What does the integral of a delta distribution even mean?

Question

Formally, we define $\delta(\phi)=\phi(0)$ where $\phi$ comes from a suitable class of test function. Based on this, the expression $\int_{-\infty}^{\infty} \delta(x) dx$ seems completely meaningless and I'm unsure how to attribute meaning to integrals involving $\delta$. I feel like I'm missing something important here - help!!

Answer 1

For a function $f\in L^1_{\text{loc}}(\Bbb R)$, we can associate with it the distribution $T_f\in \mathcal D'(\Bbb R)$ given by $$ \langle T_f, \varphi\rangle := \int_{-\infty}^\infty f(x)\varphi(x) \,dx. $$

The Dirac distribution is defined as $$ \langle \delta, \varphi\rangle := \varphi(0) $$ and is not associated with any $f\in L^1_{\text{loc}}(\Bbb R)$ (but can be associated with an atomic measure). However, some people (the physicists) prefer to abuse the notation and write $$ \langle \delta, \varphi\rangle =: \int_{-\infty}^\infty \delta(x)\varphi(x) \,dx $$ anyway, as if $\delta$ is a function. Hence we have $$ \int_{-\infty}^\infty \delta(x)\,dx = \langle \delta, 1\rangle = 1. $$ Note that we can use the constant function $1$ as an input because $\delta$ is not only a distribution, but a distribution with compact support, hence it acts on any $\psi\in C^\infty(\Bbb R)$.

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To add more details about my final remark, we have the following inclusion $$ \mathcal D \subset \mathcal S \subset \mathcal E $$ where $\mathcal D$ is the space of smooth and compactly supported test functions, $\mathcal S$ the space of Schwartz functions and $\mathcal E$ the space of smooth functions (each with its appropriate topology). Their dual spaces can be identified with subspaces of $\mathcal D'$, i.e. $$ \mathcal E' \subset \mathcal S' \subset \mathcal D' $$ where $\mathcal E'$ is the space of distributions with compact support and $\mathcal S'$ the space of tempered distributions.

The Dirac distribution $\delta$ belongs to $\mathcal E'$ and hence $\mathcal S'$, which is the "correct" space to do Fourier transform. This is what @Andreas meant in his comment about $\delta$ interacting well with Fourier transform.