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I'm having trouble with an integral for my QM senior thesis

  1. Aug 1, 2012 #1
    I have been working through a method of solving the Time-Dependent Schrodinger Equation. I'm trying to learn this techniques before I actually jump into my project, however, I have come to an integral that I have been trying to do but can't think of a way to approach it. Mathematica could not solve it either. I'm going to post it below as a word doc instead of trying to use the latex, and hopefully someone can point me in the right direction, or give me something to look up on my own.

    Attached Files:

  2. jcsd
  3. Aug 5, 2012 #2

    Simon Bridge

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    It's often considered bad manners to post restricted format documents online ... you are implicitly restricting your audience only to people with MS Word sufficiently close to yours that the formatting will come out the same. Hopefully you tested it out in, say, google docs, so you know everyone has a shot at reading it.

    eg. In OpenOffice.org 3 the equation renders as an image, with a slice missing out the top.

    It looks to me like:
    \int_1^{e^{At}} \frac{x^2 dx}{\big [

    \big ]^2}
    [/tex]...A,B are constants from a previous calculation and K,H are known constants.

    This right?
  4. Aug 5, 2012 #3
    Do you know beforehand that it has a closed form solution? If so then I'd first get rid of all that superfluous stuff. It just gets in the way. I'd work on:

    [tex]\int_1^a \frac{x^2 dx}{\left[(b+cx)\cos(dx)+(f+gx)\sin(hx)\right]^2}[/tex]

    You said Mathematica can't do the original one. How about that one? Probably not. I'm curous, can it do half of it? Like just the cosine one. Won't help you with the original problem but just the act of trying things might lead to something that could solve it. How about without the square? How about without one of the coefficients or how about with some of the constants set to zero or one? Which one can Mathematica solve explicitly and what change to it renders it unsolvable by Mathematica?
  5. Aug 5, 2012 #4
    how big are you planning on letting e^At get?
  6. Aug 5, 2012 #5
    I rather doubt this has a closed-form solution; but, you can, at least, simplify things such that you only have two constants in your integrand, rather than 4. If you make the substitution [itex]u = \frac{k}{H} x[/itex], the integral becomes
    [tex]\frac{H}{k^3} \int_{\frac{k}{H}}^{\frac{k}{H}e^{At}}\frac{u^2 du}{[(8A+3Bu) \cos(u)+(3B+8Au) \sin(u)]^2}.[/tex]
  7. Aug 5, 2012 #6
    Thank you for rewriting my integral Simon! What you have written is correct, except for the upper limit of the integrand is actually e^-Ht.

    As for a closed form solution, I am not quite sure. The original paper I am trying to work off does a more complicated problem in which H and k are not constants, but also time-dependent, and they got a solution for this part that involves spherical bessel functions (Jn) and spherical harmonics (Yn) (however, these are just combinations of trig functions). There were actually two regions in time the calculation was done, and the other region was a little less complicated, involving another rational function with linear combintations of sines and cosines in both top and bottom.

    jackmell, thanks for the suggestions. I did try a couple of the other forms you suggested and mathematica still could not do them. I was able to get an extremely complicated solution for [itex]\int[/itex]x2dx/((acos(x)+bsin(x))2
    so I assume the x's in the coefficient terms on the bottom are causing the problem.
  8. Aug 5, 2012 #7

    Simon Bridge

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    Pleased to be of service ... so adjusted:

    \int_{1}^{e^{-Ht}} \frac{x^2 dx}{\big [
    (8AH+3Bkx) \cos(\frac{kx}{H}) + (3BH+8Akx) \sin(\frac{kx}{H})
    \big ]^{2}}
    [/tex]...A,B are constants from a previous calculation and K,H are known constants.

    It may help to include a reference to the paper you are working off or to know more about what you are attempting. I'll wait a bit - don't want to get in the way of the others :)

    You are, of course, urged to learn LaTeX ... to help, I have written the markup in a structured way so you can see how it was done (use the "quote" button on this reply). It will show you how to do the integral sign, powers, fractions, and trig functions. That should cover most things and it is much easier than using MS Word's equation editor.
  9. Aug 6, 2012 #8
    Ok, that's where I would go next. Study that solution and maybe that work can help you with this one.
    Last edited: Aug 6, 2012
  10. Aug 6, 2012 #9

    Simon Bridge

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    Of course:
    If k<<H then you could use the par-axial approximation.
    If t and H are both always positive, then x will be in (0,1]... tempting us to look for some sort of series approximation.

    Really depends on what you want to achieve.
  11. Aug 9, 2012 #10
    I figured out the problem, and it was a little bitter-sweet. The above integral was to get a c3(t) coefficient. I already solved for the c1(t) and c2(t) previously, and they all contained the same A and B unknowns. I also had the initial condition of c1(0)=0, so I was able to solve for A in terms of B. Since the unknowns were the same in each equation, I was able to plug in the A=...B into the above integral. When I gave this back to Mathematica, it could now do the integral!

    Bottom line, all three c's were related to each other by a previous function that produced the A and B, and for some reason, Mathematica could not solve with both A and B, but could with A in terms of B!

    Thank you all for your help!
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