Proposition

Let $I(k) = \frac{1}{2\pi} \int_{0}^{2\pi} e^{ikt} dt$.

Show that $I(0) = 1$.
Show that $I(k) = 0$ if $k$ is a nonzero integer.
What is $I(1/2)$?

Solution

1

$I(0) = \frac{1}{2\pi}\int_{0}^{2\pi} 1dt = 1$.

2

\[\begin{align*} I(k) &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{ikt} dt \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} \cos(kt) + i\sin(kt) dt \\ &= \frac{1}{2\pi} [\int_{0}^{2\pi} \cos(kt) dt + i\int_0^{2\pi} \sin(kt) dt] \\ &= \frac{1}{2\pi} [\frac{\sin(kt)}{k} \Big\vert^{2\pi}_0 - i\frac{\cos(kt)}{k} \Big\vert^{2\pi}_0] \\ &= \frac{1}{2\pi} \cdot 0 \\ &= 0. \end{align*}\]

This argument does not work when $k = 0$ because $\frac{1}{0}$ does not make sense.

3

By the argument above, it suffices to calculate $\sin(t/2)\Big\vert^{2\pi}_0 - i\cos(t/2)\Big\vert^{2\pi}_0$. Since it equals $2i$, the answer is $2i/\pi$.