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

  • $\begingroup$ 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 $\int_{\mathbb{R}} f(x)\delta(x-y){\rm d}x = f(y)$. $\endgroup$
    – Winther
    Oct 22, 2018 at 13:14
  • $\begingroup$ There is no way you can write $\delta$ as 'function'. Dirac delta is different from Kronecker delta. $\endgroup$ Oct 22, 2018 at 13:24
  • $\begingroup$ 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 $\delta_y$ $\endgroup$
    – Masacroso
    Oct 22, 2018 at 13:53

3 Answers 3


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 $f$.

  • $\begingroup$ 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... $\endgroup$
    – user601023
    Oct 22, 2018 at 13:10
  • $\begingroup$ @user601023 That’s not the right definition of the delta function. $\endgroup$
    – Paul
    Oct 22, 2018 at 13:11
  • $\begingroup$ @Paul: I modified my previous comment. $\endgroup$
    – user601023
    Oct 22, 2018 at 13:13
  • $\begingroup$ @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... $\endgroup$
    – denklo
    Oct 22, 2018 at 13:22
  • $\begingroup$ I think you're thinking of the Kronecker delta: mathworld.wolfram.com/KroneckerDelta.html $\endgroup$
    – Chris
    Oct 22, 2018 at 13:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .