ(t) is in some sense a probability density.
In the limit the Dirac Delta Function looks like
lim δ (t) = δ (t) = (∞ t = a
ǫ→0 a,ǫ a 0 t 6= a
or sometimes
(
δ(t −a) = ∞ t=a
0 t 6= a
δa(t) also has the constraint that
Z ∞δa(t) = 1
0
and so is also a probability density. δ (t) is shown in Figure 2.
a
δ (t)
a
δ (t) = (∞ t = a
a 0 otherwise
a t
Figure 2: The Dirac Delta Function δ (t)
a
2.1 Integrals involving δ (t)
a,ǫ
δ (t) has all kinds of interesting properties. One of them involves the integral of the product δ (t) with
a,ǫ a,ǫ
some function g(t). Here we would like to evaluate integrals of the form
Z ∞δa,ǫ(t)g(t)dt (1)
0
where g(t) is continuous on the interval [a,a + ǫ].
2
The Mean Value Theorem for Integrals [2] tells us that
Z bg(t)dt = (b−a)g(c) (2)
a
where the point c lies in the interval [a,a+ǫ]. Now, since we know that δa,ǫ(t) is zero everywhere except on
the interval [a,a + ǫ] we can rewrite the improper integral in Equation 1 as the proper integral
Z a+ǫδa,ǫ(t)g(t)dt
a
Here we can notice that δ (t) = 1 on the interval [a,a + ǫ] so we can rewrite our integral as
a,ǫ ǫ
Z a+ǫ 1g(t)dt = 1 Z a+ǫg(t)dt
a ǫ ǫ a
1
Nowwecanuse Equation 2, the Mean Value Theorem for Integrals , by setting b = a+ǫ and a = a so that
b −a = ǫ. Then
1 Z a+ǫ 1 h i 1
ǫ a g(t)dt = ǫ (a+ǫ)−a g(c) = ǫǫg(c) = g(c)
| {z }
b−a
where c ∈ [a,a+ǫ].
Finally, if we look at what happens to c as ǫ → 0 we see that lim c = a (sorry about the notation abuse) so
that ǫ→0
Z ∞δa(t)g(t)dt = g(a) (3)
0
Essentially δ (t) pulls out the value of g at a, that is, g(a).
a
Another way to look at this [1] is to notice that the integrand of
Z ∞δ(t−a)g(t)dt
0
is zero everywhere except where t = a, so we can rewrite our integral as R∞ δ(t−a)g(a)dt = g(a)R∞δ(t−a)dt
R∞ 0 0
(since g(a) doesn’t depend on t). Then since by definition 0 δ(t−a)dt = 1 we get
Z ∞δ(t−a)g(t)dt = g(a)
0
1This is where the δa,ǫ(t) form of the delta function comes in handy.
3
3 The Laplace Transform
Westart by defining the integral transform of some function f(t).
Definition 3.1. Integral Transform: If a function f(t) is defined on [0,∞) then we can define an integral
transform to be the improper integral
F(s) = Z ∞K(s,t)f(t)dt
0
If the improper integral converges then we say that F(s) is the integral transform of f(t). The function
−st
K(s,t) is called the kernel of the transform. When K(s,t) = e the transform is called the Laplace
Transform.
Definition 3.2. Laplace Transform: The Laplace Transform of a function f(t) is defined to be
L{f(t)} = F(s) = Z ∞e−stf(t)dt (4)
0
The Laplace Transform will turn out to be useful when solving ordinary differential equations (ODEs).
Interestingly, the Laplace Transform of the Dirac Delta Function turns out to be
L{δa(t)} = R∞e−stδa(t)dt #Equation 4 with f(t) = δa(t)
0
= R∞g(t)δ (t)dt #set g(t) = e−st
0 a
= g(a) #byEquation 3
−sa −sa
= e #g(a)=e
3.1 The Linearity Property of the Laplace Transform
Definition 3.3. Linearity Property: L{af(t)+bg(t)} = aL{f(t)}+bL{g(t)}
Interestingly, L is what is called a linear operator in vector space parlance [5].
All good, but why does L have this property? Here’s one way to think about it:
R∞ −st
L{af(t)+bg(t)} = 0 e af(t)+bg(t) dt #definition of the Laplace Transform (Definition 3.2)
R∞ −st R∞ −st
= 0 e af(t)dt+ 0 e bg(t)dt #bythe linearity of improper integrals [3]
= aR∞e−stf(t)dt+bR∞e−stg(t)dt #neither a nor b depends on t
0 0
= aL{f(t)}+bL{g(t)} #definition of the Laplace Transform (Definition 3.2)
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