Computing Fourier sum for infinitely differentiable functions Let $f\in C^{\infty}(\mathbb{R})$ be a periodic function of period $2L$. I want to show that $$f(x)=\sum_{n=-\infty}^\infty \left(\dfrac{1}{2L}\int_{-L}^Lf(y)e^{-in\pi y/L}dy\right)e^{i\pi nx/L}$$
The sum on the right is equal to 
$$\dfrac{1}{2L}\sum_{n=-\infty}^\infty \left(\int_{-L}^Lf(y)e^{in\pi (x-y)/L}dy\right)$$ I can't see how this sum should be equal to $f(x)$.
 A: There is an entire theory about Fourier series.  Here is one article about when a Fourier expansion of f converges to f. 
  http://en.wikipedia.org/wiki/Convergence_of_Fourier_series.  There are entire books on the subject.
There is no reason why you should immediately grasp a substantial amount of analysis simply by looking at the basic equation.
The general idea is that a periodic function may be (and often is) expressable as the sums of sin(nx) and cos(nx) with suitable coefficients.  It is possible that the sum is finite, but more often to get an exact equality you need an infinite series.
You really do have sines and cosines here.  They are buried in the fact the $e^{ix} = cos(x) + i sin(x)$. 
This kind of expansion is enormously useful in all sorts of problems.  We know a lot about sines and cosines.  If the Fourier series converges to the function, then a finite sum may be a good approximation, which is often easier to deal with than the original function.  There are also many cases where simply seeing what the Fourier coefficients are provides a solution.
