I am asked to show that $$\frac{1}{2\pi}\left[-\frac{x}{in}e^{-inx}\bigg|^\pi_{-\pi} - \int\limits_{-\pi}^\pi-\frac{1}{in}e^{-inx}dx\right] = \frac{1}{2\pi} \left[-\frac{2\pi}{in}e^{in\pi}\right]$$ However, I seem to get $0$, so clearly I am doing something wrong?
Here is my attempt
\begin{align}\frac{1}{2\pi}\left[-\frac{x}{in}e^{-inx}\bigg|^\pi_{-\pi} - \int\limits_{-\pi}^\pi-\frac{1}{in}e^{-inx}dx\right] &= \frac{1}{2\pi}\left[\frac{-\pi}{in}e^{-in\pi} + \frac{\pi}{in}e^{in\pi} +\frac{1}{in}\int\limits_{-\pi}^\pi e^{-inx}dx\right] \\ &= \frac{1}{2\pi}\left[\frac{-\pi}{in}(-1)^n + \frac{\pi}{in}(-1)^n + \frac{1}{in}\int\limits_{-\pi}^\pi e^{-inx}dx\right] \\ &= \frac{1}{2\pi}\left[\frac{1}{in}\int\limits_{-\pi}^\pi e^{-inx}dx\right] \\ &= \frac{1}{2\pi}\left[\frac{1}{n^2}(e^{-in\pi} - e^{in\pi})\right] \\ &= \frac{1}{2\pi}\left[\frac{2i}{n^2}sin(n\pi)\right] \\ &= 0\end{align}
Where am I going wrong?
Let's write it better term by term. The first one:
$$\frac{1}{2\pi}\left[-\frac{x}{in}e^{-inx}\bigg|_{-\pi}^{\pi}\right]$$
$$\frac{1}{2\pi i n}(-\pi e^{-in\pi} - \pi e^{in\pi})$$
$$\frac{-1}{2in}(e^{-in\pi} + e^{in\pi})$$
$$\frac{-1}{in}\cos(n\pi)$$
$$\frac{-1}{in}(-1)^n$$
$$\frac{(-1)^{n+1}}{in}$$
Which is exactly what you have at the right side of your initial statement, since the integral is zero.
Indeed
$$\frac{1}{2\pi}\left[-\frac{2\pi}{in}e^{inx}\right] = -\frac{1}{in}e^{inx} = -\frac{1}{in}(-1)^n = \frac{(-1)^{n+1}}{in}$$