How do I show $\frac{1}{\sqrt L}e^{\frac{2\pi inx}{L}}$ is an orthonormal set for $L^2[a, b]$ (where $|b-a|=L$)?

Question

Let $$f_n=\frac{1}{\sqrt L}e^{\frac{2\pi inx}{L}}$$ To check that they're orthonormal I calculate their scalar product: $$(f_n, f_k)=\int_a^b \bar f_n(x)f_k(x)dx$$ Of course if n=k the scalar product equals to 1. My issue is with $n\not=k$, in which case we have: $$(f_n, f_k)=\int_a^b \frac{1}{\sqrt L}e^{\frac{-2\pi inx}{L}}\frac{1}{\sqrt L}e^{\frac{2\pi ikx}{L}}dx=\frac{1}{L}\int_a^b e^{\frac{2\pi i(k-n)x}{L}}=\frac{1}{2\pi i(k-n)}\left( e^{\frac{2\pi i(k-n)b}{L}}-e^{\frac{2\pi i(k-n)a}{L}} \right)$$ Firstly I suppose that $k-n$ doesn't cause any problems since it's just an integer. Something that would work is the following: $$e^{\frac{2\pi i(k-n)a}{L}}=\left(e^{2\pi i(k-n)}\right)^\frac{a}{L}=1^\frac{a}{L}=1$$ And same for $b$. Since $b/L$ and $a/L$ are real numbers though, I don't think I can do that, since we are doing the root with a real index of a complex number. I found a lot of pdfs and posts online talking about the case where $L=2\pi$, but I can't find a more general explanation, so I hope someone can help me.

Answer 1

You may do it the hard way. By definition, for any real $b\in\mathbb{R}$ we have $e^{bi}=\cos(b)+i\sin(b)$, so your integral becomes:

$\displaystyle\frac{1}{L}\int_a^b e^\frac{2πi(k-n)x}{L}dx=\displaystyle\frac{1}{L}\left(\int_a^b \cos\left(\frac{2π(k-n)x}{L}\right)dx+i\int_a^b\sin\left(\frac{2π(k-n)x}{L}\right)dx\right)$

if $y=\frac{2\pi(k-n)(x-a)}{L}$ we get

$C\left(\displaystyle\int_0^{2\pi(k-n)} \cos(y+c)dy+i\int_0^{2\pi(k-n)} \sin(y+c)dy\right)$

where $C$ and $c$ are constants.

It should be clear now that this are zero.