# A question about definite integral

When I was doing revision for definite integral, I got confused with one step.

Here is the statement:

For all continuous functions $$f$$ which are continuous at $$0$$:

$$\lim_{ε\to 0}\int_{-ε}^ε f(x)·D(x,ε)dx = f(0)·\lim_{ε\to 0}\int_{-ε}^ε D(x,ε)dx$$

where $$D(x,ε)$$is the Dirac Delta function.

But I don't really understand why is this true. Could anyone tell me the steps between them?

Rewritten as a teaching exercise:

Please explain why for all functions $$f$$ continuous at $$0$$:

$$\lim\limits_{\epsilon \to 0} \int\limits_{-\epsilon}^\epsilon f(x) \delta(x)\ dx = f(0) \lim_{\epsilon \to 0}\int\limits_{-\epsilon}^\epsilon \delta(x)\ dx = f(0).$$

where $$\delta$$ is a Dirac delta function.

Here is the definition of Dirac delta function

$$\lim_{\epsilon \to 0}\delta(t;\epsilon)=0\quad\forall t\neq0$$ and $$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \delta(t;\epsilon)dt=1$$

• Take $f=1$ a constant function. Then the integral equals $2\epsilon$ which tends to $0$ when $\epsilon \rightarrow 0$. Apr 19, 2020 at 19:24
• @ANOOB: You're new here (as revealed by your reputation), so I urge you to learn from experts here on how to ask question. "When I was doing revision for definite integral" is both ungrammatical ("doing revision"??), but also unnecessary and hence a waste of your readers' time. Also, the fact that you're "confused" is implicit in the fact that you're asking a question. You don't need to restate it. You also have many errors in MathJax, which is an added burden on readers. "Here is the statement" is redundant: just give the statement. I'll post a better question below, so you see. Apr 19, 2020 at 19:49
• What is your definition of $D(x,\varepsilon)$ ? Apr 19, 2020 at 19:50
• @DavidG.Stork Thank you so much for the correction! This is my first time asking a question so sorry about the errors I have made! Apr 20, 2020 at 5:12
• @Tuvasbien I have added the definition, thank you! Apr 20, 2020 at 5:28

## 1 Answer

Consider the function $$f(x)=6$$. Then, as you can see easily the integral $$\int_{-\epsilon}^{\epsilon}f(x)=0$$ , as $$\epsilon$$ goes to 0, not $$f(0)$$.