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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!

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

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  • 1
    $\begingroup$ It's actually easier if you leave your functions as complex exponentials and use the integral of $\exp$ to find your answer. $\endgroup$ – WetSavannaAnimal Oct 14 '14 at 7:04
  • $\begingroup$ 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. $\endgroup$ – Zheng Liu Oct 14 '14 at 7:15
  • $\begingroup$ 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 $\endgroup$ – eric Oct 14 '14 at 7:55
  • $\begingroup$ 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. $\endgroup$ – Qmechanic Oct 15 '14 at 13:56