Show that the integral of the Dirac delta function is equal to 1

In summary, the conversation discusses the example of Dirac delta function in Quantum Mechanics and how to show that the limit of its integral equals to 1. The conversation explores using L'Hospital's rule and integration to solve the problem, with the ultimate solution being to take the limit as epsilon approaches 0. The conversation also mentions the importance of using LaTex syntax for equations.
  • #1
Doitright
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0
Hi,

I am reading the Quantum Mechanics, 2nd edition by Bransden and Joachain. On page 777, the book gives an example of Dirac delta function.
$\delta_\epsilon (x) = \frac{\epsilon}{\pi(x^2 + \epsilon^2)}$

I am wondering how I can show $\lim_{x\to 0+} \int_{a}^{b} \delta_\epsilon (x) dx$ equals to 1. I've thought about using L'Hospital's rule, but there are two variables, $x$ and $\epsilon$, so seems I cannot use the rule directly. I've been stuck in this for some time. Will be grateful if some one can point the direction for me. Thanks.
 
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  • #2
Have you tried integrating that function?
 
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  • #3
I've tried integration, and get $\frac{\arctan(\frac{x}{\epsilon})}{\pi}$. However, seems there is still no way out.
 
  • #4
Doitright said:
I've tried integration, and get $\frac{\arctan(\frac{x}{\epsilon})}{\pi}$. However, seems there is still no way out.
I think you shouldn' t make the limit for x-->0+ but for epsilon-->0+.

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  • #5
Doitright said:
I've tried integration, and get $\frac{\arctan(\frac{x}{\epsilon})}{\pi}$. However, seems there is still no way out.

You need to put two dollar-signs around an equation to render it in LaTex:

\$\$\frac{\arctan(\frac{x}{\epsilon})}{\pi}\$\$

renders as:

$$\frac{\arctan(\frac{x}{\epsilon})}{\pi}$$

You basically have the answer there:

[itex]\int_{a}^{b} \frac{\epsilon dx}{\pi (\epsilon^2 + x^2)} = \frac{1}{\pi} arctan(\frac{x}{\epsilon})|_a^b = \frac{1}{\pi} (arctan(\frac{a}{\epsilon}) - arctan(\frac{b}{\epsilon}))[/itex]

In the limit as [itex]\epsilon \rightarrow 0[/itex], [itex]\frac{a}{\epsilon} \rightarrow \pm \infty[/itex] and [itex]\frac{b}{\epsilon} \rightarrow \pm \infty[/itex]

[itex]arctan(+\infty) = \frac{\pi}{2}[/itex]
[itex]arctan(-\infty) = -\frac{\pi}{2}[/itex]

So if [itex]a > 0[/itex], you get [itex]arctan(\frac{a}{\epsilon}) \rightarrow +\frac{\pi}{2}[/itex]
If [itex]a > 0[/itex], you get [itex]arctan(\frac{a}{\epsilon}) \rightarrow -\frac{\pi}{2}[/itex]
So if [itex]b > 0[/itex], you get [itex]arctan(\frac{b}{\epsilon}) \rightarrow +\frac{\pi}{2}[/itex]
If [itex]b > 0[/itex], you get [itex]arctan(\frac{b}{\epsilon}) \rightarrow -\frac{\pi}{2}[/itex]
 
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  • #6
Note also that for any ##\epsilon## you have ##\int_{-\infty}^{+\infty} \delta_{\epsilon}(x)dx =1##.
 
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  • #7
I know how to prove it now. Thanks for the help.
 

1. What is the Dirac delta function?

The Dirac delta function, denoted by δ(x), is a mathematical function that is defined to be zero everywhere except at the origin, where it is infinite. It is used in mathematics and physics to model point-like or localized phenomena.

2. How is the integral of the Dirac delta function defined?

The integral of the Dirac delta function is defined as the limit of a sequence of functions, known as mollifiers, which approximate the Dirac delta function. It is often written as ∫δ(x)dx and is equal to 1.

3. Why is the integral of the Dirac delta function equal to 1?

The integral of the Dirac delta function is equal to 1 because it represents the area under the curve of the delta function. Since the delta function has an infinite height but an infinitesimal width, the area under the curve is equal to 1.

4. Can the integral of the Dirac delta function be evaluated using standard integration techniques?

No, the integral of the Dirac delta function cannot be evaluated using standard integration techniques. This is because the delta function is not a continuous function and therefore cannot be integrated in the traditional sense. Instead, it is evaluated using the limit definition of the integral.

5. What is the significance of the integral of the Dirac delta function being equal to 1?

The integral of the Dirac delta function being equal to 1 is significant because it reflects the normalization property of the delta function. This means that when the delta function is used in equations and calculations, the total probability or amount will always be equal to 1, making it a useful tool in many areas of mathematics and physics.

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