Confused by a summation inside the integral: $\int_0^1 (v-u)^2 \sum_{i=0}^n \frac{u_i}{n}\delta_{i/n}(v) dv$

Question

I came across the following integral:

$\int_0^1 (v-u)^2 \sum_{i=0}^n \frac{u_i}{n}\delta_{i/n}(v) dv$,

where the deltas represent the Dirac delta distribution. How would one evaluate this integral? I got confused by the summation inside the integral. Any help would be grateful

Answer 1

You know that the "delta function" isn't a funcion, right? It is a Schwartz distribution, or generalized function.

The formulation $$ \int_0^1 (v-u)^2 \sum_{i=0}^n \frac{u_i}{n}\delta_{i/n}(v) dv $$ involving the distribution $\delta$ should be taken as short-hand for: $$ \int_0^1 (v-u)^2 \sum_{i=0}^n \frac{u_i}{n}\delta_{i/n}(v) dv = \sum_{i=0}^n \frac{u_i}{n}\int_0^1 (v-u)^2 \delta_{i/n}(v) dv = \sum_{i=0}^n \frac{u_i}{n} \left(\frac{i}{n}-u\right)^2 $$ where the last one no longer involves distruibutions.

Maybe.

There is still the question about the values $i=0$ and $i=n$, where $\frac{i}{n}$ is an endpoint of the interval $[0,1]$. Does your text have a convention about delta at an endpoint? Maybe take half of those two terms?