Find $\int_{-1}^{1}f(x) \delta(x)dx $ and $\int_{1}^{\infty}f(x)\delta (x)dx$, where $\delta$ is the Dirac delta

Question

Let $f(x) = \exp(-x^{2})$, calculate $$\int_{-1}^{1}f(x) \delta(x)dx $$ and $$\int_{1}^{\infty}f(x)\delta (x)dx.$$

I know that $\int_{-\infty}^{\infty}f(x) \delta (x)dx = f(0)$, is it the same if we change the integration limits to $[-1, 1]$ seeing as $0$ is in that interval?

Answer 1

Since$$\int_a^bg(x)dx=\int_{-\infty}^\infty g(x)\chi_{[a,\,b]}(x)dx=\int_{-\infty}^\infty g\chi_{(a,\,b]}dx=\int_{-\infty}^\infty g\chi_{[a,\,b)}dx=\int_{-\infty}^\infty g\chi_{(a,\,b)}dx,$$the choice $g(x)=f(x)\delta(x)$ lets you evaluate $\int_a^bgdx$ as $f(0)$ provided $0\in(a,\,b)$, or $0$ provided $0\notin[a,\,b]$. If $0\in\{a,\,b\}$ the integral is undefined, since the result wouldn't be the same for all four choices of the indicator function.