Given that $$\sum_n e^{ikan}=C\sum_m \delta(k-2\pi m/a)\tag{1}$$ for $k\in\mathbb R,a\neq0\in\mathbb R,n\in \mathbb Z, m\in\mathbb Z$, I want to prove that $C=2\pi/|a|$.
Proof
First integrate both sides of $(1)$ over the interval $[-\pi/a,\pi/a]$: $$\int_{-\pi/a}^{\pi/a}\mathrm dk\sum_n e^{ikan}=C.\tag{2}$$ Now we can switch the sum and the integration symbol $$\sum_n\int_{-\pi/a}^{\pi/a}\mathrm dk\,e^{ikan}=C.\tag{3}$$ Next evaluate the integral $$\int_{-\pi/a}^{\pi/a}\mathrm dk\,e^{ikan}=\frac{2}{an}\sin(n\pi)\tag{4}$$ For $n\neq 0$ this last expression is zero. For $n=0$ we can say $$\frac{2}{an}\sin(n\pi)=\frac{2}{a}\lim_{x\rightarrow0}\frac{\sin(x \pi)}{x}=\frac{2\pi}{a}.\tag{5}$$ Summarized $$\int_{-\pi/a}^{\pi/a}\mathrm dk\,e^{ikan}=\begin{cases}\frac{2\pi}{a} & n=0 \\0, & n \neq 0.\end{cases}\tag{6}$$
Finally take the sum \begin{align}C&=\sum_n\int_{-\pi/a}^{\pi/a}\mathrm dk e^{ikan}\tag{7}\\ &=\sum_n \begin{cases}\frac{2\pi}{a}& n=0,\\0&n \neq 0.\end{cases}\\ &=\frac{2\pi}{a} \end{align} I'm especially uncertain about the steps where I
- Switch summation and integration,
- Equate $\sin (n\pi)/n$ to its limit when $n=0$,
- I'm supposed to get $|a|$ but instead I just get $a$.