Integral Sin(ax)^2 Between Infinity and 0

1. Jun 12, 2012

Nick789

Need result for integral
Sin(ax)*Sin(ax) Between Infinity and 0

Cant find this anywhere but there is a standard result with a in it.

2. Jun 12, 2012

micromass

That integral won't exist in general.

3. Jun 12, 2012

Nick789

Should do need it for a normalisation problem

have to square the wavefunction then integrate

wavefuction form: sin(ax)

so need to integrate sin(ax)^2 over all space

problem is part of infinite square well limits should between infinity and 0.
V=0 for x< a
v= infinity for x>a

Maybe I'm thinking of the wrong limits.

should probably be between a and -a ?

Last edited: Jun 12, 2012
4. Jun 12, 2012

Whovian

Think about it. It's periodic and always nonnegative. Assuming a≠0, every period will have finite area. So the sum of the areas of the infinite periods ...

5. Jun 12, 2012

algebrat

the wave function is zero where the potential is infinite

6. Jun 12, 2012

Nick789

Yeah my limits are wrong because the well is bound between a and -a

so need integral between -a and a for sin(ax)^2

7. Jun 12, 2012

Millennial

You mean $\displaystyle \int_{-a}^{a}\sin^2(ax)dx$ I presume.
Did you try the half-angle identity and u-substitution?

8. Jun 12, 2012

rbj

Nick, take a look at http://en.wikipedia.org/wiki/Wikipedia:Math and learn (it's very easy) a little math-symbol paste-up, like LaTeX. perhaps there is a better description somewhere.

just remember that $\sin^2(x)$ has an average value of 1/2 and if you integrate any non-zero constant over anything to $\infty$, you will get an infinite number. and i am wondering if the limits should be from -1/a to +1/a ? or should it be a 1/a in the sin() argument?

9. Jun 13, 2012

Nick789

yeah thanks its done now