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Solving integral - gaussian distribution of cos

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data

    I have to prove:

    ∫(-infinity:infinity) cos(pi*v/2L)*e^-((L-L_av)^2/sqrt(2pi)*sigma^2) dL proportional to
    cos(pi*v/2L_av)*e^-(t/tau)^2

    tau is some constant, and sigma << L_av.

    3. The attempt at a solution

    i can change the integral to 0:infinity, since sigma << L_av. Then i have to look up some integral solution, probably:

    ∫(0:infinity) cos(bx)*e^-ax^2 dx

    I assume i have to do some trick like 1/L approximately L/L_av^2 - but how can i justify that?
     
  2. jcsd
  3. Sep 14, 2012 #2
    Can you use complex numbers? If so, you can rewrite your cosine in term of complex exponentials which will make things quite easy.
     
  4. Sep 14, 2012 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    It is very unclear what your original problem is, because you do not use brackets. You have something written as L_av. Does this mean Lav or does it mean Lav? If you mean the first, write L_{av} or L_(av), and if you mean the second, either write a*L_v or (L_v)a. Better still, use the "X2" button in the menu above the input panel. Anyway, setting L_av = c, some constant, it is still not clear whether you mean
    [tex] \cos \left( \frac{\pi v}{2L} \right) \text{ or } \cos \left( \frac{\pi v}{2} L \right) [/tex] in the integrand. If you mean the first, write cos(pi*v/(2L)), but if you mean the latter, write cos((pi*v/2)L).

    Similarly, when you write e^-ax^2, you are literally writing ##e^{-a} x^2## if we read it using standard rules and conventions. I guess you mean ##e^{-ax^2},## which is e^{-ax^2} or e^(-ax^2) in plain text, or better still, e-ax^2 or e-ax2, using the "X2" button in the menu at the top of the input panel.

    RGV
     
  5. Sep 15, 2012 #4
    Okay, I have to prove:

    ∫(-infinity:infinity) cos(pi*v/(2L))*e-((L-L_{av})^2/(2*sigma^2)) dL proportional to
    cos(pi*v/(2L_{av}))*e-(t/tau)^2

    sigma << L_{av} - both positive
     
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