Solving for A in 1 = A2∫-2(x/a)2sin2(kx)

[jocdrummer]

This is actually part of a homework question for my Quantum Mechanics course but it is purely a math question.

1 = A²\inte^-2(x/a)2sin²(kx)

Note: for some reason the integral is showing up as a psi.

where A, a, and k are constants and the integral is from -inf to inf (or 0 to inf with a constant 2 multiplying because the integrand is symmetric)

What I am trying to do is solve for A. For the integral, there aren't any common forms that it matches up with that I am aware of. I've tried IBP but it seems to just get more and more complicated as it goes along. Any suggestions would be awesome.

 

[lanedance]

i'm gussing the limits are -inf to inf, this should help simplify things as
\int_{-\infty}^{\infty} dx.e^{-x^2} = \sqrt{\pi}

I think a few IBPs is the right way to though it will be pretty messy

 

[Dick]

Don't do parts. Use that sin(kx)=(exp(i*k*x)-exp(-i*k*x))/(2i). Expand everything and complete the squares. Then do a change of variables on each integral to convert everything to real integrals.