I am stuck in this question given here because I am facing a dilemma whether to insert a dirac-delta function within the integral (or not) and solve it, or there is some other method to do this problem.
[edit:- I still can't solve it. So, kindly can anyone let me know how to solve this question. I am really interested to know its solution.]
On
What you should show is that for every $\varphi\in C^\infty_c(\mathbb{R})$ the following limit holds: $$ \lim_{\epsilon\to 0} \int_{-\infty}^{\infty} \frac{1}{\epsilon} f(\frac{x}{\epsilon}) \varphi(x) \, dx = \varphi(0) . $$
Using the substitution $x=\epsilon y$ we have $$ \int_{-\infty}^{\infty} \frac{1}{\epsilon} f(\frac{x}{\epsilon}) \varphi(x) \, dx = \int_{-\infty}^{\infty} \frac{1}{\epsilon} f(y) \varphi(\epsilon y) \, \epsilon\,dx = \int_{-\infty}^{\infty} f(y) \varphi(\epsilon y) \, dx . $$
Here $\varphi(\epsilon y)\to\varphi(0)$ as $\epsilon\to 0.$ If $f\in L^1(\mathbb R)$ then, by dominated convergence theorem, $$ \lim_{\epsilon\to0} \int_{-\infty}^{\infty} f(y) \varphi(\epsilon y) \, dx = \int_{-\infty}^{\infty} \lim_{\epsilon\to0} f(y) \varphi(\epsilon y) \, dx = \int_{-\infty}^{\infty} f(y) \varphi(0) \, dx = \int_{-\infty}^{\infty} f(y) \, dx \, \varphi(0) = \varphi(0). $$
I think the notion is that if the integral converges without any sort of regularization scheme, $f(x)$ must go to zero at $\pm \infty$. Then, you can show that the transformed function goes to zero at any $x \ne 0$ and it still integrates to unity. Those are the two defining characteristics of a Dirac delta function.
If you did have a regularization scheme like taking the integral over reals to be the limit of $\int_{-a}^{a}$ as $a \rightarrow \infty$ then I don't think the result is true any more, at least not without some further regularization.