Are the functions $\{e^{int}\}$ orthogonal on $[0,2\pi]$? Are the functions $\{e^{int}\}$ orthogonal on $[0,2\pi]$?
For example, $e^{it}e^{i2t}=(\cos t+i \sin t) (\cos 2t+i \sin 2t)$, and then it is easy to get that $ \int_{0}^{2\pi}e^{it}e^{i2t} dt = 0 $. But $ \int_{０}^{２\pi}(e^{int})^2 dt = 0 $. I guess something is wrong here.
 A: This is a fairly straight-forward integral. For any $k,l \in \mathbb{Z}, k + l \neq 0$, we have the following:
$$\int_0^{2\pi} e^{ikt} e^{ilt} dt = \int_0^{2\pi} e^{i(k+l)t} = \left. \frac{e^{i(k+l)t}}{i(k+l)} \right|_0^{2\pi} = \frac{e^{2\pi (k+l) i} - e^0}{i(k+l)} = 0. $$
On the other hand, if $k+l = 0$:
$$\int_0^{2\pi} e^{ikt} e^{ilt} dt = \int_0^{2\pi} e^{i(k+l)t} = \int_0^{2\pi} 1 dt = 2 \pi. $$
So, if the inner product is defined as
$$\langle f, g \rangle = \int_0^{2\pi} f(t) g(t)^* dt, $$
with $g(t)^*$ denoting the complex conjugate of $g(t)$, then for any $k, l \in \mathbb{Z}$,
$$ \left\langle e^{ikt}, e^{ilt} \right\rangle = \int_0^{2\pi} e^{ikt} \left(e^{ilt}\right)^* dt = \int_0^{2\pi} e^{ikt} e^{-ilt} dt = \begin{cases} 0, & \text{if }k \neq l \\ 2\pi, & \text{if } k = l \end{cases}.$$
So yes, the functions are orthogonal. They are, however, not orthonormal, since $\langle e^{ikt}, e^{ikt} \rangle \neq 1$. In order to (ortho)normalize the functions, you should simply divide them by their length, which is $\sqrt{\langle e^{ikt}, e^{ikt}\rangle} = \sqrt{2\pi}$.
Note that it is not unusual to define the inner product as
$$\langle f, g \rangle = \frac{1}{2\pi} \int_0^{2\pi} f(t) g(t)^* dt, $$
in which case these functions would be orthonormal.
