Show that the functions $e^{in\pi x/l}$, n = 0, ±1, ±2, ..., are a set of orthogonal functions on $(-l, l)$


$A(x)$ 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, Void Oct 14 '14 at 12:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Neuneck, Danu, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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