I don't understand one step in an answer sheet.
Exercise: Let the function $f: [-\pi,\pi]\to \mathbb{R}$ be defined by
$$f(x) = \begin{cases} \dfrac{\sin(x)}x, & x\neq 0 \\[0.3cm] 1, &x=0 \end{cases}$$
and for $x$ outside $[-\pi,\pi]$, $f(x)$ is the periodic extension.
Show that $f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx}$
where $\displaystyle c_n = \frac{1}{2\pi} \int^{(n+1)\pi}_{(n-1)\pi}\frac{\sin(x)}{x}dx$.
And then the answer:
\begin{align*} c_n &= \frac{1}{2\pi} \int^{\pi}_{-\pi}\frac{\sin(x)}{x}e^{-inx}dx\\[0.3cm] &= \frac{1}{2\pi}\int^{\pi}_{-\pi}\frac{e^{ix(1-n)}-e^{-ix(1+n)}}{2ix}\, dx\\[0.3cm] &=\frac{1}{2\pi}\int^{\pi}_{-\pi}\frac{1}{2ix}e^{-ix(n-1)}dx- \frac{1}{2\pi} \int^{\pi}_{-\pi}\frac{1}{2ix}e^{-ix(n+1)}dx\\[0.3cm] &= \frac{1}{2\pi}\int^{(n-1)\pi}_{-(n-1)\pi}\frac{1}{2iz}e^{-iz}dz - \frac{1}{2\pi}\int^{(n+1)\pi}_{-(n+1)\pi}\frac{1}{2iz}e^{-iz}dz \end{align*}
(and here finally is the step i don't understand:)
$$=\frac{1}{2\pi}\int^{-(n-1)\pi}_{-(n+1)\pi}\frac{1}{2iz}e^{-iz}dz - \frac{1}{2\pi}\int^{(n+1)\pi}_{(n-1)\pi}\frac{1}{2iz}e^{-iz}dz $$
You have gone from integral over the Green interval - the integaral over the red interval to the the two blue intervals
$\int_{-(n-1)\pi}^{(n-1)\pi} f(z) dz - \int_{-(n+1)\pi}^{(n+1)\pi} f(z) dz\\ \int_{-(n-1)\pi}^{(n-1)\pi} f(z) dz - \int_{-(n+1)\pi}^{-(n-1)\pi} f(z) dz-\int_{-(n-1)\pi}^{(n-1)\pi} f(z) dz- \int_{(n-1)\pi}^{(n+1)\pi} f(z) dz\\ - \int_{-(n+1)\pi}^{-(n-1)\pi} f(z) dz- \int_{(n-1)\pi}^{(n+1)\pi} f(z) dz$
Should I have a sign flip?