Integral Sin(ax)^2 Between Infinity and 0

[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.

 

[micromass]

That integral won't exist in general.

 

[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 ?

 

[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 ...

 

[algebrat]

the wave function is zero where the potential is infinite

 

[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

 

[Millennial]

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

 

[rbj]

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?

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?

 

[Nick789]

yeah thanks its done now