Why $\int_0^\infty x\delta (y-x)dx=y$? Why $$p(y)=\int_0^\infty x\delta (y-x)dx=y\ \ ?$$
For me, $$p(y)=\int_0^\infty x\delta (y-x)dx=\int_{\{y\}}xdx=0.$$
If it would be written $\int_0^\infty xd\delta _y$, then I would be agree with the answer. But here it's written $$\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$$
what I interpret as $\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$.
 A: Mathematical approach.  The Schwartz distribution $\delta(y-x)\;dx$ is a measure.  It is the unit point mass at the point $y$.  Let's write $\epsilon_y$ for that unit mass.  So we have
$$
\int_0^\infty x\delta(y-x)\;dx \qquad\text{is notation meaning}\qquad \int_0^\infty x\;\epsilon_y(dx)
$$
If $y>0$, we get answer $y$.  If $y<0$ we get the answer $0$.  If $y=0$, then this is ambiguous.  For the integral from $0$ to $\infty$ do we mean the set $[0,\infty)$ or the set $(0,\infty)$ ??  A mathematican would write either
$$
\int_{[0,\infty)} x\;\epsilon_y(dx)\qquad\text{or}\qquad \int_{(0,\infty)} x\;\epsilon_y(dx)
$$
instead of something ambiguous.
A: The $\delta_y$ of Dirac is a measure that is defined as
$$
\delta_y(A)=\begin{cases}1,& y\in A\\ 0,&y\notin A\end{cases}
$$
for any subset $A$ of the measure space (in your case the measure space seems to be $(0,\infty)$ or $\Bbb R$). In your case you have
$$
\int_0^\infty x\delta_y(dx)=\int_{\{y\}\cap (0,\infty)}x\delta_y(dx)+\underbrace{\int_{(0,\infty)\setminus\{y\}}x\delta_y(dx)}_{=0}\\=\delta_y\big(\{y\}\cap(0,\infty)\big)\,x|_y=\begin{cases}y,& y\in(0,\infty)\\0,&y\notin (0,\infty)\end{cases}
$$ 
The reasons why this is the value of the integral can be understood knowing the theory of the Lebesgue integral (or Bochner integral) for arbitrary measures.

The integral $\int_X f(x)\mu(dx)$ for some arbitrary measure space $(X,\mu,\Bbb R)$ is defined as the limit of a sequence of integrals of simple functions that approximate the value of the integral of $f$ in $X$ in some sense.
A simple function is a finite sum of weighted characteristic functions of measurable subsets of $X$ with finite measure (similarly to Riemann sums), a simple function have the form
$$
s(x):=\sum_{k=0}^m \chi_{A_k}(x)\,c_k\tag1
$$
where $\mu(A_k)<\infty$ for each $k=1,\ldots,m$ and the $c_k$ are constants. 
Let $(f_j)$ a sequence of simple functions. If $(f_j)\to f$ point-wise almost everywhere then
$$\int_X f(x)\mu(dx):=\lim_{j\to\infty}\int_X f_j(x)\mu(dx)\tag2$$
And for a simple function the Lebesgue integral is defined by
$$\int_X f_j(x)\mu(dx):=\sum_{k=0}^{n_j}\mu(A_{k,n_j})c_{k,n_j}\tag3$$
In your case we have the sequence of simple functions $f_j(x):=\sum_{k=0}^{n_j}\chi_{A_{k,j}}(x)c_{k,j}$ defined by
$$
A_{k,j}:=[jk2^{-j},j(k+1)2^{-j}),\quad\text{for }n_j:=2^j,\quad\text{and }c_{k,j}=jk2^{-j}\tag4
$$
Then we find that $\lim_{x\to\infty}f_j(x)= x$ for each $x\in(0,\infty)$ and
$$
\int_0^\infty f_j(x)\delta_y(dx)=\sum_{k=0}^{2^j}\delta_y(A_{k,2^j})\, jk2^{-j}=\begin{cases}jm2^{-j}, &y\in A_{m,2^j}\text{ for some }m\\0,&\text{otherwise}\end{cases}\tag5
$$
With a bit of work it can be shown that $\int_0^\infty x\delta_y(dx)=\lim_{j\to\infty}\int_0^\infty f_j(x)\delta_y(dx)=y$ whenever $y\in(0,\infty)$.
A: By the definition of the dirac delta function,
$$
\int_{0}^\infty f(x) \delta(y-x) dx = f(y)
$$
for all $y \in [0, \infty)$ and arbitrary function $f$.
