Fuente TeX:

\frac{e^{i(n+1)t}}{n+1}\left|_0^{2\pi}\right. = \frac{e^{i(n+1)2\pi}}{n+1} - \frac{e^{i(n+1)0}}{n+1} = \frac{1}{n+1} - \frac{1}{n+1} = 0