# Dirac delta function proof help

• Raze2dust
In summary, Dirac delta function is a function that is undefined at x=0, and it has a limit as the series of functions that get more and more peaked near zero.f

#### Raze2dust

[SOLVED] Dirac delta function

## Homework Statement

Prove that $$\delta(cx)=\frac{1}{|c|}\delta(x)$$

## The Attempt at a Solution

For any function f(x), $$\int_{-\infty}^{\infty}f(x)\delta(cx) dx = \frac{1}{c}\int_{-\infty}^{\infty}f(t/c)\delta(t) dt$$

where I have used t=cx.
$$=\frac{1}{c}f(0)$$
This is fine and matches RHS for c>0. But how do we get the mod sign for c<0. Why isn't the above procedure valid for c<0 as well?

You have implicitly assumed that c>0 when you changed variables. Either that, or you forgot what else needs to be changed...

## Homework Statement

Prove that $$\delta(cx)=\frac{1}{|c|}\delta(x)$$

## The Attempt at a Solution

For any function f(x), $$\int_{-\infty}^{\infty}f(x)\delta(cx) dx = \frac{1}{c}\int_{-\infty}^{\infty}f(t/c)\delta(t) dt$$

where I have used t=cx.
$$=\frac{1}{c}f(0)$$
This is fine and matches RHS for c>0. But how do we get the mod sign for c<0. Why isn't the above procedure valid for c<0 as well?

What happens to the limits of integration when you make the change of variable and c is negative?

What happens to the limits of integration when you make the change of variable and c is negative?

thanks nrged!
wonder how it didn't strike me after pulling my hair off for one hour

Consider a spring-mass system with a time-dependent force f(t) applied to the mass.
The situation is modeled by the second-order differential equation
mx"(t) + cx0'(t) + kx(t) = f(t)
where t is time and x(t) is the displacement of the mass from equilibrium. Now suppose
that for t greater than or equal to 0 the mass is at rest in its equilibrium position, so x'(0) = x(0) = 0. At
t = 0 the mass is struck by an “instantaneous” hammer blow. This situation actually occurs
frequently in practice—a system sustains an forceful, almost-instantaneous input. Our goal
is to model the situation mathematically and determine how the system will respond.
In the above situation we might describe f(t) as a large constant force applied on a very
small time interval. Such a model leads to the forcing function

The Dirac delta function is a special kind of object - it is not a
function in the usual sense, although we sometimes think of it that
way. Basically, it can be thought of as the "function" d(x) such that:

d(x) = 0 for x <> 0
= Infinity for x = 0

and the "size" of the infinity is such that the integral of d(x) with
respect to x from a to b is zero if a and b are of the same sign, and
the integral is exactly unity if a < 0 < b. More precisely, it can be
thought of as the limit of a series of functions that get more and
more peaked near zero, and whose area is always unity.

It is always to be implicitly used inside an integral, even though we
carry out operations such as:

f(x) = d(x) + d(x-2) (when integrated, this has area 2)

Last edited:

Consider a spring-mass system with a time-dependent force f(t) applied to the mass.
The situation is modeled by the second-order differential equation
mx"(t) + cx0'(t) + kx(t) = f(t)
where t is time and x(t) is the displacement of the mass from equilibrium. Now suppose
that for t greater than or equal to 0 the mass is at rest in its equilibrium position, so x'(0) = x(0) = 0. At
t = 0 the mass is struck by an “instantaneous” hammer blow. This situation actually occurs
frequently in practice—a system sustains an forceful, almost-instantaneous input. Our goal
is to model the situation mathematically and determine how the system will respond.
In the above situation we might describe f(t) as a large constant force applied on a very
small time interval. Such a model leads to the forcing function
f (t) = $$\left\{\stackrel{^{\frac{A}{\epsilon},}}{0,else}$$ 0 $$\leq$$ t $$\leq$$ $$\epsilon$$

Here A is some constant and $$\epsilon$$ is a “small” positive real number. When $$\epsilon$$ is close to zero the applied force is very large during the time interval 0 $$\leq$$ t $$\leq$$ $$\epsilon$$
In this case it’s easy to see that for any choice of $$\epsilon$$ we have
$$\int$$$$\stackrel{\infty^}{f (t) dt = A}$$
$$-\infty$$

Last edited:

k..thanx for the info Horbacz !
:)

your welcome raze 2 dust. and by the way, I am only 13...