Evaluate fourier coefficient of $f(t)=t$.

Question

Evaluate the Fourier coefficient of $f(t)=t$.

$$\hat{f}(n) = \frac{1}{2\pi}\int_0^{2\pi} te^{-int}dt$$

I'd be glad for help with this calculation. My integration skills need an improvement.

My Try: (following the hint)
$$ \int_0^{2\pi} te^{-int}dt = \frac{te^{-int}}{-in}|_0^{2\pi} - \int_0^{2\pi} \frac{e^{-int}}{-in} = \frac{2\pi e^{-2\pi in}}{-ni} - \frac{e^{-int}}{-n^2}|_0^{2\pi} \\= \frac{2\pi e^{-2\pi in}}{-ni} - \left( \frac{e^{-2\pi in}}{-n^2}- \frac{e^0}{-n^2} \right)$$

Am I on the right path?

Answer 1

HINT: Integrate it by parts

$$\int_0^{2\pi} te^{-int}\, dt={i\over n}\left(te^{-int}\bigg|_0^{2\pi} - \int_0^{2\pi}e^{-int}\, dt\right)$$

Answer 2

Take \begin{alignat}{2} u&=t &\qquad dv&=e^{-int}\,dt\\ du&=dt &\qquad v&={e^{-int}\over -in}. \end{alignat} Integrating by parts and simplifying you get \begin{align} {1\over 2\pi}\int_0^{2\pi}te^{-int}\,dt&={1\over 2\pi}\left[-{te^{-int}\over in}\Bigg|_{t=0}^{t=2\pi}-\int_0^{2\pi}{e^{-int}\over -in}\,dt\right]\\ &={1\over 2\pi}\left[-{te^{-int}\over in}+{e^{-int}\over n^2}\right]_{t=0}^{t=2\pi}\\ &={i\over n}. \end{align}