Evaluate $\int_{-\pi}^{\pi} te^{-int} dt $

Question

Evaluate $\int_{-\pi}^{\pi} te^{-int} dt $

Using integration by parts:

$$\int_{-\pi}^{\pi} te^{-int} dt = t\frac{e^{-int}}{-in}|_{-\pi}^\pi - \int_{-\pi}^\pi \frac{e^{-int}}{-in}dt $$

Let's evaluate first term:

$$t\frac{e^{-int}}{-in}|_{-\pi}^\pi = -\frac{\pi}{in} \left( e^{-in\pi} + e^{in\pi} \right) = \frac{-2\pi}{in}(-1)^n$$

The integral on the RHS turns to be $0$.

BUT, the correct answer is: $$\frac{2\pi i}{n}(-1)^n$$

Where is my mistake?

Answer 1

$$ \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} =-i $$

Answer 2

$$\frac{-2\pi}{in}(-1)^n=\frac{-2\pi i}{i^2n}(-1)^n=\frac{-2\pi i}{-n}(-1)^n=\frac{2\pi i}{n}(-1)^n$$