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

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SUMMARY

The integral of the Dirac delta function, represented as $\delta_\epsilon (x) = \frac{\epsilon}{\pi(x^2 + \epsilon^2)}$, converges to 1 as $\epsilon$ approaches 0. The correct approach involves evaluating the integral $\int_{a}^{b} \delta_\epsilon (x) dx$ and applying the limit as $\epsilon \rightarrow 0$. The result shows that the arctangent function approaches $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ depending on the limits of integration, confirming that the integral equals 1 when evaluated over the entire real line.

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  • Understanding of the Dirac delta function and its properties
  • Familiarity with limits and L'Hospital's rule
  • Basic knowledge of integration techniques in calculus
  • Proficiency in LaTeX for rendering mathematical expressions
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  • Learn about the application of limits in integral calculus
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Students and professionals in physics, particularly those studying quantum mechanics, as well as mathematicians and educators looking to deepen their understanding of the Dirac delta function and its applications in integrals.

Doitright
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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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Have you tried integrating that function?
 
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I've tried integration, and get $\frac{\arctan(\frac{x}{\epsilon})}{\pi}$. However, seems there is still no way out.
 
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+.

--
lightarrow
 
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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:

\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}))

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

arctan(+\infty) = \frac{\pi}{2}
arctan(-\infty) = -\frac{\pi}{2}

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

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