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## 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!

\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!