I am having difficulties with this problem. I don't really know where to start, I suspect there is something I am supposed to know or "see" that I am missing.
$u$ is a $2\pi$-periodic function defined by
$$u(x)=\begin{cases} \frac{\sin x}{x} & 0 \lt |x| \lt \pi \\ 1 & x = 0 . \end{cases}$$
Show that the Fourier coefficients of $u$ are given by
$$c_n = \frac{1}{2\pi} \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x} dx$$
and use this to compute the improper integral
$$\int_{0}^{\infty} \frac{\sin x}{x}.$$
Trying to compute the Fourier coefficients the way I normally would is messy since $\frac{sinx}{x}$ doesn't "disappear" when doing partial integration. Also, the bounds are confusing me.
Then I thought maybe I could use that
$$u(0) = 1 = \sum_{-\infty}^{\infty} c_n,$$ and $$u(\pi) = \frac{\sin \pi}{\pi} = \sum_{-\infty}^{\infty} c_n e^{in\pi},$$
but after that I am stuck again.
Hint. The given function is even, then you just have to consider $$ \int_0^\pi\frac{\sin x}{x}\:\cos (nx)\:dx, \quad n\ge0, $$ one may recall that $$ \sin a \cos b=\frac12\left(\sin(a+b)+\sin(a-b) \right), $$ then use a change of variable to get $$ \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x} dx. $$