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I can't understand this integral solution (Solution given)

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    I've been studying a scientific paper for my dissertation and in it there is an integral that I cannot work out how its been derived in any way shape or form, so here goes.

    Importantly, kθ0>>1 and θ0<<1


    2. Relevant equations
    So the equation is ∫θθ0 θ-2e-2kθ

    (just so you know, the integral is from θ→θ0

    and in the paper this apparently becomes after the integration has been performed,

    e-2kθθ-2/2k

    Yet I have no idea how this is possible, although it seems like the θ-2 has been treated as a constant. Any rules I need to use or help would be greatly appreciated.
     
  2. jcsd
  3. Apr 9, 2013 #2
    You sure it integrates in ##\theta## something which has ##\theta## as extremal???
     
  4. Apr 9, 2013 #3
    Yes, the paper states, 'this integral can only be solved in extreme cases of drag, kθ0'
     
  5. Apr 9, 2013 #4
    may you post the paper (attached or some reference)??
     
  6. Apr 9, 2013 #5
  7. Apr 9, 2013 #6
    I'm so sorry... I can't see it... I actually even do not see how it goes from (17) to (18)... even assuming he simply integrates, it should be integrated in ##t##, not in ##\theta##...
    What seems strange to me in (18) is that he is integrating in one variable, and the same variable appears outside the integral and in the integration limits. Also he tells you that (18) gives you ##\dot{\theta}(t)##, but I do not really see where the time dependence should come out... unless of course you are integration in time instead than in ##\theta##... or unless he gives the same exact name to many different things, which wouldn't be so smart in case.

    Anyway sorry really, let's see if someone else finds out something which I missed :cry:
     
  8. Apr 9, 2013 #7
    The rest of it makes complete sense to me, infact I've derived pretty much the rest of the paper except that one integral. Just looking at the answer I know he's either done something I've never done in maths before or made it up haha. My tutor believes it's a numerical approximation or something but I've explored this angle also and still not come up with his solution. My theory was that since e-2kθ→0 much faster than θ-2, θ-2is treated as a constant but if this is the case then it goes against everything I've ever been taught in mathematics.
     
  9. Apr 9, 2013 #8
    I eventually was thinking something a little different... but it all depends on the values ##\theta## itself can have... indeed if you assume that ##\theta## is not too different from ##\theta_0##, then the integration domain is small and you may assume ##\theta^{-2}## as constant there... not the same for the exponential because if ##k\theta_0\gg 1## and at the same time ##\theta_0\ll 1## then it is ##k## which is quite big; in this case the variation (not the value) of the exponential is big (as it is essentially ##k##) and therefore even for a small domain you can't assume it to be constant...

    But I repeat it all depends on how different can be ##\theta## and ##\theta_0##...
     
  10. Apr 9, 2013 #9
    Just to let you know to get from (17) to (18) you need to multiply through by [itex]\dot{\theta}e^{-2k\theta}[/itex] and then notice that if you differentiate [itex]d/dt \,\dot{\theta}^2e^{-2k\theta}/2 = (\dot{\theta}/2)(-2ke^{-2k\theta}\dot{\theta}^2+2\ddot{\theta}e^{-2k\theta})[/itex]

    Which using the chain rule can be integrated with respect to θ.

    My tutors idea was similar to yours, however I was quite lost when he started going through it.

    I kind of get what he's trying to get at though, does the fact that θ0 has to be << 1 have an effect?
     
    Last edited: Apr 9, 2013
  11. Apr 9, 2013 #10
    Oh... yes you are right, it does work :smile:... Thank you for the hint...
    as for your problem, the only solution I can think about is the one I told you... as you know the problem far better than me having studied it, try to see if the values of ##\theta## are such that they permit you to do this trick... otherwiseI do not know
     
  12. Apr 9, 2013 #11
    Do you have any idea what the techniques called or what kind of calculus it is so I can have do some background research on it, since I've never studied it I think I'll be expected to provide some sort of reference or background theory.

    The Functional Integration article on Wiki seems to do something quite similar could this be it?
     
  13. Apr 9, 2013 #12
    It could be some kind of Calculus of Variations, but I am not really sure... try ask to some mathematicians if you have some at hand :wink:
     
  14. Apr 9, 2013 #13
    I appreciate the help, will do some research into it and let you know if I get anywhere.
     
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