Dirac delta function approximation

Hi,

I'm not sure if I have calculated task b correctly, and unfortunately I don't know what to do with task c?



I solved task b as follows

##\displaystyle{\lim_{\epsilon \to 0}} \int_{- \infty}^{\infty} g^{\epsilon}(x) \phi(x)dx=\displaystyle{\lim_{\epsilon \to 0}} \int_{\infty}^{\epsilon} 0 \phi(x)dx + \displaystyle{\lim_{\epsilon \to 0}} \int_{- \epsilon}^{\epsilon} \frac{1}{2 \epsilon} \phi(x)dx +\displaystyle{\lim_{\epsilon \to 0}} \int_{\epsilon}^{\infty} 0 \phi(x)dx=\displaystyle{\lim_{\epsilon \to 0}} \int_{- \epsilon}^{\epsilon} \frac{1}{2 \epsilon} \phi(x)dx = \displaystyle{\lim_{\epsilon \to 0}} \frac{\varphi(\epsilon) - \varphi(- \epsilon)}{2 \epsilon}= \phi(0)##

Does task c mean the following
##\int_{- \epsilon}^{\epsilon} \frac{1}{2 \epsilon} \phi(x) dx=\displaystyle{\lim_{\epsilon \to 0}} \sum\limits_{x= -\epsilon}^{\epsilon} \frac{1}{2 \epsilon} \phi(x)##

task b) looks correct to me though the equality in the very last step is not so obvious and you have to show the intermediate steps (but I guess maybe in your book there is a sub problem that does that ). Also it is a bit confusing that you are using two different representations of the greek letter phi to denote a continuous function and its antiderivative.

Hold on while I process task c).

What you wrote in c) doesn't even make sense cause you seem to consider a sum over a continuous variable x, while the sum/series i know are over discrete variables from the set of integers.

What task wants you to is to write the integral as the limit of a Riemann sum and then take the limit for epsilon and then change the order with which you take the limits.

I mean the limit of a Riemann sum is something like $$\lim_{n\to\infty} \sum_{i=1}^{n} f(a+i\frac{b-a}{n})\frac{b-a}{n}$$ and then take the limit of this with epsilon tending to zero , setting first ##a=-\epsilon ,b=\epsilon## (##f(x)=g^e(x)\phi(x)##) and then first compute the limit as epsilon tending to zero and then in what is computed take the limit n tending to infinity. You ll find that you get infinity and not the desired result by doing it that way.

For b I would not work with the antiderivative explicitly. Instead, I would apply the mean value theorem for integrals
$$
\int_a^b f(x) dx = (b-a) f(c)
$$
for some ##c## such that ##a<c<b##. This means that
$$
\frac 1{2\epsilon} \int_{-\epsilon}^\epsilon \phi(x)dx
= \phi(x^*_\epsilon)
$$
where ##|x^*_\epsilon | < \epsilon## such that ##\lim_{\epsilon \to 0} \phi(x^*_{\epsilon}) = \phi(0)##. It just feels cleaner to me.

Orodruin said:

It just feels cleaner to me.

The whole idea seems nice and neat except that it is not so clear to me that $$\lim_{\epsilon \to 0} x_{\epsilon}^{*}=0$$.

To the OP:
Forget what i said in post #3 about the limit of the Riemann sum ( i ll leave the post there though and not delete it cause it might seem interesting to you)

What c) wants you to do is to compute $$\int_{-\infty}^{\infty}\lim_{\epsilon \to 0}g^{\epsilon}(x)\phi(x)dx$$

Delta2 said:

The whole idea seems nice and neat except that it is not so clear to me that $$\lim_{\epsilon \to 0} x_{\epsilon}^{*}=0$$.

If you choose ##\epsilon > 0## then for any ##\delta < \epsilon##: ##|x^*_{\delta}-0| < \epsilon## so ##x^*_\epsilon## converges to zero by definition.

Orodruin said:

If you choose ##\epsilon > 0## then for any ##\delta < \epsilon##: ##|x^*_{\delta}-0| < \epsilon## so ##x^*_\epsilon## converges to zero by definition.

Ye ok what confused me is the way you write it it seems to be a sequence with a continuous variable as index so you know very confusing but it is in fact a real function $$x_{\epsilon}^{*}=h(\epsilon)$$ such that $$|h(\epsilon)|<\epsilon$$ from which it follows that that limit is zero indeed.

Thank you Delta2 and Orodruin for your help

I have now tried to calculate the task c

##\int_{-\infty}^{\infty} \displaystyle{\lim_{\epsilon \to 0}} g^{\epsilon}(x)\phi(x)dx=\int_{-\epsilon}^{\epsilon}\displaystyle{\lim_{\epsilon \to 0}} \frac{1}{2 \epsilon}\phi(x)##

But if I now form the limit value, I get ##\displaystyle{\lim_{\epsilon \to 0}} \frac{1}{2 \epsilon}=\infty## or am I misinterpreting the term in the integral?

Lambda96 said:

Thank you Delta2 and Orodruin for your help

I have now tried to calculate the task c

##\int_{-\infty}^{\infty} \displaystyle{\lim_{\epsilon \to 0}} g^{\epsilon}(x)\phi(x)dx=\int_{-\epsilon}^{\epsilon}\displaystyle{\lim_{\epsilon \to 0}} \frac{1}{2 \epsilon}\phi(x)##

But if I now form the limit value, I get ##\displaystyle{\lim_{\epsilon \to 0}} \frac{1}{2 \epsilon}=\infty## or am I misinterpreting the term in the integral?

You cannot take the integral out of the limit. Its boundaries depend on ##\epsilon##!

Orodruin said:

You cannot take the integral out of the limit. Its boundaries depend on ##\epsilon##!

That's correct but i also kinda did what the OP did at #9 and i thought we get ##+\infty## in this case. But i see now its wrong. So I cant understand what task c is supposed to mean.

Thank you Orodruin and Delta2 for your help

Unfortunately, I also don't understand what is meant by commuting the integral and the limit. Isn't the integral (1) a Riemann integral, why does the problem specifically state that it should now be interpreted as one?

Lambda96 said:

Thank you Orodruin and Delta2 for your help

Unfortunately, I also don't understand what is meant by commuting the integral and the limit. Isn't the integral (1) a Riemann integral, why does the problem specifically state that it should now be interpreted as one?

I had the exact same question to myself . That is a Riemann definite integral anyway it isnt a Lebesque integral. The only thing I see is that the task means to view it as the limit of a Riemann sum (as i describe in post #3) and not as an antiderivative by using the fundamental theorem of calculus.

Ah hm , I think afterall what i say in post #6 is what the task c wants us to do but we didnt calculate it correctly.

I think that ##\lim_{\epsilon\to 0}g^{\epsilon}(x)=0## think about it and tell me your thoughts. So that integral is 0.