My question is about the $\delta$ function. It has following property: $$\int_{-\infty}^\infty f(x)\delta (x-t) \,\mathrm{d} x = f(t) $$ What's the meaning of the equation? Why not directly calculate $f(t)$?

  • $\begingroup$ It turns out this formalism has a lot of applications, particularly in solving differential equations. $\endgroup$
    – Simon S
    Jun 19, 2015 at 22:08
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    $\begingroup$ This is something called the sifting property, and is used to calculate $f(t)$. $\endgroup$
    – HDE 226868
    Jun 19, 2015 at 22:10
  • $\begingroup$ If you've studied vector calculus, you might be interested to note that it pops up naturally here: $$\nabla\cdot\left(\frac{\hat{r}}{\|r\|^{2}}\right)=4\pi \delta^{3}(\vec{r})$$ And this is extremely useful in deriving Coulomb's law for a pointlike particle. $\endgroup$ Jun 19, 2015 at 22:12
  • $\begingroup$ The Dirac's delta $\delta$ is defined in terms of area. The equation you wrote is the sampling property of $\delta$, very useful in signal processing. You can learn more about this function in mathworld.wolfram.com/DeltaFunction.html $\endgroup$ Jun 19, 2015 at 22:12

2 Answers 2


Aegon's answer gives an idea of the meaning. I'd like to say a little bit about why you would want such a thing. The basic application is to solutions of differential equations, and the idea is the following abstract calculation. Suppose you want to be able to solve the differential equation


where $L$ is some differential operator, $u$ is the unknown function, and $f$ is a given function. We'll say it's on the full space to avoid boundary issues, although this is useful in boundary value problems as well.

When studying this, it would be nice if we had a single, unified approach, so that we could solve the equation for one $f$ and then get a representation of the solution for every other $f$. The way that we do this is based on your equation, which can be abstractly written as

$$\delta * f = f$$

where $*$ denotes convolution. This means that if we can find $g$ such that


then we can convolve with $f$ on both sides to get


Finally if we can argue that $Lg*f=L(g*f)$, then we've solved the problem, and $g*f$ is our solution. This $g$ is called the Green's function or the fundamental solution for $L$. Like the Dirac delta, $g$ is never strictly speaking a function, it is always a distribution, although it is often a function away from $0$. It can be found explicitly in a number of very important examples, including linear ODEs with constant coefficients, and the three "classic" PDE, i.e. the Laplace, heat, and wave equations.

  • $\begingroup$ @Aegon@MarianoSuárez-Alvarez Good explain. My question is from MDP(Markov Decision Process). They use that equation to calculate the transition function $f(x_{k+1}|x_k,a_k)$ with noise $w_k$, whose density is $f_k ^w(w_k)$. The model is like this $x_{k+1}=h(x_k, a_k, w_k)$. Then they use the following equation:$$f(x_{k+1}|x_k,a_k)= \int \delta (x_{k+1}-h(x_k, a_k, w_k))\cdot f_k ^w(w_k)\, \mathrm{d}w_k$$ So I can't figure out why they do that. If we know $f_k ^w(w_k)$ already. Why don't we calculate $f(x_{k+1}|x_k, a_k)=f_k ^w(x_{k+1}- h(x_k, a_k))$ instead. $\endgroup$
    – Kun
    Jun 20, 2015 at 13:35
  • $\begingroup$ @Micheal That formula should be understood as "integrate $f_k^w(w_k)$ only over the set given by $x_{k+1}=h(x_k,a_k,w_k)$". So you don't integrate over all values of $w_k$. You can think of it as just like restricting the limits of integration, except that it generalizes in the right way to surfaces. (For example, integrating against $\delta(x^2+y^2-1)$ over $\mathbb{R}^2$ amounts to integrating over the circle.) In practice you use the co-area formula to perform these calculations if you're doing them by hand. $\endgroup$
    – Ian
    Jun 21, 2015 at 0:09
  • $\begingroup$ I can get the idea of what you talked about. But the point is why not directly calculate it and why we have to use the $\delta$ function. $\endgroup$
    – Kun
    Jun 21, 2015 at 0:24
  • $\begingroup$ @Micheal I would think of it as an intermediate step in the calculation. Like if $(X,Y)$ have joint pdf $f$, and you want $E(X|h(Y)=z)$, then this is $\int_{-\infty}^\infty \int_{-\infty}^\infty x f(x,y) \delta(h(y)-z)$. Then if I needed to actually calculate this I would use the co-area formula. $\endgroup$
    – Ian
    Jun 21, 2015 at 0:44

A Dirac-delta function, $\delta(x)$, is strictly speaking not a function. It is a distribution and the equation you have above is actually a defining property of the Dirac-delta function - it only makes mathematical sense under an integral.

What the equation intuitively

$\int_{-\infty}^{\infty} f(x) \delta(x - t) dx = f(t)$

means is that $\delta(x)$ vanishes identically everywhere expect at the origin, where it is infinitely peaked.

The equation isn't used to find $f(x)$; rather, it tells you how a $\delta$-function affects $f(x)$ when integrated against it.

$\delta$-functions are extremely useful and show up everywhere in physics and mathematics, for instance, when solving certain differential equations.

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    $\begingroup$ I think it might be better to say that the equation intuitively means that. Rigorously it means exactly what it says, nothing more, nothing less. We can use approximate identities to make that intuition precise, but at the end of the day the definition is what it is. I stress this because in my experience the "infinite peak" idea doesn't actually help you calculate anything. $\endgroup$
    – Ian
    Jun 19, 2015 at 22:40
  • $\begingroup$ That's right, I should make that explicit. $\endgroup$
    – Aegon
    Jun 19, 2015 at 22:41
  • $\begingroup$ It is not true that «it only makes mathematical sense under an integral». The integral appearing in the question is not an integral in any sensible sense. $\endgroup$ Jun 19, 2015 at 22:50
  • $\begingroup$ @MarianoSuárez-Alvarez That's a bit pedantic. The basic examples of distributions are integration against reasonable functions. We think of distributions which are not functions as "integration" against some "function" which is not strictly speaking a function. This formalism is actually rather productive; it informs us of how to define the convolution of two distributions, and how to define the Fourier transform of a tempered distribution. $\endgroup$
    – Ian
    Jun 19, 2015 at 22:52
  • $\begingroup$ No, it is not pedantic a all. The evaluation of a disribution at a test function is most certainly not an integration. Things are set up so that it behaves like an integration, of course, but that does not make it an integration. Integration against functions does define distributions, but that has nothing to do with this. The integral sign appearing in the OP's question is a convenient notation, not an integral. $\endgroup$ Jun 19, 2015 at 22:54

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