$\int_{-L}^{+L}h(z)\,dz = 2 \sum_{-\infty}^{+\infty}\frac {a_n}{n} \sin (nL)$ is it possible to find a formula for $a_n$ from 
$$\int_{-L}^{+L}h(z)\,dz = 2 \sum_{-\infty}^{+\infty}\frac {a_n}{n} \sin (nL)$$
For $n=0$ the series is $0$

Thanks
 A: If we have a function $h: [-L,L]\to \mathbb{R}$ then (under some conditions) we can expland it in a Fourier series
$$h(x) = \sum a_n \sin(2\pi x/L) + b_n \sin(2\pi x/L)$$
where the coeficients $a_n,b_n$ are given by some well known relations.
In your case you don't have a function to expand in a Fourier series (which it seems like you are trying to do). You only have a single value
$$I = \int_{-L}^Lh(x)dx$$
which you want to write as
$$I = \ldots + \frac{2a_{-1}\sin(-L)}{-1} + 2La_0 + \frac{2a_{1}\sin(L)}{1} + \frac{2a_{2}\sin(2L)}{2} + \ldots$$
This is one single equation with an infinite amount of unknowns ($a_0,a_1,\ldots$) and therefore it has an infinite number of solution. One example is $a_0 = \frac{I}{2L}$ with the rest of the terms being zero. Another (random example) is $a_0 = 7$ and $a_2 = \frac{I-14L}{\sin(2L)}$ and the rest zero.
A: I can offer another way to see what Winther is getting at, and maybe provide a different direction you can take this problem. 
Using the fact that sine is an odd function, you know that for all $n \in \Bbb{N}$ that $$a_{-n}\frac{\sin(-nL)}{-n} = a_{-n}\frac{-\sin(nL)}{-n} = a_{-n}\frac{\sin(nL)}{n}$$ and hence $$a_{-n}\frac{\sin(-nL)}{-n}+a_n\frac{\sin(nL)}{n} = (a_n+a_{-n})\frac{\sin(nL)}{n}$$ You have already said that $a_{n}\frac{\sin(nL)}{n} \equiv 0$, so you can rewrite your sum as $$2 \sum_{-\infty}^{+\infty}\frac {a_n}{n} \sin (nL) = 2\sum_{n=1}^\infty (a_n+a_{-n})\frac{\sin(nL)}{n}$$ But now you have a coefficient of $(a_n+a_{-n})$ on each term in your sum. Let's say you know for certain that $a_1 =a_{-1}=\frac{1}{2}$. Well, obviously then $a_1 +a_{-1}=1$. But now it stands to reason, why not just do away with $a_{-1}$? If we set $a_{-1}=0$ and $a_1 =1$, then we'll get the same result for the quantity $(a_1 +a_{-1})$. Next you can extend this reasoning for all $n \in \Bbb{N}$ to the quantity $(a_n +a_{-n})$. I believe you are best suited to define $b_n = \frac{a_n +a_{-n}}{2}$ and ultimately work with the sum $$2\sum_{n=1}^\infty (a_n+a_{-n})\frac{\sin(nL)}{n} = \sum_{n=1}^\infty b_n\frac{\sin(nL)}{n}$$ Now that your sum has been condensed into a nicer form, you may be able to get a bit farther, although I'm guessing you will still have the problem that Winther points out in the last paragraph of their answer. Hope my answer can help clear things up a little bit, and I can only suggest looking into some orthogonal relationships of the trig functions to solve for $b_n$. It may end up being similar to solving the coefficients of a Fourier Series.
