homework and exercises - What intermediate steps of the Dirac Delta Function and Fourier Series am I missing in finding a solution to the Kronig-Penney Model?

We're looking at the Kronig-Penney model in class and one of the conundrums is related to the Kronig-Penney potential for a chain of $N$ atoms. I'm supposed to squeeze out some expression for the Fourier components, $U_G$, but I don't end up with the right expression, although I think I did the right stuff.

The following is given for the delta potential function,

$$\begin{equation} \tag 1 U(x)=A\sum\limits_{n=-\infty}^\infty \delta(x-na), \end{equation}$$ of which I have to show that the Fourier components are given by $$\begin{equation} \tag 2 U_G=\frac{A}{a}, \qquad G\in\mathbb{Z}, \end{equation}$$ using the hint that $\sum^N_{n=0}\cos(Gna)=N$.

---

I'll start with the generalized form of $U_G$ (I am not sure whether this expression is correct),

$$\begin{equation} \tag 1 U_G=\frac1a\int_a U(x)e^{-iGx}\mathrm{d}x, \end{equation}$$ in which I can substitute my potential function $U(x)$ ($U(x)=A\sum\limits_{n=-\infty}^\infty \delta(x-na)$), to find $$\begin{equation} \tag 2 U_G=\frac1a\int_a A\sum\limits_{n=-\infty}^\infty \delta(x-na)e^{-iGx}\mathrm{d}x. \end{equation}$$ $A$ is just a constant and the sum and integral signs can be swapped, so $$\begin{equation} \tag 3 U_G=\frac{A}a\sum\limits_{n=-\infty}^\infty \int_a\delta(x-na)e^{-iGx}\mathrm{d}x. \end{equation}$$ Now by the definition of the delta function, $$\begin{equation}\tag 4 \int_a\delta(x-na)e^{-iGx}\mathrm{d}x=\int_a\delta(x-na)f(x)\mathrm{d}x=f(na)=e^{-iGna} \end{equation}$$ Back to our original equation, $$\begin{align} \tag 5 U_G&=\frac{A}a\sum\limits_{n=-N}^Ne^{-iGna}\\ \tag 6 &=\frac{2A}a\sum\limits_{n=0}^N\frac{e^{-iGna}+e^{iGna}}2\\ \tag 7 &=\frac{2A}{a}\sum\limits_{n=0}^N\cos(Gna)\\ \tag 8 &=\frac{2A}{a}N. \end{align}$$ ...which doesn't add up?

---

The wikipedia article on this derivation simply skips a few steps but I'm having some trouble filling those in:

$$\begin{align} \tag 9 U_G&=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}\\ \tag {10} &=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}\\ \tag {11} &=\frac A a \end{align}$$

See this link for more info.

Answer

From this step,

$$U_G=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}$$

note that the summation is an impulse train with spacing $a$. Since the integral is from $-\frac{a}{2}$ to $\frac{a}{2}$, just the impulse at $x = 0$ is integrated over so only the $n=0$ term in the summation contributes to the integral:

$$U_G = \frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx} = \frac1a\int_{-a/2}^{a/2}\mathrm{d}xA\cdot \delta(x)e^{-iGx} = \frac{Ae^{-iG0}}{a} = \frac{A}{a}$$