Show that the functions $e^{in\pi x/l}$, n = 0, ±1, ±2, ..., are a set of orthogonal functions on $(-l, l)$
using:
$A(x)$ and $B(x)$ are orthogonal on $(a,b)$ if
$\int^b_a A^*(x)B(x)dx = 0$
where $A^*(x)$ is the complex conjugate of $A(x)$.
I'm assuming you rewrite the function as:
$\cos(n\pi x/l) + i\sin(n\pi x/l)$,
so the complex conjugate would be
$\cos(n\pi x/l) - i\sin(n\pi x/l)$
This would make the integral
$\int^l_{-l} \cos(n\pi x/l) - i\sin(n\pi x/l)(\cos(n\pi x/l) + i\sin(n\pi x/l))dx = 0$
I think that this is correct so far and that I'm pretty close, just struggling with this integral and how it proves the claim. Any guidance would be greatly appreciated, Thanks!
In the definition of orthogonality functions $A$ and $B$ must be different; for the system $\left \lbrace e^{\frac{in\pi x}{l}} \right\rbrace_{n\in\mathbb{Z}}$ write $$\int\limits_{-l}^{l}{A_{n}^{*}(x)B_{k}(x)\ dx}=\int\limits_{-l}^{l}e^{\frac{ik\pi x}{l}}e^{-\frac{in\pi x}{l}}\ dx, \;\;k \ne n$$ and then integrate: $$\int\limits_{-l}^{l}e^{\frac{ik\pi x}{l}}e^{-\frac{in\pi x}{l}}\ dx=\int\limits_{-l}^{l}e^{\frac{i(k-n)\pi x}{l}}\ dx.$$