Set of orthotogonal complex functions [closed]

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:

$$\cos\left(\frac{n\pi x}{l}\right) + i\sin\left(\frac{n\pi x}{l}\right),$$

so the complex conjugate would be

$$\cos\left(\frac{n\pi x}{l}\right) - i\sin\left(\frac{n\pi x}{l}\right)$$

This would make the integral

$$\int^l_{-l} \left(\cos\left(\frac{n\pi x}{l}\right) - i\sin\left(\frac{n\pi x}{l}\right)\right)\left(\cos\left(\frac{n\pi x}{l}\right) + i\sin\left(\frac{n\pi x}{l}\right)\right)dx = 0$$

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!

closed as off-topic by John Rennie, Neuneck, Danu, ACuriousMind♦, VoidOct 14 '14 at 12:27

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• It's actually easier if you leave your functions as complex exponentials and use the integral of $\exp$ to find your answer. – WetSavannaAnimal Oct 14 '14 at 7:04
• Hi. Orthogonal means $e^{in\pi x/l}$ and $e^{im\pi x/l}$ are orthogonal, when $m\neq n$. So in your integration, one part is m and one part is n. Then you can directly do the integration and show it is 0. – Zheng Liu Oct 14 '14 at 7:15
• Thank you both!, I'm too tired to try it tonight but I'll let you know how it goes tomorrow I think I see how to do it – eric Oct 14 '14 at 7:55
• Hi eric. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Oct 15 '14 at 13:56