I have a function
(1) $E(u) = S(u) + u^2 = u(1-u) + u^2 = u$. We can compute $E(u)$ using the given integral:
$E(u) = \int_{-\infty}^{\infty} v^2(1-u)\delta_0(v)dv + \int_{-\infty}^{\infty} v^2u\delta_1(v)dv = (1-u)\int_{-\infty}^{\infty} v^2\delta_0(v)dv + u\int_{-\infty}^{\infty} v^2\delta_1(v)dv$.
I know that $\int_{-\infty}^{\infty} \delta(v)dv =1$ and we can also use the fact that $\int_{-\infty}^{\infty} f(v)\delta(v)dv = f(v)\int_{-\infty}^{\infty} \delta(v)dv = f(0)$.
However I do not know how to proceed from this integral to get the same answer as in (1). I have never worked with an integral containing two "different" dirac delta functions. Since $f(0) = 0$ in both parts of the integral, I get $E(u) = 0$ instead of $E(u) =u$.
Any tips or any explanation would be gratefull. Thanks in advance.
$$ \begin{array}{rcl} E(u) &=& \displaystyle (1-u)\int_\mathbb{R}v^2\,\delta_0(v)\,\mathrm{d}v + u\int_\mathbb{R}v^2\,\delta_1(v)\,\mathrm{d}v \\ &=& \displaystyle (1-u) \cdot \left.v^2\right|_{v=0} + u \cdot \left.v^2\right|_{v=1} = u \end{array} $$