The Dirac Delta function $\delta(x)$ is very cool in the sense that
$$ \delta(x) = \begin{cases}
+\infty, \, & x =0 \\
0, \, & x \ne 0
\end{cases}
$$
Its unique characteristics do not end there though, because when integrating the Dirac Delta function we would get
$$ \int_{-\infty}^\infty \delta(x) dx = 1$$
Or, if we have another function $f(x)$ multiplied to the Dirac Delta function and integrating them we would get
$$ \int_\infty^\infty f(x) \delta(x) \, dx = f(x) \int_{-\infty}^\infty \delta(x) \, dx = f(0) $$
$$\\$$
Since
$$ \int_{-\infty}^\infty \delta(x) \, dx = \begin{cases}
0, \, & x \ne 0 \\
1, \, & x = 0 \end{cases}$$
Therefore in the previous integral we would have
$$\int_{-\infty}^\infty f(x)\delta(x) \, dx = \int_{-\infty}^\infty f(0)\delta(0) \, dx = f(0)$$.
What if we have $\delta(x-a)$? It's the same thing! The only thing here is we need to satisfy the condition $x-a = 0$ such that $\int_{-\infty}^\infty \delta(x-a) \, dx = 1$.
If the bounds of our integral though is not infinity, we need to make sure that the if we let $x = a$, $a$ would be in the bounds of the integral or else the integral would evaluate into zero.
$$\int_{-b}^b f(x)\delta(x-a) \, dx = \begin{cases}
0, & \text{if $b<a$ or $-b >a$ such that we cannot let $x \ne a$} \\
f(0), & \text{if $-b<a<b$ such that we can let $x = a$}
\end{cases}$$.
What if we have $\int_{-\infty}^\infty f(x) \delta'(x-a) \, dx$? Let's then compute it by integration by parts.
$$\int_{-\infty}^\infty f(x) \delta'(x-a) \, dx = f(x)\delta(x-a)\bigg|_{-\infty}^\infty - \int_{-\infty}^\infty \delta'(x-a)f(x) \, dx = -f'(a)$$
Or in general
$$\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) \, dx = (-1)^n f^{(n)}(a)$$.
You have to use the property
$$\delta(g(x)) = \sum_i \frac{\delta(x-c_i)}{\vert g'(c_i) \vert},\tag{1}$$
where $c_i$ are the roots of $g$. This may look strange but the proof is not very difficult.
Also, the relation that you wrote is not correct. Once you expressed the delta as shown in (1), then you use:
$$\int_a^b f(x)\delta(x-x_0) dx = \begin{cases}
f(x_0) & \text{if } a<x_0<b,\\
0 & \text{otherwise.}
\end{cases}$$
Your zeros are clearly $-1, 1, 2$.
Best Answer
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