Proving the integral of the Dirac delta function is 1 Was wondering if my solution is mathematically accurate enough:
The question in the book yields:

Derive
  $$
1=\int_{-\infty}^{\infty} \delta(x-x_i)\ dx_i
$$
  From
  $$
f(x)=\int_{-\infty}^{\infty} f(x_i)\delta(x-x_i)\ dx_i 
$$
  [Hint: let $f(x)=1$]

My method is:

$$
f(x)=\int_{-\infty}^\infty f(x_i)\delta(x-x_i)\ dx_i
$$
$$
f(x)=1
$$
so
  $$ 
1=\int_{-\infty}^{\infty} \delta(x-x_i)\ dx_i
$$

I think this is not correct. Does someone know is this is correct, or how to do it better?
Cheers
 A: The error is that in the definition mentioned, f(x) has to be compactly supported and making it constant equal 1 is not correct. 
A: I think you can look at it as a definition.
You can also look at it as the limit of any function that is symmetric around the y axis, which integral is 1, and you make it narrower and narrower, forcing its peak to go to infinity. This function can be a gaussian, like shown in the wikipedia article gif image, or a uniform function, as shown in this Khan Academy video.
A: https://en.wikipedia.org/wiki/Dirac_delta_function
assumming the following limit as an accepted definition for delta
$$
{\displaystyle \delta (x)=\lim _{b\to 0}{\frac {1}{|b|{\sqrt {\pi }}}}e^{-(x/b)^{2}}}
$$
by letting
$$
{T(x) = {\frac {1}{|b|{\sqrt {\pi }}}}e^{-(x/b)^{2}}}
$$
we can
$$
 \int\limits_{-\infty}^{\infty} T(x)\ \mathrm dx\ = 1
$$
then we can take the limit as b -> 0 and as we can see the the value if the integration is independant of the limit so we can so we get delta in the side of the integration and one on the left side
A: It looks fine. We know that $$\delta_a(t-t_0)=\frac{1}{2a},\;\; \text{when}\;\;|t-t_0|<a\;\; \text{and}\;\;\delta_a(t-t_0)=0,\;\; \text{when}\;\;|t-t_0|\geq a$$ Here when $a$ tends to zero, the resulted expression which is not a function at all, as you noted above, is: $$\lim\delta_a(t-t_0)=\delta(t-t_0)$$ Using $$\int_{-\infty}^{\infty} \delta_a(t-t_0)\ dt=1$$ we can characterized two peraperties below for that: $$\delta (t-t_0)=\infty,\;\; \text{when}\;\;t=t_0\;\; \text{and}\;\;\delta (t-t_0)=0,\;\; \text{when}\;\;t\neq t_0 $$ and $$\int_{-\infty}^{\infty} \delta(t-t_0)\ dt=1$$
A: As it was mentioned in other answers and comments, the Dirac- $\delta$ is not a function, and it is not even defined pointwise. It is defined as element of the dual of the space of $C^{\infty}$ functions with compact support.
$$
\langle \delta, \phi\rangle = \phi(0), \quad \phi \in C^{\infty}_c
$$
When $f$ is locally integrable, it can be identified with the distribution
$$
\langle T_f, \phi\rangle \int_{\mathbb{R}} f(x) \phi(x) \,dx,
$$
but this is not the case with the Dirac - $\delta$. Although there are axiomatic definitions of distributions that allow the definition of value of a distribution, and therefore also the concept of integral, this is not very usual. Regarding distributions, what you can easily compute are derivatives:
$$
\langle T', \phi\rangle  = - \langle T, \phi'\rangle. 
$$
In particular, this means that any locally integrable function as derivative (in fact, derivatives of any order) in the sense of distributions:
$$
\langle T_f', \phi\rangle = - \int_{\mathbb{R}} f(x) \phi'(x)\, dx.
$$
So, although it is sometimes useful to identify the Dirac-$\delta$ with a function andf reason about its integral, this is mostly symbolic and should be taken as a definition, and not something that you can demonstrate using integration/measure.
A: If f(x) is a function of x, then it cannot be removed from the integral. Therefore this proof only works if f(x) is a constant.
