How do I show $e^{2\pi ikx/(b-a)}$ is orthogonal on $(a,b)$ I know how to show the orthogonality on $(-1/2,1/2)$ by integration, but I could not solve this one one an arbitrary interval $(a,b)$.
The solution for interval $(-1/2,1/2)$ is as follows. Hope someone can help. Thank you.

 A: $$\int_a^b e^{\frac{2 \pi  i k x}{b-a}} e^{-\frac{2 \pi  i m x}{b-a}} \, dx=\int_a^b e^{\frac{2 \pi  i x (k-m)}{b-a}} \, dx=
\frac{(b-a)}{2 \pi  i (k-m)}\left(e^{\frac{2 i \pi  b (k-m)}{b-a}}-e^{\frac{2 i \pi  a (k-m)}{b-a}}\right)
$$
Then just use the fact that $\frac{b}{b-a}=\frac{a}{b-a}+1$, so
$$e^{\frac{2 i \pi  b (k-m)}{b-a}}-e^{\frac{2 i \pi  a (k-m)}{b-a}}=e^{\frac{2 i \pi  a (k-m)}{b-a}}\left(e^{2 i \pi (k-m)}-1\right)=
e^{\frac{2 i \pi  a (k-m)}{b-a}}e^{ i \pi (k-m)}\left(e^{ i \pi (k-m)}-e^{- i \pi (k-m)}\right).
$$
The rest is as in your initial case.
A: Let $f_k(x) = e^{2\pi i k/(b - a)}$, for all $x\in (a,b)$ and $k\in \Bbb N$. If $k\neq m$,
\begin{align}\int_a^b f_k^*(x)f_m(x)\, dx &= \int_a^b e^{2\pi i(m - k)x/(b - a)}\, dx \\
& = \frac{b - a}{2\pi i(m - k)}e^{2\pi i(m - k)x/(b - a)}\bigg|_{x = a}^b\\
& = \frac{(b - a)[e^{2\pi ib/(b - a)} - e^{2\pi i a/(b - a)}]}{2\pi i(m - k)} \\
& = \frac{(b - a)[e^{2\pi i(1 + a/(b - a))} - e^{2\pi ia/(b - a)}]}{2\pi i(m-k)}\\
& = \frac{(b - a)[e^{2\pi ia/(b - a)} - e^{2\pi ia/(b - a)}]}{2\pi i(m- k)}\\
& = 0.
\end{align}
