Fourier series and uniform convergence for a given function. The Fourier series of $x$ is $$\sum_{n=1}^{\infty} 2(-1)^n\frac{\sin(nx)}{n}$$
I can see that the series does not converge pointwise to x in $x=\pi$
My question is:
what happen in $[0,a]$ when $a<\pi$ ,my intution says that we should get uniform convergence, but i cant prove that is true.
For general, what can we say about uniform convergence of Fourier series of a function which is continuous, the derivative continuous but $f(\pi) \neq f(-\pi)$ 
in $[0,a]$ when $a<\pi$ ?
What about this specific case? I did not succeed to determine if there is a   uniform convergence. 
I am interested in this question because I tried to 
calculate the Gibs phenomenon in the function $-1+x$ at $[0,\pi] $and $1+x$ at $ [-\pi,0]$ We know that the Gibs phenomenon in the function 1 occur at $\pi /2n $ so because $x$ is a nice function, I think that the Gibs phenomenon of the given function above occur in  $\pi /2n $.  So, if the Fourier series of x  uniform convergence, then that should be true.

Let $a\in \mathbb C$, $|a|\le 1$ and $a\ne -1$. Then (integration below is complex along the interval $[0,a]\subset\mathbb C$)
$$
\sum_{k=1}^n (-1)^k\frac{a^k}{k}=\int_0^a \left(\sum_{k=1}^n (-t)^{k-1}\right) dt
=-\int_0^a \frac{1-(-t)^n}{1+t} \,dt\\ =-\log (1+a)+(-1)^n\int_0^a \frac{t^n}{1+t} \,dt=
-\log (1+a)+\delta_n
$$
where $\log z$ is the principal branch defined in $\mathbb C\setminus(-\infty,0]$ and
$$
|\delta_n|\le\int_0^a \frac{|t^n|}{|1+t|} \,|dt|
\le \frac{1}{n(|1+a|)}.
$$
Hence, for every $x\ne (2k+1)\pi i$
$$
\left|\sum_{k=1}^n (-1)^k\frac{\mathrm{e}^{ikx}}{k}+\log(1+\mathrm{e}^{ix})\right|\le
\frac{1}{n(|1+\mathrm{e}^{ix}|)},
$$
and in particular, the series $\sum_{n=1}^\infty (-1)^n\frac{\mathrm{e}^{inx}}{n}$
converges uniformly in any closed interval of $x$ not containing a $(2k+1)\pi$, and so does its imaginary part
$$
\sum_{n=1}^\infty (-1)^n\frac{\sin (nx)}{n}
$$
