- #1
hnicholls
- 49
- 1
(Note: although arising in QM, this is essentially a calculus question)
Ѱ (x) = A sin (n╥x/a)
1 = ∫ l Ѱ (x) l^2 dx with limits of integration a to 0
1 = ∫ A^2 sin^2 (n╥x/a) dx with limits of integration a to 0
Indefinite integral ∫ sin^2 x dx = x/2 - sin2x/4
I know this integral should reduce to 1 = A^2 a/2
But what I get is
I = A^2 [a/2 - sin (2n╥a/4a) - 0/2 - sin (2n╥0/4a)]
Third and fourth term within bracket are 0, but second term is 1
i.e., sin (n╥/2) = 1
I assume I'm missing something, but not sure what it is.
Thanks
Ѱ (x) = A sin (n╥x/a)
1 = ∫ l Ѱ (x) l^2 dx with limits of integration a to 0
1 = ∫ A^2 sin^2 (n╥x/a) dx with limits of integration a to 0
Indefinite integral ∫ sin^2 x dx = x/2 - sin2x/4
I know this integral should reduce to 1 = A^2 a/2
But what I get is
I = A^2 [a/2 - sin (2n╥a/4a) - 0/2 - sin (2n╥0/4a)]
Third and fourth term within bracket are 0, but second term is 1
i.e., sin (n╥/2) = 1
I assume I'm missing something, but not sure what it is.
Thanks