What does the integral of a delta distribution even mean? 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!!
 A: 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 $\psi\in C^\infty(\Bbb R)$.

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.
