If we have the following integral containing Dirac delta functions:
$\int_{0}^{v}(v-x)^2\delta_1(x)dx - \int_{0}^{v}(v-x)^2\delta_0(x)dx$,
where $\delta_1(x)$ evaluates to 1 if $x=1$ and $\delta_0(x)$ evaluates to 1 if $x=0$.
Is there a (linear) transformation such that we can transform the integral domain from $(0,v)$ to $(-a,a)$. Or is there another way to solve this integral?
Any help would be grateful. Thanks in advance.
What I tried: By symmetry:
$\frac{1}{2} \int_{-v}^{v} (v-x)^2\delta_1(x)dx - \frac{1}{2} \int_{-v}^{v} (v-x)^2\delta_0(x)dx = \frac{1}{2}(v-1)^2 - \frac{1}{2}(v-0)^2 = \frac{1}{2} - v$.
However, I was told that symmetry is not the way due to the Dirac function.
And how can we then solve the following: I got $a(u)=1-2u$, however the textbook says $a(u)=-u$
$a(u) = \iint_{\hat{v}<v}(v-\hat{v})^2F(u,v) \frac{\partial}{\partial u} F(u,\hat{v}) d\hat{v} dv$ where $F(u,v) = (1-u)\delta_0(v)+u\delta_1(v)$ and $\frac{\partial}{\partial u} F(u, \hat{v}) = -\delta_0(\hat{v})+\delta_1(\hat{v})$
Symmetry doesn't really help in that case. Let's recall the "definition" of the Dirac delta : $$ \int_a^b f(x)\,\delta_c(x) \,\mathrm{d}x = \left\{ \begin{array}{cl} f(c) & \mathrm{if}\; c \in (a,b) \\ \frac{1}{2}f(c) & \mathrm{if}\; c=a \;\mathrm{or}\; c=b \\ 0 & \mathrm{otherwise} \end{array} \right. $$ Note that the middle line is a "limit case", whose convention can differ among references. In consequence, the first integral evaluates as $$ \int_0^v (v-x)^2\delta_1(x) \,\mathrm{d}x = \left\{ \begin{array}{cl} (v-1)^2 & \mathrm{if}\; v > 1 \\ 0 & \mathrm{otherwise} \end{array} \right. $$ while the second one gives $$ \int_0^v (v-x)^2\delta_0(x) \,\mathrm{d}x = \left\{ \begin{array}{cl} v^2 & \mathrm{if}\; v > 0 \\ 0 & \mathrm{otherwise} \end{array} \right. $$