## Integral Sin(ax)^2 Between Infinity and 0

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.

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 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 ?

## Integral Sin(ax)^2 Between Infinity and 0

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 ...

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

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

 yeah thanks its done now