Valid Representation of Dirac Delta function

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RJLiberator
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Homework Statement



Show that this is a valid representation of the Dirac Delta function, where ε is positive and real:

[tex]\delta(x) = \frac{1}{\pi}\lim_{ε \rightarrow 0}\frac{ε}{x^2+ε^2}[/tex]

Homework Equations



https://en.wikipedia.org/wiki/Dirac_delta_function

The Attempt at a Solution



I just need help on how to start this one. I'm sure I can bring it to it's logical conclusion, but I have no starting point known.

I've tried looking for different definitions, but to no avail. Is there any hint that you can give me to start this?
 
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RJLiberator said:

Homework Statement



Show that this is a valid representation of the Dirac Delta function, where ε is positive and real:

[tex]\delta(x) = \frac{1}{\pi}\lim_{ε \rightarrow 0}\frac{ε}{x^2+ε^2}[/tex]

Homework Equations



https://en.wikipedia.org/wiki/Dirac_delta_function

The Attempt at a Solution



I just need help on how to start this one. I'm sure I can bring it to it's logical conclusion, but I have no starting point known.

I've tried looking for different definitions, but to no avail. Is there any hint that you can give me to start this?
For a given value of x not equal to zero, what is ##\delta(x)##? If x = 0, what is ##\delta(x)##? Also, can you evaluate this integral? ##\int_{-\infty}^\infty \delta(x)dx##?
 
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If x = 0, the dirac delta function evaluates to infinity. If x =/= 0 then dirac delta function evaluates to 0.

So then, we let x = 0 and see that the limit of 1/ε goes to infinity.

Ok, that makes sense, but if we let x = 10, we get ε/(100+ε^2) which goes to 0. So it features the similar form of the Dirac Delta function.
 
RJLiberator said:
If x = 0, the dirac delta function evaluates to infinity. If x =/= 0 then dirac delta function evaluates to 0.

So then, we let x = 0 and see that the limit of 1/ε goes to infinity.

Ok, that makes sense, but if we let x = 10, we get ε/(100+ε^2) which goes to 0. So it features the similar form of the Dirac Delta function.
Yes, but as Mark44 mentioned, you still need to verify that the integral over all space is unity.
 
Oh, that's easy. I see now that it equals 1.

So, that's all the requirements needed then for the Dirac Delta representation. I see.

Thank you.
 
RJLiberator said:
Oh, that's easy. I see now that it equals 1.

So, that's all the requirements needed then for the Dirac Delta representation. I see.

Thank you.

No, not yet!

If ##\delta_{\epsilon}(x)## is your function before taking the limit, you still need to verify that
$$\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(x) f(x) \, dx = f(0)$$
for an appropriate class of test functions ##f(x)## or that
$$\lim_{\epsilon \to 0} \int_{-a}^b \delta_{\epsilon}(x) f(x) \, dx = f(0) $$
for any "reasonable" continuous function and any ##a, b > 0##.
 
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Thanks for the catch -- I saw your conversation earlier today (10am my time) and checked it with my homework. Thanks for going the extra mile to help me out here.