Periodic Dirac Delta Functions

Consider the functions

\begin{eqnarray}
P_1(t) &=& \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})} \\
P_2(t) &=& \frac{\cos[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})} \\
\end{eqnarray}

$P_1(t)$ is periodic with a period $2\pi$.

$P_2(t)$ is an odd function, $P_2(t)=P_2(-t)$, independently of $N$. Therefore

\begin{eqnarray}
\int_{-A}^{A}\frac{\cos[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}dt =0
\end{eqnarray}

for any $A \in \mathbb{R}$.

The integral of $P_1(t)$ for a single period is independent of $N$

\begin{eqnarray}
\frac{d}{dN}\int_{-\pi}^{\pi}\frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}dt &=& (N+\frac{1}{2}) \int_{-2\pi}^{2\pi}\frac{\cos[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}dt \\
&=& (N+\frac{1}{2}) \int_{-2\pi}^{2\pi} P_2(t)dt =0
\end{eqnarray}

which follows from the oddness of $P_2(t)$.

Theorem:

\begin{eqnarray}
\lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})} = 2\pi \sum_{k=-\infty}^{\infty} \delta(t-2\pi k)
\end{eqnarray}

 

Proof: As P_1 (t) is periodic with period $2\pi$, it is sufficient to prove the theorem for a single period.  For the interval $ -\pi<t<\pi$ the above theorem reduces to

\begin{eqnarray}
\lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})} = 2\pi \delta(t)
\end{eqnarray}

Let $f(t)$ be a smooth and bounded function for $ -\pi<t<\pi$. Foe $-\pi<a<0<b<\pi$ consider the integral

\begin{eqnarray}
\int_a^b \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt
\end{eqnarray} +

we can separate this integral into three parts:

\begin{eqnarray}
\int_a^{-\epsilon} \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt + \int_{-\epsilon}^{\epsilon} \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt +\int_{\epsilon}^b \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt
\end{eqnarray}

• The first and third integrals are multiplications of a smooth function $\frac{f(t)}{\cos(\frac{t}{2})}$ with an infimitely fast oscillation $\lim_{N \rightarrow \infty}\sin[(N+\frac{1}{2}) t]$. By the Riemann-Lebesgue theorem, the result is zero.
• The second integral is taken in a very narrow region around zero. In this region
  — $f(t) \approx f(0)$
  —$\sin(\frac{t}{2}) \approx \frac{t}{2}$

therefore

\begin{eqnarray}
\int_{-\epsilon}^{\epsilon} \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt = f(0)\int_{-\epsilon}^{\epsilon} \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\frac{t}{2}} dt
\end{eqnarray}

but we have derived above that

\begin{eqnarray}
\lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\frac{t}{2}} = 2 \pi \delta(t)
\end{eqnarray}

therefore

\begin{eqnarray}
\int_{-\epsilon}^{\epsilon} \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt = f(0)\int_{-\epsilon}^{\epsilon}2\pi  \delta(t) dt = 2\pi f(0)
\end{eqnarray}

and

\begin{eqnarray}
\int_{a}^{b} \lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})}f(t)dt = 2\pi f(0)
\end{eqnarray}

This indicates that

\begin{eqnarray}
\lim_{N \rightarrow \infty} \frac{\sin[(N+\frac{1}{2}) t]}{\sin(\frac{t}{2})} = 2\pi \delta(t)
\end{eqnarray}

on a single period. When we extend this result to all the periods, we prove the theorem.

Theorem: