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Difficult Integral

  1. Feb 15, 2005 #1
    "Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster from the observed (two-dimensional) density that can be analyzed from a photograph. Suppose that that in a spherical cluster of radius [tex]R[/tex] the density of stars depends only on the distance [tex]r[/tex] from the center of the cluster. If the perceived star density is given by [tex]y(s)[/tex], where [tex]s[/tex] is the the observed planar distance from the center of the cluster, and [tex]x(r)[/tex] is the actual density, it can be shown that

    [tex]y(s) = \int _s ^R \frac{2r}{\sqrt{r^2 - s^2}}x(r)\: dr[/tex]

    If the actual density of stars in a cluster is

    [tex]x(r)=\frac{1}{2}(R-r)^2[/tex]

    find the perceived density [tex]y(s)[/tex]."


    I've tried working with the aid of a technical computing software, but it doesn't make things clear to me right now---although I have the solution:

    [tex]y(s) = \int _s ^R \frac{2r}{\sqrt{r^2 - s^2}}\: x(r)\: dr = \left. \frac{1}{3}\sqrt{r^2 - s^2} \left( r^2 - 3rR + 3R^2 + 2s^2 \right) -Rs^2 \ln \left( r + \sqrt{r^2 - s^2} \right) \right] _s ^R[/tex]

    [tex]y(s)=\frac{1}{3}\sqrt{R^2 - s^2}\left( R^2 + 2s^2 \right) - Rs^2 \ln \left( R + \sqrt{R^2 - s^2} \right) - \left( -Rs^2 \ln s \right) [/tex]

    [tex]y(s)=\frac{1}{3}\sqrt{R^2 - s^2}\left( R^2 + 2s^2 \right) + Rs^2 \ln s - Rs^2 \ln \left( R + \sqrt{R^2 - s^2} \right) [/tex]

    I couldn't find any reference in my table of integrals to help me understand this better. Maybe you guys can explain me how this integration is carried out. Any help is highly appreciated.
     
    Last edited: Feb 15, 2005
  2. jcsd
  3. Feb 15, 2005 #2
    Use the subtitution

    [tex] r = s sec\theta[/tex]

    you will end up with bunch of integrals of the form



    [tex] \int \frac{1}{cos^n \theta} d\theta[/tex] , with n = 2,3 and 4

    and these are standard integrals.
     
  4. Feb 15, 2005 #3
    Use the subtitution

    [tex] r = s sec\theta[/tex]

    you will end up with bunch of integrals of the form



    [tex] \int \frac{1}{cos^n \theta} d\theta[/tex] , with n = 2,3 and 4

    and these are standard integrals.
     
  5. Feb 15, 2005 #4

    dextercioby

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    Use the Hyperbolic [itex] \cosh x [/itex] as a substitution and the result will come uot nicely.

    Daniel.
     
  6. Feb 15, 2005 #5

    saltydog

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    If I had to guess, I'd think in practice, y(s) is obtained through experiment and then x(s) needs to be determined? I don't know, just think that's how it would be. In that case, you'd handle the equation as a Volterra type integral equation right?

    [tex]y(s) = \int _s ^R F(r,s)x(r)dr[/tex]

    Can someone here comment about this? Or maybe Thiago can comment about it too.

    Thanks,
    Salty
     
  7. Feb 15, 2005 #6

    krab

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  8. Feb 16, 2005 #7

    dextercioby

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    Yes,Salty,it's the other way around indeed.Knowing the perceived density,one must find the "real" one,which means solving for "x" from the integral equation you posted.
    But in this case,being an elementary one,the physics is turned upside down for the likes of simple mathematics...

    Daniel.
     
  9. Feb 16, 2005 #8
    Well, thank you guys for the help!!

    What I can see is that there are two ways to do it:

    1. Hyperbolic substitution with [tex]\cosh x[/tex]
    2. Trigonometric substitution with "[tex]\mbox{ssec } \theta[/tex]". By the way, don't you mean [tex]\sec \theta[/tex]?

    I gotta review some of these techniques so that I can do it on my own. Anyway, I now know what direction I should take. Thanks
     
    Last edited: Feb 16, 2005
  10. Feb 16, 2005 #9

    dextercioby

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    That was definitely a typo.And it came out twice,since he double-posted the message.It's [itex] \sec \theta [/itex],as it should be...

    Daniel.
     
  11. Feb 16, 2005 #10

    saltydog

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    Wait a minute. We're not done. Can you give some "realistic" perceived data so that I can work on the integral equation (I mean, once I figure out how to solve just the integral with the subs. that were suggested)?

    What about a real reference with real astronomical data? Are you familiar with Chandrasachar's study of star brightness in the Milky Way? He solves a beautiful PIDE by infinitely folding an integral! Beautiful. It's ok if you don't want to pursue it. I have time.

    Salty
     
  12. Feb 16, 2005 #11

    saltydog

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    Thanks a bunch Krab, I'd have jumped into my integral equations books and found a world of hurt. Your reference is a great place to start.

    Salty
     
  13. Feb 16, 2005 #12

    saltydog

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    Just though I'd report the equations I was referring to. Not trying to show off. Just beautiful math and perhaps others will think so too. Here's the original Chandrasachar equation:

    [tex]g(u,e)+\frac{\partial g}{\partial u}+\frac{\partial g}{\partial e}=\int_0^1 g(\frac{u}{q},e) \Psi(q) \frac{dq}{q}[/tex]

    Actually, he makes some simplifying assumptions in order to reduce it to an IDE:

    [tex]f(u)+\frac{df}{du}=\int_0^1 f(\frac{u}{q})\Phi(q)dq [/tex]

    The solution to the second problem is very interesting to follow (took me several weeks, I ain't proud).
     
  14. Feb 16, 2005 #13

    It was not a typo. It is [tex] r = s sec\theta[/tex]. You need 's' so that when you factor out 's' you will end up with [tex]s\sqrt{sec^2\theta -1} = s\tan\theta[/tex]

    Have a look at the integral again.


    [tex]y(s) = \int _s ^R \frac{2r}{\sqrt{r^2 - s^2}}x(r)\: dr[/tex]


    Regards.



    Note: I accidently double posted it yesterday. How do I delete one if that happens in the future? Thanks.
     
  15. Feb 16, 2005 #14

    Astronuc

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    Send PM to moderator to delete post.
     
  16. Feb 16, 2005 #15

    dextercioby

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    You can delete it immediately (actually it goes as an edit,so no more than 24hrs).

    Daniel.
     
  17. Feb 16, 2005 #16

    dextercioby

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    And one more thing:
    [tex] s\cdot \sec\theta [/tex] makes as difference compared to what u've written.And if u still don't like the multiplicative dot,u can use the [tex] \times [/tex] sign...

    Daniel.
     
  18. Feb 16, 2005 #17

    saltydog

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    Wasn't easy for me.

    Consider:

    [tex]\int_x^R \frac{r(R-r)^2}{\sqrt{r^2-x^2}} dr [/tex]

    Using [itex] r=xSec(\theta))[/itex] then the integral becomes:

    [tex] \int x\sec^2(\theta)(R-x\sec(\theta))^2 d\theta [/tex]

    This actually turn out to be manageable yielding:

    [tex] xR^2\tan(\theta)-Rx^2[\ln(\sec(\theta)+\tan(\theta))+\sec(\theta)\tan(\theta)]+x^3[\tan(\theta)+\frac{1}{3}tan^3\(\theta)]
    [/tex]

    Using a right-triangle to convert [itex]\theta[/itex] back to oposite, adjacent, and hypotnuse and back substituting and then a lot of rearranging, I got (with difficulty) the answer reported by Thiago (just used x to avoid confusion with sec). Where are you anyway, we're not done yet . . . what about a plot? What about the integral transform?
     
    Last edited: Feb 16, 2005
  19. Feb 16, 2005 #18
    Salty: "Wait a minute. We're not done. Can you give some "realistic" perceived data so that I can work on the integral equation (I mean, once I figure out how to solve just the integral with the subs. that were suggested)?

    What about a real reference with real astronomical data? Are you familiar with Chandrasachar's study of star brightness in the Milky Way? He solves a beautiful PIDE by infinitely folding an integral! Beautiful. It's ok if you don't want to pursue it. I have time."


    Sorry, but the information in the first post of this thread is all that there is to the problem. Unfortunately, there isn't any real data to look into. In fact, this is just a problem from my calculus book, which only asks for the integral (no plot or transform is required).

    I'm not familiar with Chandrasachar's study of star brightness in the Milky Way, but it's always fun to go ahead and learn a bit more. If you have any reading suggestions, please let me know.

    Thank you
     
  20. Feb 16, 2005 #19

    saltydog

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    That's Ok Thiago, really just kidding about you doing anything further. I intend to pursue it further but I'll post the transform in the Cosmology group (once I work with it a bit) as it's more appropriate there. Interesting. Will try to find real data. I'm into Astronomy.

    Thank You,
    Salty
     
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