1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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

3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $f$.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.