Dirac's delta integration

Question

What about the following integral?

$$\int_0^a x^3 \delta(x-1) dx$$

If $a$ is more or less than 1 it's all clear, but what if $a=1$.

Is the integral is equal to $1/2$ ?

Edit: this is my motivation, assume f(x) is even, $a>0$

$$\int_{-a}^a f(x) \delta(x) dx = \int_{-a}^0 f(x) \delta(x) dx + \int_{0}^a f(x) \delta(x) dx = -\int_{a}^0 f(-x) \delta(-x) dx + \int_{0}^a f(x) \delta(x) dx = 2\int_{0}^a f(x) \delta(x) dx = f(a)$$ implies $$ \int_{0}^a f(x) \delta(x) dx = \frac{f(0)}{2}$$

It's certainly true for even functions, what displaced odd functions puzzle me.

Answer 1

The function is not defined at $a=1$, or better, you have to specify what do you mean by that integral. Indeed write: $$\int_0^1x^3\delta(x-1)dx=\int_\mathbb{R}x^3\delta(x-1)\chi_{[0,1]}(x)dx=x^3\chi_{[0,1]}(x)|_{x=1}=1$$ but at the same time, if you write: $$\int_0^1x^3\delta(x-1)dx=\int_\mathbb{R}x^3\delta(x-1)\chi_{[0,1)}(x)dx=x^3\chi_{[0,1)}(x)|_{x=1}=0$$

Answer 2

The answer depends on the precise definition of things so heavily that it'd say this isn't well-defined.

But, just for entertainment, you can view the $\delta$-distribution as a measure, with the property that for $X \subset \mathbb{R}$, $\delta(X) = 1$ if $0 \in X$ and $\delta(X) = 0$ otherwise. Your integral then becomes (assuming you mean $x^3$ where you wrote $t^3$) $$ \int_0^a x^3\delta(x - 1) \,dx = \int_{-1}^{a-1} (x+1)^3 \delta(x) = \int_{-1}^{a-1} (x+1)^3 \,d\delta(x) \text{.} $$

So the question becomes, how does one deal with the integration bound for not absolutely continuous measures? For absolutely continuous measures, we know that single points have measure zero, so it doesn't matter much whether we assume the boundaries are included in the integral interval or not. But we don't have that liberty here, and so the result heavily depends on how we define this.

Let's start with a common way to define the distribution function of a measure $\mu$ (if it exists), which is requiring that $$ \mu\left((a,b]\right) = F_\mu(b) - F_\mu(a) \text{.} $$ This then makes distribution functions right-continuous. But note that this is simply a convention - we could just as well make all distribution function left-continuous by swapping the open and closed end of the interval.

Let's add to that the requirement that $$ \int_a^b 1 \,d\mu = F_\mu(b) - F_\mu(a) \text{,} $$ which seems sensible. Since we clearly want that $\int_A 1\,d\mu = \mu(A)$, we must then define $$ \int_a^b f\,d\mu = \int_{(a,b]} f \,d\mu \text{,} $$ i.e. decide that the upper bound is included, while the lower bound is excluded.

With those things in place, it follows that $$ \int_{-1}^0 (x+1)^3 \,\delta(x) = 1 $$ and that $$ \int_0^1 (x+1)^3 \,\delta(x) = 0 \text{.} $$

But this is highly arbitrary! For integrals with distributions in the integrald, my advice is to always use the syntax $\int_{I}$ where $I$ is some interval, instead of writing $\int_a^b$, because the former defined the precise set we're integrating over.