Show that set is an orthonormal set in PC(0,l)

Main Question or Discussion Point

Show that set $$\{ (2/l)^{1/2} sin(n-1/2)(\pi x/l)\}^\infty_1$$ is an orthonormal set in PC(0,l).
Of course, I need to show that $$<\phi_m, \phi_n> = \delta_{mn}$$
$$<\phi_m, \phi_n> = (2/l) \int_0^l sin(m-1/2)(\pi x/l) sin(n-1/2)(\pi x/l)\, dx \\ = 1/l \int_0^l cos(m-n)(\pi x/l) cos(m+n+1)(\pi x/l)\, dx \\ = \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \left( \frac 1 l \right) \frac {l} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l \\ = \frac {1} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \frac {1} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l$$

I used the trig identity:
$$2 sin \theta sin \psi = cos(\theta - \psi) - cos(\theta + \psi)$$

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mathman
What is your question? The expression you have (you should check it out) = 0 for $m\ne n$ and =1 for $m=n$.

What is your question? The expression you have (you should check it out) = 0 for $m\ne n$ and =1 for $m=n$.
Thanks for the reply. When I looked at it, I could swear it's zero in both cases. Let me check again.

$$\\ = \frac {1} {\pi(m-n)} sin(m-n)(\pi x/l) \Big|_0^l - \frac {1} {\pi(m-n)} sin(m+n+1)(\pi x/l)\ \Big|_0^l$$

When you evaluate at x = l and x = 0, the argument for sine is either zero or a multiple of pi. When m = n, there's an indeterminate form in the common term.

mathman
When m=n, I suggest you don't use the cos identity, but use the original form, where the integral is $2/l\int_0^lsin^2((m-1/2)(\pi x/l))dx$.