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Solving integral - gaussian distribution of cos

  • Thread starter poul
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  • #1
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Homework Statement



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.

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?
 

Answers and Replies

  • #2
Can you use complex numbers? If so, you can rewrite your cosine in term of complex exponentials which will make things quite easy.
 
  • #3
Ray Vickson
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Homework Statement



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.

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?
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
 
  • #4
17
0
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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