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Consider the collection $\left\{e^{2\pi ikx/(b-a)}\,:\, k=0,\pm 1, \pm2,\ldots \right\}\,,$ on $(a,b)$. Verify that this system is orthogonal in $L^2$.

I have attempted the two most obvious things: (1) integrating with exponentials, (2) integrating with trig functions. Both are messy, and neither are working out. I don't want to write out a lot of work, because my notes are a tedious nightmarescape, but I basically keep ending up with the following:

Set $\theta=\frac{2\pi}{b-a}$. Take $e^{2\pi ikx/(b-a)}$ and $e^{2\pi ikx/(b-a)}$. Taking their inner product, I get something like $\frac{i}{\theta(k-n)}\left[e^{\theta(k-n)bi}-e^{\theta(k-n)ai}\right]\,.$

I feel like I'm going crazy, but it is not at all apparent that this will end being zero....

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    The precise value of $\theta$ is vital. Substitute it back in.2011-02-25

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Note that $\displaystyle{\frac{b}{b-a} = \frac{a}{b-a} + 1}$, so those exponential terms differ by a factor of $e^{2\pi(k-n)i}$.

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    @Bey: I'm not sure which part is unclear. The factor $e^{2\pi(k-n)i}$ is $1$, because $k-n$ is an integer, so $e^{\theta(k-n)bi}=e^{\theta(k-n)ai}$, meaning that the integral is $0$ (when $k\neq n$). In case it is the relevance of the first part that is unclear, note that $\theta(k-n)bi=2\pi(k-n)\frac{b}{b-a}i$ and $\theta(k-n)ai=2\pi(k-n)\frac{a}{b-a}i$.2011-02-28
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You should consider the cases $n=k$ and $n\not=k$ separately (before you integrate).