# Integral of dirac delta function times another another function

Dirac delta function $$\delta(x)$$ is defined with two properties:
1) At $$x=0$$ its value is $$\infty$$ and everywhere else it is $$0$$
2) Area under the curve is $$1$$

How does above definition result in $$\int\limits_{-\infty}^{\infty}\delta(x)f(x)dx = f(0)$$ ? Shouldn't the integral evaluate to $$\delta(0)f(0)$$?

• That's just by definition, so there is not a really a "should". Why do you expect it to be other value? Commented Nov 20, 2019 at 13:39
• Your definition of Dirac delta is wrong. It is a "baby" definition for those who do not know much mathematics. That integral is closer to the real definition, although still needs some explanation. Commented Nov 20, 2019 at 13:41
• en.wikipedia.org/wiki/Dirac_delta_function Commented Nov 20, 2019 at 13:43
• Suppose $f(x)=1$. Then $\int \delta(x)=1=f(0)$, so there’s really no way to have any additional dependence on $\delta(x).$ Also, 1) is heuristic not a rigorous definition. Commented Nov 20, 2019 at 13:47
• ...a function equal to zero everywhere except for zero and whose integral over the entire real line is equal to one ... In fact, there is no such function. Read down further on that Wiipedia page to Definitions. Commented Nov 20, 2019 at 13:49

I suggest reading about distributions (generalised functions), e.g. https://en.wikipedia.org/wiki/Distribution_(mathematics) . The point is, Dirac's $$\delta$$-function is not a function at all, at least not in the usual sense. What you have given as a "definition" is merely an intuition for it.

What $$\delta$$-function is is this: it is a linear functional that maps a certain class of functions (infintely smooth real functions that are nonzero only on a bounded set of $$\mathbb R$$) into real numbers, the following way:

$$\delta(f)=f(0)$$

Now, the right side looks like your right-hand side, but what about the left-hand side? One should note that, if $$g:\mathbb R\to\mathbb{R}$$ is a locally integrable function, then you can use it to define another functional of similar type:

$$T_g(f)=\int_{-\infty}^{\infty}g(x)f(x)dx$$

... and because for a given $$T_g$$ the function $$g$$ can be proven to be uniquely determined (up to a set of zero measure), then you can, in a sense, identify $$T_g$$ with $$g$$.

Now, in a common abuse of notation, we imagine that the functional $$\delta$$ is also identified with a fictional function $$\delta(x)$$ such that:

$$\delta(f)=\int_{-\infty}^{\infty}\delta(x)f(x)dx$$

so that is where the left-hand side is coming from. This is an abuse of notation because no such function $$\delta(x)$$ exists.