Prove certain integral equals $0$ I've been trying to solve some problems from my Partial Differential Equations course. I got the idea to solve this one but there's one step at the end I'm not sure about. It goes like this:

Given  $$G(x)=\int_{x_0}^xg^*(y)dy,$$ where $g^*$ is a $2l$-periodic
odd function, and $x_0$ is fixed. Prove $G$ is $2l$-periodic.

The question was a bit more complex in context (about wave equation) but I've simplified it so anyone can understand it (making it just a real analysis problem). So what I need to prove is that $G(x)=G(x+2l)$ for any $x\in\mathbb R$. I've seen that
$$G(x+2l)=\int_{x_0}^{x+2l}g^*(y)dy = \int_{x_0}^xg^*(y)dy + \int_x^{x+2l}g^*(y)dy = G(x) + \int_x^{x+2l}g^*(y)dy,$$
so proving $G(x+2l)=G(x)$ would be equivalent to proving that
$$\int_x^{x+2l}g^*(y)dy=0, \text{ }\forall x\in\mathbb R.$$
To prove this, I obviously need to use the fact that $g^*$ is both odd and $2l$-periodic. I've tried doing:
$$\int_x^{x+2l}g^*(y)dy=\int_{-l}^{l}g^*(y+x+l)dy,$$
what would be equal to zero if $g^*(y+x+l)$ was still an odd function (because it's integral limits are $-l$ and $l$ after that change). I'm not sure if changing what's inside the function $g^*$ makes it stop being odd or if it just does not matter.
Is it still odd? If it is, is it trivial or do I need to prove it? If it's not, how can I see then that the integral equals zero?
Another way to prove it that came to my mind was using that the antiderivative of a periodic function is also a periodic function of the same period, but I'm not sure if it's true (I know the other way it's true, derivative of periodic is periodic, but my intuition tells me that antiderivative of periodic may not necessarily be periodic of the same period, don't know why).
Any help, hint or comment will be appreciated, thanks in advance.
 A: If $f$ is periodic with a period of $p$, then $$\int_a^{a+p} f(x)dx$$ is independent from $a$. This can be understood intuitively as you are, no matter what $a$ is, always integrating over the entire interval over which $f$ is defined. See below for a rigorous proof *.
This can help you, since you can use the fact that $g$ is odd which gives you $$\int_{-l}^l g(x)dx = 0,$$ and therefore, the statement above tells you that the interval of $g$ over any interval of length $2l$ has the same value, which is $0$.

*For a rigorous proof, let's prove that $$\int_a^{a+p} f(x)dx=\int_0^p f(x)dx.$$
The statement can be proven quite quickly. First of all, you can notice that $$\int_a^b f(x)dx = \int_{a+kp}^{b+kp} f(x)dx$$ for every $k\in\mathbb Z$. That is, you can always shift any integral by a multiple of the period in any direction. The equality can be proven by introducing a new cariable $u=x+kp$ which gives
$$\int_a^b f(x)dx = \int_{a+kp}^{b+kp} f(u-kp)du = \int_{a+kp}^{b+kp} f(u)du$$
where we used the fact that the function is periodic with period $p$ to get $f(u-kp)=f(u)$.
Once you have this "lemma", you can see that, without loss of generality, you can assume that $a\in [0,p)$ in the original statement.
Now, you can write
$$\int_a^{a+p} f(x)dx = \int_a^p f(x)dx + \int_p^{a+p} f(x)dx$$
and apply the shifting of the period (the "lemma") again (with $k=-1$) on the second summand. The lemma tells us that
$$\int_p^{a+p} f(x)dx = \int_0^a f(x)dx.$$
Plugging this into the previous equation, you get
$$\int_a^{a+p} f(x)dx = \int_a^p f(x)dx + \int_p^{a+p} f(x)dx = \int_a^p f(x)dx + \int_0^{a} f(x)dx = \int_{0}^p f(x)dx$$
