- #1

Shackleford

- 1,666

- 2

Of course, I need to show that [tex] <\phi_m, \phi_n> = \delta_{mn} [/tex]

[tex] <\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

[/tex]

I used the trig identity:

[tex] 2 sin \theta sin \psi = cos(\theta - \psi) - cos(\theta + \psi)[/tex]