Question about $\frac{1}{\pi} \int_{-\pi}^{\pi} te^{int} dt$ to $\frac{-2i(-1)^{n}}{n}$.

Question

How do you get

$\frac{1}{\pi} \int_{-\pi}^{\pi} te^{int} dt$

to $\frac{-2i(-1)^{n}}{n}$.

The furthest i come to is:

$\frac{1}{\pi}((\frac{\pi e^{in\pi}+\pi e^{-in\pi}}{in}) -\frac{1}{in}(\frac{e^{in\pi}-e^{-in\pi})}{in}))$

Answer 1

$$I=\frac{1}{\pi}\int_{-\pi}^{\pi}t e^{int} dt$$ Integration by parts leads to $$I=\frac{1}{\pi}\left (t \frac{e^{int}}{in}+\frac{e^{int}}{n^2}\right)_{-\pi}^{\pi}$$ $$I=\frac{\pi}{in\pi}(e^{in\pi}+e^{-in\pi})+\frac{1}{n^2 \pi} (e^{in\pi}-e^{-in\pi})$$ $$I=\frac{2\cos(n\pi)}{in}+\frac{2i\sin n\pi}{n^2\pi}=\frac{-2i(-1)^n}{n}.$$ As $\cos n\pi=(-1)^n, \sin n\pi =0$