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Integral Sin(ax)^2 Between Infinity and 0

 
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Jun12-12, 02:07 PM   #1
 

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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That integral won't exist in general.
 
Jun12-12, 02:33 PM   #3
 
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 ?
 
Jun12-12, 02:56 PM   #4
 

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 ...
 
Jun12-12, 03:06 PM   #5
 
the wave function is zero where the potential is infinite
 
Jun12-12, 03:07 PM   #6
 
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
 
Jun12-12, 03:35 PM   #7
 
You mean [itex]\displaystyle \int_{-a}^{a}\sin^2(ax)dx[/itex] I presume.
Did you try the half-angle identity and u-substitution?
 
Jun12-12, 07:14 PM   #8
rbj
 
Quote by Nick789 View Post
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 View Post
You mean [itex]\displaystyle \int_{-a}^{a}\sin^2(ax)dx[/itex] 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 [itex] \sin^2(x) [/itex] has an average value of 1/2 and if you integrate any non-zero constant over anything to [itex]\infty[/itex], 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?
 
Jun13-12, 08:02 AM   #9
 
yeah thanks its done now
 
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