Limiting Case of Dirac Function: delta(x) = lim_{a->0} (1/sqrt[pi*a])*e^(-x^2/a), |x|<a

[mario99]

Show that the following function define the Dirac δ-function as a limiting case.

$$\delta(x) = \lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}, |x| < a$$
If I solve the limit, does it mean that I have shown what is required?

 

[blamocur]

Show that the following function define the Dirac δ-function as a limiting case.

$$\delta(x) = \lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}, |x| < a$$
If I solve the limit, does it mean that I have shown what is required?

Not sure what you mean by solving the limit in this case, but I would start with the definition of the Dirac delta function.
Also, I am not sure how the formula defines $\delta(x)$ for arbitrary $x$ since it stipulates $|x|<a$. Is this a complete statement of the problem?

 

[topsquark]

Show that the following function define the Dirac δ-function as a limiting case.

$$\delta(x) = \lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}, |x| < a$$
If I solve the limit, does it mean that I have shown what is required?

The delta "function" is not really a function in the technical sense. If you could take this limit, you would get an infinite "spike" at x= 0 and 0 everywhere else. This does not give you anything you can work with.

What you need to do with this expression is to apply it:
$\displaystyle \int _a^b f(x) \delta (x - a) \, dx = \begin{cases} f(x_0) & , \, a \leq x_0 \leq b \\ 0 &, \, \text{else} \end{cases}$
and other properties may be derived by using this limit.

-Dan

 

[mario99]

Thank you blamocur and topsquark for helping.

Not sure what you mean by solving the limit in this case, but I would start with the definition of the Dirac delta function.
Also, I am not sure how the formula defines $\delta(x)$ for arbitrary $x$ since it stipulates $|x|<a$. Is this a complete statement of the problem?

This limit $$\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}$$
The definition of Dirac delta function is simple. It is has an impulse of magnitude 1 at the origin and 0 every where else.
As topsquark has said, this limit is infinite at x = 0 and zero everywhere else. This might means we have to take the limit of right and left of a which will restrict x to be bounded between them.

The delta "function" is not really a function in the technical sense. If you could take this limit, you would get an infinite "spike" at x= 0 and 0 everywhere else. This does not give you anything you can work with.

What you need to do with this expression is to apply it:
$\displaystyle \int _a^b f(x) \delta (x - a) \, dx = \begin{cases} f(x_0) & , \, a \leq x_0 \leq b \\ 0 &, \, \text{else} \end{cases}$
and other properties may be derived by using this limit.

-Dan

Do you mean that I have to apply Dirac function in expression and then to apply the limit in the same expression to get the same result?

 

[blamocur]

The delta "function" is not really a function in the technical sense. If you could take this limit, you would get an infinite "spike" at x= 0 and 0 everywhere else. This does not give you anything you can work with.

What you need to do with this expression is to apply it:
$\displaystyle \int _a^b f(x) \delta (x - a) \, dx = \begin{cases} f(x_0) & , \, a \leq x_0 \leq b \\ 0 &, \, \text{else} \end{cases}$
and other properties may be derived by using this limit.

-Dan

I am pretty sure you meant $\delta(x-x_0)$ instead of $\delta(x-a)$ in the integrand. Also, there is already variable $a$ in the original post, and "recycling" it for the integration limits would be confusing.

 

[blamocur]

The definition of Dirac delta function is simple. It is has an impulse of magnitude 1 at the origin and 0 every where else.

It might be simple but not particularly useful. What does "impulse of magnitude 1" mean when applied to a "function"? I.e., I like @topsquark 's definition much more than the "simple" one

 

[topsquark]

Do you mean that I have to apply Dirac function in expression and then to apply the limit in the same expression to get the same result?

In order to apply known properties of the delta function, no. But if you are trying to prove those properties, or are in a new theoretical situation such as deriving the Heaviside function, then you would need to apply some limit form of the delta function to derive the result.

-Dan

 

[blamocur]

Do you mean that I have to apply Dirac function in expression and then to apply the limit in the same expression to get the same result?

I am not familiar with the specifics of your course and the expectations of your teacher, but I would think that you might be expected to prove, for example, that
$$\lim_{a\rightarrow \infty} \int_c^d f(x) \frac{1}{\sqrt{\pi a}}e^{-x^2/a} dx = f(0)$$for any $c<0,d>0$ and any $f(x)$ continuous at 0.

 

[mario99]

I am not familiar with the specifics of your course and the expectations of your teacher, but I would think that you might be expected to prove, for example, that
$$\lim_{a\rightarrow \infty} \int_c^d f(x) \frac{1}{\sqrt{\pi a}}e^{-x^2/a} dx = f(0)$$for any $c<0,d>0$ and any $f(x)$ continuous at 0.

This will be hard a little bit.

What about the other way around. To prove this:

$$\int _c^d f(x) \delta (x) \ dx = f(0)$$
Or this

$$\int _c^d f(x) \delta (x - a) \ dx = f(a)$$
Or may be even this

$$\int _c^d f(x - a) \delta (x) \ dx = f(a)$$

In order to apply known properties of the delta function, no. But if you are trying to prove those properties, or are in a new theoretical situation such as deriving the Heaviside function, then you would need to apply some limit form of the delta function to derive the result.

-Dan

Eventually, I would need to derive everything.

What do you think of what blamocur said? Will it be enough to show what is required?

 

[topsquark]

This will be hard a little bit.

What about the other way around. To prove this:

$$\int _c^d f(x) \delta (x) \ dx = f(0)$$
Or this

$$\int _c^d f(x) \delta (x - a) \ dx = f(a)$$
Or may be even this

$$\int _c^d f(x - a) \delta (x) \ dx = f(a)$$



Eventually, I would need to derive everything.

What do you think of what blamocur said? Will it be enough to show what is required?

I have no idea what you mean by "the other way around". You have the definition of the delta function. You have access to, I should imagine, the usual five or so ways to approximate it using limits. To prove the theorems we use one limiting form or another. How do you reverse that?

Blancmour is saying quantitatively what I am saying qualitatively. We are both saying the same thing. You may need to use one of the other forms to do the specific integrals, but that's how you do the actual calculation if you are working with the theory. If you are simply applying it, then you can say that
$\displaystyle \int_0^5 \dfrac{sin^{5/4}(x) \, e^{cos(5x/2)}}{log(tan(4x+3))} \delta (x-2) \, dx = \dfrac{sin^{5/4}(2) \, e^{cos(5)}}{log(tan(11))}$

But you need to know how to do the basic groundwork in order to understand why this is so.

-Dan

 

[blamocur]

What about the other way around.

To second @topsquark's post, I don't see the other way around in this problem. As a side note I'll point out that the whole statement of the problem, i.e., $\delta(x) = \lim...$ is somewhat misleading: while you can show that the limit is 0 for $x\neq 0$ you cannot really compute $\delta(0)$ there because saying $\delta(0) = \infty$ does not really define it. In particular, note the scaling by $\frac{1}{\sqrt{\pi a}}$ instead of a simpler $\frac{1}{\sqrt{a}}$ in your post of the problem. It is needed to satisfy the definition of $\delta(x)$ in post #3.

 

[mario99]

Thank you for both of you. I think that I am getting it now.

The definition of Dirac Delta function is:

$$\delta (x) = \begin{cases} 0 , \, x \neq 0 \\ \infty, x = 0 \end{cases}$$
One of the properties of Direc Delta is:

$$\displaystyle \int _{-\infty}^{\infty} f(x) \delta (x) \ dx = f(0)$$
If I can show

$$\displaystyle \int _{-\infty}^{\infty} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx = f(0)$$
I can say

$$\delta(x) = \lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}$$
I will let

$$u = \frac{x}{\sqrt{a}}$$
$$\displaystyle \int _{-\infty}^{\infty} \lim_{a\rightarrow 0} f(u\sqrt{a}) \left(\frac{1}{\sqrt{\pi}}e^{-u^2}\right) \ du$$
$$\lim_{a\rightarrow 0} f(u\sqrt{a})\displaystyle \int _{-\infty}^{\infty} \left(\frac{1}{\sqrt{\pi}}e^{-u^2}\right) \ du$$
This is a famous integral in statistics

$$\int _{-\infty}^{\infty} \left(\frac{1}{\sqrt{\pi}}e^{-u^2}\right) \ du = \sqrt{\pi}$$
The resulting is

$$\lim_{a\rightarrow 0} f(u\sqrt{a})\displaystyle \frac{\sqrt{\pi}}{\sqrt{\pi}}$$
$$\lim_{a\rightarrow 0} f(u\sqrt{a}) = f(0)$$
This was done for

$$-\infty<x<\infty$$
Since

$$\int _{-\infty}^{\infty} f(x) \delta (x) \ dx = \int _{-a}^{a} f(x) \delta (x) \ dx = f(0)$$
This implies

$$\displaystyle \int _{-a}^{a} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx = f(0)$$
Which shows

$$\delta(x) = \lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}, |x| < a$$
topsquark and blamocur, what do you think of what I have done?

Even if this proof lacks precision, at least now I have the idea.

 

[blamocur]

No, I don't see this as a valid proof. While I have questions for half of your steps, I'll point out that moving $f(u\sqrt{a})$ from inside the intergral to the outside (after "I will let") is completely illegitimate. If you don't understand why then you really need to brush up on your prerequisites, integrals in particular.

 

[mario99]

No, I don't see this as a valid proof. While I have questions for half of your steps, I'll point out that moving $f(u\sqrt{a})$ from inside the intergral to the outside (after "I will let") is completely illegitimate. If you don't understand why then you really need to brush up on your prerequisites, integrals in particular.

While it is correct that moving f(ua^0.5) "alone" outside the integral violates the rules, but its limit is just a constant.

Tell me what steps do you have questions about?

Also one more thing that I could not edit in my post was the integral of statistics. It should have been written

$$\int _{-\infty}^{\infty} e^{-u^2} \ du = \sqrt{\pi}$$
Or

$$\int _{-\infty}^{\infty} \left(\frac{1}{\sqrt{\pi}}e^{-u^2}\right) \ du = 1$$

 

[blamocur]

While it is correct that moving f(ua^0.5) "alone" outside the integral violates the rules, but its limit is just a constant.

"f(ua^0.5)" depends on 'u', so moving outside the integral is wrong, limits or no limits. BTW, I would not put the limit inside the integral to begin with, this why I wrote my version of the problem in post #8.

 

[blamocur]

As topsquark has said, this limit is infinite at x = 0 and zero everywhere else.

This is not a definition of $\delta(x)$. For example, how would you distinguish $\delta(x)$ and $2\delta(x)$ ? Would you consider them equal?

 

[mario99]

"f(ua^0.5)" depends on 'u', so moving outside the integral is wrong, limits or no limits. BTW, I would not put the limit inside the integral to begin with, this why I wrote my version of the problem in post #8.

I agree that I should have done what you suggested.

This is not a definition of $\delta(x)$. For example, how would you distinguish $\delta(x)$ and $2\delta(x)$ ? Would you consider them equal?

What makes you think that it is not the definition?

 

[topsquark]

What makes you think that it is not the definition?

What makes you think that it is?

The Dira delta "function" is defined using a non-standard Lebesgue integral of a continuous compactly supported function f(x), as a measure $\delta (dx)$ with the property
$\displaystyle \int _a^b f(x) \, \delta (dx) = \begin{cases} f(0) & , \, a \leq 0 \leq b \\ 0 & , \, \text{else} \end{cases}$

which reduces to the expression I gave you above when we consider a Riemann integral over an interval on the real line.

Sometimes it is more loosely defined as a "function" with the property
$\displaystyle \int _a^b f(x) \, \delta (x) \, dx = \begin{cases} f(0) & , \, a \leq 0 \leq b \\ 0 & , \, \text{else} \end{cases}$

for all smooth functions f(x), with the constraint
$\displaystyle \int _{-\infty} ^{\infty} \delta (x) \, dx = 1$

though this is not strictly correct.

The statement
$\delta (x) = \begin{cases} {\infty} & , x= 0 \\ 0 & , \, \text{else} \end{cases}$

is somewhat intuitive, but is not a valid definition of any function on the real numbers, and is not precise enough to define it properly anyway. There is too much wiggle room. As blamocur asks, how would you tell the difference between $\delta (x)$ and $2 \delta (x)$?

Please check your source.

-Dan

 

[mario99]

It is defined like that in the book.

What are you trying to say?

$$\displaystyle \int _{-\infty}^{\infty} f(x) \delta(x) \ dx = \displaystyle \int _{-\infty}^{\infty} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$
But

$$\displaystyle \int _{-a}^{a} f(x) \delta(x) \ dx \neq \displaystyle \int _{-a}^{a} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$
Or

$$\displaystyle \int _{-\infty}^{\infty} f(x) \delta(x) \ dx \neq \displaystyle \int _{-\infty}^{\infty} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$
$$\displaystyle \int _{-a}^{a} f(x) \delta(x) \ dx \neq \displaystyle \int _{-a}^{a} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$

 

[topsquark]

It is defined like that in the book.

What are you trying to say?

$$\displaystyle \int _{-\infty}^{\infty} f(x) \delta(x) \ dx = \displaystyle \int _{-\infty}^{\infty} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$
But

$$\displaystyle \int _{-a}^{a} f(x) \delta(x) \ dx \neq \displaystyle \int _{-a}^{a} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$
Or

$$\displaystyle \int _{-\infty}^{\infty} f(x) \delta(x) \ dx \neq \displaystyle \int _{-\infty}^{\infty} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$
$$\displaystyle \int _{-a}^{a} f(x) \delta(x) \ dx \neq \displaystyle \int _{-a}^{a} f(x) \left(\lim_{a\rightarrow 0} \frac{1}{\sqrt{\pi a}}e^{-x^2/a}\right) \ dx$$

It is defined like which definition in your book? I gave you two.

Since you seemed to be confused about the definition, I gave you the definition. Your post here is talking about applying a limiting form. The first line is correct, so long as the integral exists.

-Dan