$\text{y is a 2$\pi$-periodic function defined as:}$
$y(x) = \begin{cases} \dfrac{\sin(x)}{x} & \text{, $0\le|x|\le \pi$} \\ 1 & \text{, $x=0$} \end{cases}$
$\text{I want to show that the fourier coefficients of y are equal to:}$ $$c_{n}=\frac{1}{2\pi}\int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x} dx $$
My attempt (using Euler's formula):
$$c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\sin x}{x}e^{-inx} dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{e^{ix(1-n)}-e^{-ix(1+n)}}{2xi} dx$$
Not sure how to continue though, also am I allowed to integrate from $-\pi$ to $\pi$? Since y is defined differently at $x=0$?
We have \begin{align} \int_{-\pi}^{\pi} \frac{\sin x}{x}e^{-inx} \,dx &= \int_{-\pi}^{\pi} \frac{\sin x}{x}\cos(nx) \,dx -i \underbrace{\int_{-\pi}^{\pi} \frac{\sin x}{x}\sin(nx) \,dx}_{0 \text{ (odd integrand)}} \\ &= \int_{0}^{\pi} \frac{\sin((n+1)x)}{x} \,dx + \int_{0}^{\pi} \frac{\sin((1-n)x)}{x} \,dx \\ &= \int_0^{(n+1)\pi} \frac{\sin x}{x} \,dx - \int_{0}^{(n-1)\pi} \frac{\sin x}{x} \,dx \\ &= \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x} \,dx \end{align}