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Integration of wave functions problem.

  1. May 31, 2013 #1
    Note: I am completely self taught. I have only gone as far as Algebra and regular physics in high school. But I took the initiative to teach myself advanced physics and up to multivariable calculus from reading a book on the subject. So I may not be completely informed on everything present. I just need help with my work. Thank you! I'm also not sure how to make the icons appear as though they are over the other.

    Okay, I was given a problem and I got through most of it. But there is one part where I'm not completely sure how to do the integration.

    2[itex]\int^∞_0[/itex][itex]e^{-3r/2a_0}[/itex][itex]r^2[/itex]dr + λ[itex]\int^∞_0[/itex][itex]e^{-3r/2a_0}[/itex][itex]r^3[/itex]dr=0

    I did solve the problem in the end (solving for λ) in a very unorthodox way. But I am curious as to how it should be solved. I consulted a table of integrals and gave a shot at the improper integrals. But ended up with the wrong coefficients but everything else right. I'm curious as to why my coefficients were wrong. In this specific problem, the coefficients are infact negligible, so, again, I got the answer correct. But I would at least like to have the knowledge to do it correctly.
     
  2. jcsd
  3. May 31, 2013 #2
    If your constant [itex]a_0[/itex] is positive, then indeed, those improper integrals converge. What you want is a technique referred to as Integration by Parts. This will allow you to find antiderivatives of functions of the form [itex]f(r)=r^n e^{-cr}[/itex] among others, where [itex]c[/itex] is a positive constant (so in your case, [itex]c=\frac{3}{2a_0}[/itex]).

    Integration by parts is essentially the product rule backwards. Typically this is one of the first techniques taught when beginning integral calculus. See http://en.wikipedia.org/wiki/Integration_by_parts.

    Give it a try, and tell me where your coefficients go wrong. Show your work too, please.
     
  4. May 31, 2013 #3

    Mark44

    Staff: Mentor

    I would use integration by parts on each integral. For the first integral, u = r2 and dv = e-3r/(2a0dr. That would yield an integral where r's exponent is 1. I would then use int. by parts one more time, using almost the same substitutions.

    For the other integral I would do almost the same, starting with u = r3. You would need to do integration by parts two more times.

    LaTeX tip: Use one pair of tex or itex tags for the whole expression or equation, rather than a whole bunch of pairs of tags.

    This site also takes ## for itex and $$ for tex.
     
  5. May 31, 2013 #4
    I am familiar with integration by parts. And I did try to use it when I tried to solved originally. Would I have a formula at the end similar to: ?

    [itex]\int[/itex][itex]x^n[/itex][itex]e^{ax}[/itex]dx = (1/a)[itex]x^n[/itex][itex]e^{ax}[/itex]-(n/a)[itex]\int[/itex][itex]x^{n-1}[/itex][itex]e^{ax}[/itex]dx
     
    Last edited: May 31, 2013
  6. May 31, 2013 #5

    Mark44

    Staff: Mentor

    Not at the end, but at each step along the way. At each step I described you reduce the exponent on xn by 1.
     
  7. May 31, 2013 #6

    Mark44

    Staff: Mentor

    As already mentioned, you need only a single pair of itex tags. Click the Quote button to see what I did.

    ##\int x^n e^{ax}dx = (1/a) x^n e^{ax}-(n/a)\int x^{n-1}e^{ax}dx ##
     
  8. May 31, 2013 #7
    I can't seem to get it. I wrote down my work, so here's a picture. If you can help me I'd greatly appreciate it. This is only the first integral, obviously. It's supposed to equal [itex]16a^3_0/27[/itex]


    Edit: I forgot the "2" in front of the integral. And this is only the first integral, obviously. It's supposed to equal [itex]16a^3_0/27[/itex]
     

    Attached Files:

    Last edited: May 31, 2013
  9. May 31, 2013 #8

    Mark44

    Staff: Mentor

    You can check your work, you know. Take the derivative of what you have, and you should get your integrand. It's always a good idea to do this.

    As a very simple example of this,
    $$\int x^2dx = (1/3)x^3 + C$$

    To check, d/dx( (1/3)x3 + C) = 3 * (1/3)x2 + 0 = x2. That's the integrand in the original problem, so my answer checks.
     
  10. May 31, 2013 #9
    When I factor a little further and then use the limits of the improper integral I get [itex]32a^3_0/27[/itex]... which is the correct answer! So in this case, I got the correct coefficients too. Thank you!
     
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