I have question,
Each of the following distributions is defined by its action. Identify the generalized function for each of them
a) $B_a(\phi)=a^2 \phi(a)$
b) $B_b= -e^{-b}(\phi(b)+\phi^{'}(b))$
c) $B_c= \int_{a}^{b} x^2 \phi(x)dx$
For part a) I know that since a distribution has form $\int_{a}^{b} f(x) \phi(x)dx$, $f(x)$ must be $x^2 \delta (x-a)$ So that's the generalized function. But is f(x) the generalized function or is the distribution the generalized function?
I know what the answers are supposed to be.
for c) the answer is $H(x-a)(1-H(x-b))x^2$ Edit: A little search yielded this : Writing integral in terms of distributions and Nikita's comment has been most helpful
I'd appreciate knowing why Paul's answer is also correct would appreciate more understanding how to do part b
Paul seems to use the following equalities: $$ \int_a^b f(x) \, dx = \int_{-\infty}^{\infty} \chi_{[a,b]}(x) \, f(x) \, dx = \int_{-\infty}^{\infty} (H(x-a)-H(x-b)) \, f(x) \, dx , $$ where $\chi_A$ is the indicator function of the set $A$ and $H$ is the Heaviside step function.
If $a<b$ then $$ H(x-a) = \begin{cases} 0 & (x<a) \\ 1 & (x>a) \\ \end{cases} = \begin{cases} 0 & (x<a) \\ 1 & (a<x<b) \\ 1 & (b<x) \\ \end{cases} \\ H(x-b) = \begin{cases} 0 & (x<b) \\ 1 & (x>b) \\ \end{cases} = \begin{cases} 0 & (x<a) \\ 0 & (a<x<b) \\ 1 & (b<x) \\ \end{cases} $$ Thus, $$ H(x-a)-H(x-b) = \begin{cases} 0-0=0 & (x<a) \\ 1-0=1 & (a<x<b) \\ 1-1=0 & (b<x) \\ \end{cases} $$ i.e. $ H(x-a)-H(x-b) = \chi_{[a,b]}(x) $ modulo values at the end points, where it doesn't really matter when integrating.