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

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.
 

PeroK

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

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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]
 

PeroK

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Note also that for any ##\epsilon## you have ##\int_{-\infty}^{+\infty} \delta_{\epsilon}(x)dx =1##.
 
I know how to prove it now. Thanks for the help.
 

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