# Integration of delta function

## Main Question or Discussion Point

Problem arises from next situation. If we have some distribution (of mass for example) on a ring which is given by

\begin{equation}
\rho=m\delta(\phi)
\end{equation} where phi is azimuthal angle.

What is the value of integral ?

\begin{equation}
\int_0^{2\pi} \! \rho \, \mathrm{d} \phi
\end{equation}

If we use definition of delta function

\begin{equation}
\int_0^\infty \! \delta({x}) \, \mathrm{d} x=1/2
\end{equation}

result is m/2 but in this problem we integrate on the whole domain so i am not sure if it is necessary to divide by half!

Thank you!

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chiro
Hey benjamin_cro and welcome to the forums.

Hint: Sub in m*d(whatever_that_symbol_is_in_greek) into your equation and remember that Integral (a,b+c) = Integral (a,b) + Integral (b,c) where a < b < c (i.e. they are limits of the integral).

Mute
Homework Helper
For convenience, I would probably take the lower limit of integration to be infinitesimally less than zero so that you get the result

$$\int_{0^-}^{2\pi} d\phi~m \delta(\phi) = m.$$

The reason I suggest this is because if you have a ring with a particle one it (which is where the mass is concentrated), then you can really assign the angular coordinate in any way you like and since you haven't changed the physical situation, you should get the same answer no matter how you do it. So, for instance, suppose I decide to orient the ring so that the mass is still at $\phi = 0$, but I decide that $\phi$ is in the range $[-\pi,\pi)$. Then, there is no issue at all:

$$\int_{-\pi}^\pi d\phi~m\delta(\phi) = m.$$

To put it in a slightly more technical way, really what you're doing is taking the delta function defined on the real line and compressing it onto the circle. Suppose you integrate the delta function against a periodic function f(x) (of period $2\pi$, say)

$$\begin{eqnarray*} \int_{-\infty}^\infty dx~f(x) \delta(x-x_0) & = & \sum_{k=-\infty}^\infty \int_{2\pi k}^{2\pi (k+1)} dx~f(x) \delta(x-x_0) \\ &= & \int_0^{2\pi}dx~\sum_{k=-\infty}^\infty f(x - 2\pi k) \delta(x-x_0 - 2\pi k) \\ &= & \int_0^{2\pi}dx~f(x) \sum_{k=-\infty}^\infty \delta(x-x_0 - 2\pi k) \\ &= & \int_0^{2\pi}dx~f(x) \delta_{\rm circ}(x-x_0~\mbox{mod} 2\pi). \end{eqnarray*}$$

The function $\delta_{\rm circ}(x-x_0~\mbox{mod} 2\pi)$ is defined in such a way that it will give a contribution to the integral whenever $x = x_0 - 2\pi k$ for some integer k. So, even if x is outside the integration bounds, it will still contribute to the integral. So, even if $x_0 = 0$ and you take the definition that the delta function only contributes a half when the argument is zero at a boundary, the circular delta function actually contributes at both boundaries: $x=0$ and $x = 2\pi$, so you would get two halves and the overall integral would give you $m$ again.

Thank you very much for your explanation.

If we use definition of delta function

\begin{equation}
\int_0^\infty \! \delta({x}) \, \mathrm{d} x=1/2
\end{equation}
Is this definition of delta function?
Sometimes people say that the delta function is an even function and
its integral from -∞ to ∞ is 1, therefore it is 1/2 if it is integrated from -∞ to 0
or 0 to ∞.
Is this logic completely and mathematically right?

pwsnafu
Is this definition of delta function?
Sometimes people say that the delta function is an even function and
its integral from -∞ to ∞ is 1, therefore it is 1/2 if it is integrated from -∞ to 0
or 0 to ∞.
Is this logic completely and mathematically right?
No, its a heuristic. Dirac is a distribution https://en.wikipedia.org/wiki/Dirac_delta#As_a_distribution so it acts on test functions. The two most popular classes of test functions are
1. smooth functions with compact support,
2. Schwartz functions.
Given a test function, f, we define
$\int_{-\infty}^\infty \delta(x) \, f(x)\, dx = f(0)$.
But f(x)=1 is not a valid test function, so $\int_{-\infty}^\infty \delta(x) \, dx = 1$ is invalid (it's actually undefined).

And $\int_0^\infty \delta(x) \, f(x) \, dx = f(0)/2$ is also convention. It's neither even nor odd. That's a property for functions whose domain is ℝ. What people mean is by that is choose a nanset which is even and approximate. The convention came about because
$\int_{-\infty}^\infty \delta(x) f(x) dx = \int_{-\infty}^0 \delta(x) f(x) dx + \int_0^\infty \delta(x) f(x) dx$
is desirable.

Edit: I forgot something. When you are not working on R, but on a locally compact abelian group, the test function space is the Schwartz-Bruhart space. On a ring (or a torus) that's all smooth functions, including f(x)=1. So it's allowed in this case.

Another edit: while I'm here, I also want to mention that $\int_{-\infty}^\infty \delta(x) f(x) \, dx$ is not an integral in the standard sense:
As a result, the latter notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral.

Last edited:
$\int_0^t \delta(x) \, dx$
Thank you for the detailed explanation.
I just wanted to know the above integral and asked a question.
Some say it becomes 1/2 and others say it is 1.

Actually, I'm not a mathematician and found this in a book of stochastic process.