I have to solve this integration by parts question, $$\hat{f}(n) = \frac{1}{2\pi i n}\left( \int_{-\pi}^{0}(\pi + 2\theta) e^{-in\theta}d\theta + \int_{0}^{\pi}(\pi - 2\theta) e^{-in\theta}d\theta \right). $$
I solved it and my answer was, $$\begin{cases} 0 & \texttt{n even}, \\ \frac{-1(4\pi i n + 8)}{2\pi(in)^3} & \texttt{n odd. } \end{cases}$$
Even though, one of my friends solved it and the answer was,$$\begin{cases} 0 & \texttt{n even}, \\ \frac{4}{i\pi(n)^3} & \texttt{n odd. } \end{cases}$$
Could anyone participate with us in this calculation?
thanks.
Noting that
$$\int_{-\pi}^{0} (\pi+2\theta)e^{-in\theta}\,d\theta \;=\; \int_{0}^{\pi} (\pi-2\theta)e^{in\theta}\,d\theta$$
the integrals become
$$\begin{align} \int_{-\pi}^{0} (\pi+2\theta)e^{-in\theta}\,d\theta + \int_{0}^{\pi} (\pi-2\theta)e^{-in\theta}\,d\theta &\;=\; \int_{0}^{\pi} (\pi-2\theta)(e^{in\theta}+e^{-in\theta})\,d\theta \\[0.2cm] &\;=\; 2\int_{0}^{\pi} (\pi-2\theta)\cos(n\theta)\,d\theta. \end{align}$$
using $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$. Since
$$\int_{0}^{\pi}\cos(n\theta)\,d\theta \;=\;\left.\frac{\sin(n\theta)}{n}\right|_{\theta=0}^{\pi} \;=\; 0 $$
and, integrating by parts:
$$\begin{align} \int_{0}^{\pi}\theta\cos(n\theta)\,d\theta &\;=\; \left.\frac{\theta\sin(n\theta)}{n}\right|_{\theta=0}^{\pi} - \int_{0}^{\pi}\frac{\sin(n\theta)}{n}\,d\theta \\[0.3cm] &\;=\; 0 - \left.\left(-\frac{\cos(n\theta)}{n^{2}}\right)\right|_{\theta=0}^{\pi} \;=\; \frac{\cos(n\pi)}{n^{2}}-\frac{1}{n^{2}} \;=\; \frac{(-1)^{n}-1}{n^{2}}, \end{align}$$
where we used $\cos(n\pi)=(-1)^{n}$, you find that
$$\begin{align} \int_{-\pi}^{0} (\pi+2\theta)e^{-in\theta}\,d\theta + \int_{0}^{\pi} (\pi-2\theta)e^{-in\theta}\,d\theta &\;=\; \frac{4(1-(-1)^{n})}{n^{2}} \end{align}.$$
So if $n$ is even, the integral is zero (since $(-1)^{n}=1$), whereas for $n$ odd, the integral yields $\frac{8}{n^{2}}$ (since $(-1)^{n}=-1$). This will then lead to your friend's result.