What is the value of the integral $\int_{-\infty}^a \delta(x - a) dx$ and related integrals? All the Dirac delta integrals I have seen either contain the Dirac delta function strictly within the limits of the lower and upper limits of integration or do not contain the delta function. What happens when one of the limits is at the location of the Dirac delta.
Also, how does one solve $\int_{-\infty}^a f(x)\, \delta(x - a) \,\mathrm d x$? Does the sifting property work now?
 A: So the short answer is that the notation
$$
\int_{-\infty}^a \delta(x-a)\,f(x)\,\mathrm d x
$$
does not make sense. Do you have a definition for it?

Here is a longer answer. So first the Dirac delta $\delta_a(x) = \delta(x-a)$ is not a function, it is either defined as a distribution of order $0$ or as a bounded measure. As a measure, it is defined as acting on a set $A$ by
$$
\delta_a(A) = \left\{
\begin{array}{}
1 &\text{ if } a∈ A
\\
0 &\text{ if } a\notin A.
\end{array}\right.
$$
Similarly as for the Lebesgue measure $\mathrm d x$, one can then define the integration with respect to this new measure $\delta_a(\mathrm d x)$ by
$$
∫_{\mathbb{R}} \varphi(x)\, \delta_a(\mathrm{d}x) = \varphi(a)
$$
for any nice function $\varphi$. In the same way if $f∈L^1$ is a classical function, it can be identified with the measure $\mu_f(\mathrm{d}x) = f(x)\,\mathrm{d}x$ verifying
$$
∫_a^b \varphi(x)\,\mu_f(\mathrm{d}x) = ∫_a^b \varphi(x)\,f(x)\,\mathrm{d}x = ∫_{[a,b]} \varphi(x)\,f(x)\,\mathrm{d}x = ∫_{(a,b)} \varphi(x)\,f(x)\,\mathrm{d}x
$$
and this is the reason why the Dirac delta is often identified to a function. Notice however that the last identity is a special feature of locally integrable functions (coming from the fact that the integral is the same if we remove a set of measure $0$) which allows to use the notation $\int_a^b$. This is not the case of the dirac measure. Therefore we have to specify if we are looking at the integral of the dirac over $[a,b]$ or $(a,b)$. And then we have
$$
\begin{align*}
\int_{[a,b]} \varphi(x)\,\delta_a(\mathrm{d}x) &= \varphi(a)
\\
\int_{(a,b)} \varphi(x)\,\delta_a(\mathrm{d}x) &= 0.
\end{align*}
$$

So the full answer is
$$
\begin{align*}
\int_{-\infty}^a \varphi(x)\,\delta_a(\mathrm{d}x) &\text{ is not a clear notation}
\\
\int_{(-\infty,a]} \varphi(x)\,\delta_a(\mathrm{d}x) &= \varphi(a)
\\
\int_{(-\infty,a)} \varphi(x)\,\delta_a(\mathrm{d}x) &= 0.
\end{align*}
$$
