Show that the Fourier transform of a periodic potential is $0$ for all but two values of $k$

Question

I am asked to compute the Fourier transform of a periodic potential, $V(x)=\beta \cos(\frac{2\pi x}{a})$ such that, $$\tilde{V}(k)=\frac{1}{L}\int_0^LV(x)e^{-ikx}\,dx$$ and hence show that it is non-zero for only two values of $k$. I have, $$\tilde{V}(k)=\frac{\beta}{L}\int_0^L\cos(\frac{2\pi x}{a})e^{-ikx}\,dx.$$ According to Wolfram, this gives, $$\tilde{V}(k)=\frac{ia\beta e^{-ikL}(-ake^{ikL}+ak\cos(\frac{2\pi L}{a})+2i\pi\sin(\frac{2\pi L}{a}))}{L(a^2k^2-4\pi^2)}.$$ I am also given that $L=Na$, $N\in\mathbb{Z}^+$, such that the expression simplifies to, $$\frac{i\beta e^{-ikL}(-ake^{ikL}+ak)}{N(a^2k^2-4\pi^2)}.$$ From this we have, $$\tilde{V}(k)=\frac{ia\beta(ke^{-ikNa}-1)}{N(a^2k^2-4\pi^2)}.$$ However, given this, I have no idea how to show that this is not equal to $0$ for only two values of $k$. I tried setting $ke^{-ikNa}-1$ to $0$, such that, $$e^{-ikNa}=\frac{1}{k},$$ and solving for $k$, however, I'm not sure if this is the correct approach.

Edit: $N\gg 1$.

Answer 1

If $k$, $N$ and $a$ are all real numbers then $\displaystyle{\frac{1}{k}}$ is a real number and

$$\left|\mathrm e^{-\mathrm i kNa} \right| = 1$$

The only real numbers with absolute value $1$ are $\pm 1$.

If $\displaystyle{\mathrm e^{-\mathrm i kNa}=\frac{1}{k}}$ then $k = \pm 1$.

Answer 2

Write the potential in terms of complex exponentials. That is, $\cos(x)=\frac{e^{ix}+e^{-ix}}{2},$ such that in your particular case, $$\beta\cos(\frac{2\pi x}{a})=\frac{\beta}{2}(e^{i2\pi x/a}+e^{-i2\pi x/a}).$$ We can then rewrite the integral as, $$\frac{\beta}{2}(\frac{1}{L}\int_0^Le^{-ix(k-2\pi/a)}\,dx+\frac{1}{L}\int_0^Le^{-ix(k+2\pi/a)}).$$ You should recognize both these integrals as Kronecker delta functions, such that you end up with, $$\frac{\beta}{2}(\delta_{k,2\pi/a}+\delta_{k,-2\pi/a}),$$ which is non-zero for only $k=\pm\frac{2\pi}{a}$, given that $\beta\neq 0$. (I'll just assume that $\beta$ is non-zero, such that the potential itself is non-zero.)