I (perhaps incorrectly) derived the following formula: $$\sigma_0\left(x+\frac{1}{2}\right)=\frac{1}{2}\sum_{n=2}^{\infty}\left[\frac{1}{n}+\frac{i}{\pi}\ln\left(\frac{e^{\frac{i\pi}{n}(2x-n+1)}-e^{\frac{i\pi}{n}(3x-2n+2)}}{e^{\frac{i\pi}{n}(x-n+1)}-1}\right)\right]$$
Where $\sigma_0(x)$ is the number of divisors of $x$. Does this look familiar? If not, how might I go about computing values to disprove the identity?
Context: My method was to construct a sum of Fourier series for square waves of period $n$ that evaluate to $1$ between $n-1$ and $n$, and 0 everywhere else. Using some trig identities I managed to "simplify" the expression from a double infinite sum to a single one. Is this a valid approach?
More a comment than an answer, but easier to enter here.
This reminds me of Ramanujan's sum:
https://en.wikipedia.org/wiki/Ramanujan%27s_sum
In particular,
$-\sigma_0(n) =\sum_{k=1}^{\infty} \dfrac{\ln(k)}{k}c_k(n) $ where $c_k(n) =\sum\limits_{a=1,(a, k)=1}^n e^{2\pi ia n/k} $.