In a nice introductory paper about Bernoulli numbers that I found, the following claim is made (p. 5, theorem 4.3)
The Fourier series of $x$ is given by $b_n = \dots$ (not important, it is wrong in the paper as far as I see it, but I corrected the error already), $a_n$ = 0 when $n \geq 1$ (seems fine to me) and $a_0 = 1 \Rightarrow \frac{a_0}{2} = \frac{1}{2}$.
Now I don't see how to arrive at $a_0 = \frac{1}{2}$. $T = 1$ since we integrate from $-\frac{1}{2}$ to $\frac{1}{2}$ so then $a_0$ is given by
$$ a_0 = \int_{-\frac{1}{2}}^{\frac{1}{2}} x \cos(0 \cdot 2 \pi x) \, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} x \, dx = \left [ \frac{1}{2}x^2 \right ]_{-\frac{1}{2}}^{\frac{1}{2}} = 0 \neq 1 $$
The paper is riddled with errors, but I think I'm wrong on this one, since else the complete proof that follows would collapse. Any thoughts?
In the last step on page 5, where a translation from $[-\frac{1}{2}, \frac{1}{2}]$ to $[0, 1]$ is performed, the $\frac{1}{2}$ on the RHS comes implicitly from the $-\frac{1}{2}$ that is introduced on the LHS when translating. This has of course nothing do to with $a_0$ which the author explicitly states. $a_0 = 0$ obviously and not $\frac{1}{2}$.
It seems that the author copied a proof from somewhere but messed up badly.