Dirac delta function, change of variable confusion

[bob900]

The Dirac delta "function" is often given as :

δ(x) = ∞ | x $=$ 0
δ(x) = 0 | x $\neq$ 0

and ∫δ(x)f(x)dx = f(0).

What about δ(cx)? By u=cx substitution into above integral is, ∫δ(cx)f(x)dx = ∫δ(u)f(u/c)du = 1/c f(0).

But intuitively, the graph of δ(cx) is the same as the graph of δ(x)! At x=0, cx=0 so δ(cx)=∞, and everywhere else cx$\neq$0 so δ(cx)=0.

So how can it be that ∫δ(x)f(x)dx = 1/c ∫δ(cx)f(x)dx ?

 

[HallsofIvy]

The difficulty is talking about the "graph" of $\delta(x)$ at all- which is a consequence of talking about $\delta(x)$ as a function. There is, in fact, NO function of x that satisfies the conditions you give. It is a "generalized function" (also called "distribution"). A "generalized" function has to be interpreted as an operator on functions rather than a function itself.

 

[bob900]

The difficulty is talking about the "graph" of $\delta(x)$ at all- which is a consequence of talking about $\delta(x)$ as a function. There is, in fact, NO function of x that satisfies the conditions you give. It is a "generalized function" (also called "distribution"). A "generalized" function has to be interpreted as an operator on functions rather than a function itself.

Right, but then what justification do we have to do u-substitution to get the value of ∫δ(cx)f(x)dx, if u-substitution only applies to integrals of only functions, and not to integrals of functions multiplied with 'distributions'?

 

[micromass]

Right, but then what justification do we have to do u-substitution to get the value of ∫δ(cx)f(x)dx, if u-substitution only applies to integrals of only functions, and not to integrals of functions multiplied with 'distributions'?

Because that is the definition of $\delta(cx)$ it is defined exactly as the distribution such that

$$\int \delta(cx)f(x)=\frac{1}{c}f(0)$$

 

[blue_raver22]

The Dirac delta function is a mathematical tool used to represent an infinitely narrow and infinitely tall spike at a specific point. It is often used in physics and engineering to describe impulse functions, which are sudden and short-lived changes in a system.

When using the Dirac delta function, it is important to understand that it is not a traditional function in the sense that it cannot be graphed or evaluated at specific points. Instead, it is defined by its properties, such as its integral over a certain interval.

In the given example, the confusion arises from the change of variable u=cx. This change of variable results in a scaling of the x-axis, which affects the function δ(x). However, the properties of the Dirac delta function remain the same, so the integral is still equivalent to f(0).

To understand this concept intuitively, imagine a spike at x=0 on a number line. When we scale the x-axis by a factor of c, the spike will also be scaled by the same factor. However, its properties, such as its height and location, remain the same. Therefore, the integral will also be scaled by the same factor, resulting in 1/c f(0).

In conclusion, the Dirac delta function is a powerful tool in mathematics and science, but it requires a thorough understanding of its properties and how it behaves under different operations, such as change of variables. It is important to remember that the Dirac delta function is not a traditional function and should be used carefully and accurately in calculations.