Question on Dirac Delta function

In summary, the conversation discusses how to prove the equalities (1) ##\delta(x)=\delta(-x)## and (2) ## \delta(kx)=\frac{1}{|k|}\delta(x)##, with the use of variable substitution and the changing of bounds in the integral. The conversation also mentions the importance of considering the polarity of k when determining the result of (2).
  • #1
yungman
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I want to proof (1)##\delta(x)=\delta(-x)## and (2) ## \delta(kx)=\frac{1}{|k|}\delta(x)##

(1) let ##u=-x\Rightarrow\;du=-dx##
[tex]\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)[/tex]
but
[tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)[/tex]

I cannot proof (1) is equal as I don't know if ##f(u)=-f(-u)##. Please help.

(2)Let ##u=kx\Rightarrow\;\frac{du}{k}=dx##. The sign of the limit reverse depends on the polarity of k.
[tex]\int_{-\infty}^{\infty}f(x)\delta(kx)dx=\frac{1}{k}\int_{-\infty}^{\infty}f(\frac{u}{k})\delta(u)du=^+_-\frac{1}{k}f(0)[/tex]
Since sign is +ve if k is positive. Sign is -ve if k is -ve. This means the result is always ##\frac{1}{|k|}f(0)##. Am I correct?

Thanks
 
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  • #2
yungman said:
I want to proof (1)##\delta(x)=\delta(-x)## and (2) ## \delta(kx)=\frac{1}{|k|}\delta(x)##

(1) let ##u=-x\Rightarrow\;du=-dx##
[tex]\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)[/tex]
but
[tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)[/tex]

I cannot proof (1) is equal as I don't know if ##f(u)=-f(-u)##. Please help.

The bounds of the integral change too, so you would have

[tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{\infty}^{-\infty}f(-u)\delta(u)du[/tex]

intead of what you had. Does that change anything?
 
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  • #3
micromass said:
The bounds of the integral change too, so you would have

[tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{\infty}^{-\infty}f(-u)\delta(u)du[/tex]

intead of what you had. Does that change anything?

Got it. How about (2)

Thanks
 
  • #4
yungman said:
Got it. How about (2)

Thanks

You should switch the bounds in (2) too, if ##k## is negative. Aside from that, it's alright.
 
  • #5
micromass said:
You should switch the bounds in (2) too, if ##k## is negative. Aside from that, it's alright.

Thanks
 

1. What is the Dirac Delta function and why is it important in science?

The Dirac Delta function, also known as the unit impulse function, is a mathematical function that is used to model impulsive phenomena in various fields of science such as physics, engineering, and signal processing. It is important because it allows us to represent a point-like concentration of mass or energy, which is often encountered in real-world systems.

2. How is the Dirac Delta function defined mathematically?

The Dirac Delta function is defined as a distribution, rather than a conventional function, and is typically denoted by the symbol δ(x). It has the property that it is zero for all values of x except at x = 0 where it is infinite, and it integrates to 1 over the entire real line. Mathematically, it is represented by the following equation: ∫f(x)δ(x)dx = f(0), where f(x) is any well-behaved function.

3. What is the physical interpretation of the Dirac Delta function?

The Dirac Delta function can be interpreted as a spike or impulse that is localized at a specific point. In physics, it is often used to represent the concentration of mass or charge at a single point, or the force exerted by a point-like object. It can also be used to model other phenomena such as a sudden change in temperature or pressure.

4. How is the Dirac Delta function used in practical applications?

The Dirac Delta function is used in various practical applications, such as in solving differential equations, representing idealized point sources in electrostatics and fluid dynamics, and in signal processing to model abrupt changes in a signal. It is also used in probability and statistics to represent a discrete random variable with a continuous probability distribution.

5. Are there any limitations or drawbacks to using the Dirac Delta function?

One limitation of the Dirac Delta function is that it is not a true function, but rather a distribution. This means that it cannot be evaluated at a specific point, and its value can only be determined through integration with another well-behaved function. Additionally, it can lead to mathematical inconsistencies if not used carefully, and may not accurately represent physical systems that have a finite size or extent.

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