# 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 $$\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$$.

• The function ${\bf 1}_{x=y}$ is $1$ if $x=y$ and $0$ otherwise. This is quite different from the distribution $\delta(y-x)$ which is defined via $\int_{\mathbb{R}} f(x)\delta(x-y){\rm d}x = f(y)$. Oct 22, 2018 at 13:14
• There is no way you can write $\delta$ as 'function'. Dirac delta is different from Kronecker delta. Oct 22, 2018 at 13:24
• the notation $\delta(x-y)dx$ is somewhat confusing or misleading. I think it would be more clear to write $\delta_y(dx)$ or just $\delta_y$ Oct 22, 2018 at 13:53

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.

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 $$y\in(0,\infty)$$.

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 $$f$$.

• By definition ? but $\delta (y-x)=\boldsymbol 1_{x=y}$... So should be $\int_0^\infty f(x)\boldsymbol 1_{y=x}dx$... I don't get the point. It's not written $\int_0^\infty xd\delta_y$ and thus I would agree with your answer... Oct 22, 2018 at 13:10
• @user601023 That’s not the right definition of the delta function.
– Paul
Oct 22, 2018 at 13:11
• @Paul: I modified my previous comment. Oct 22, 2018 at 13:13
• @user601023 "but $\delta(x-y) = \mathbf{1}_{x=y}$" I don't think this ist true depending on the defenition of $\mathbf{1}$ you provide above... Oct 22, 2018 at 13:22
• I think you're thinking of the Kronecker delta: mathworld.wolfram.com/KroneckerDelta.html Oct 22, 2018 at 13:36