Proving the integral of the Dirac delta function is 1

Question

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

Answer 1

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 $$\int_{-\infty}^{\infty} \delta(t-t_0)\ dt=1$$

Answer 2

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.