I'll except either one. With preference given to more than one answer.
I'll except either one. With preference given to more than one answer.
I'm getting that the integral is 1 at R=1. Working on an interpretation.
Хьюстон, это Саравак, от подписания.
And that is the answer given by wolframalpha. I'll accept that of course.
The integrated function is undefined at theta=0, R=1, but it can be assigned a limiting value that is finite, and makes the integral equal to 1.
However, the integrated function is also the limit of *both* sets of functions, one set from the left all of which have an integral of zero, and one set from the right, which has a limit of integral of 2, obviously. The difference between the two is they both approach the same function at R=1, *except* at theta=0 where the left one would have a negative Dirac delta function, and the right one would have a positive Dirac delta function.
Good. I was afraid you might except it instead
Can define the integrand to be anything you like at theta=0, R=1, and the Lebesgue integral remains well defined.
That looks like it. If you take the absolute value of the integrand, then the answer is always the one for R>1. But without the absolute value and with R<1, that nasty hump near theta=0 (which is bigger, the closer R is to one) flips signs at R=1.
Хьюстон, это Саравак, от подписания.
I happen to like the Dirac delta function...
The one for R=1, but I knew what you meant.That looks like it. If you take the absolute value of the integrand, then the answer is always the one for R>1.
It all depends on whether you're touching the point on the left, or the right.But without the absolute value and with R<1, that nasty hump near theta=0 (which is bigger, the closer R is to one) flips signs at R=1.
So, what if you move to the bottom or top of it? And that's only for a super-thin shell, too.
ETA: your turn!
Arg, it's more complicated than I realised. I was thinking the solution for R>1 extended to continuity to R=1, but it doesn't.
Let me think about it, when I'm not IP banned.
Хьюстон, это Саравак, от подписания.
There aren't any Dirac functions in the solution
Хьюстон, это Саравак, от подписания.
In the solutions for R>1? I dunno, I'd have to say, taken as a whole, their limit is a delta function--and after all, that's all a delta function is really, just a limit of a bunch of ordinary functions.
What's weird about this is, for R>1, the value of the integrand at theta=0 is zero. So I agree that it's not taking the usual approach to a delta function.
To be more specific then, if we include a delta function in the integrand at R=1, then the definite integral is 2, which agrees with the 2/R^{2} values for the R > 1.
If we include a negative delta function in the integrand at R=1, then the value of the definite integral is 0, which agrees with the values for R < 1.
In both cases, the augmented integrand at R=1 is the respective limit function (functional? generalized function?) from either side.
I'm getting 2 for the definite integral at R=1 by a direct calculation.
Scratch that, i'm getting 1 (i inadvertently integrated twice previously)
I don't quite understand the problem, is it that the solution is discontinuous at R=1?
It's not my question, but my interpretation was, what is it qualitatively or intuitively about the integrand, that generates the discontinuity (in R) in the integral.
I am rather bothered by this - at first, I felt like the phenomenon was driven by the fact that the integrand is the square root of a rational function, but we need to take the negative square root over part of the range of integration for R<1. But then the R=1 case should extend by continuity from the R>1 case, and yet it clearly doesn't. (I just verified the 2/R^2 formula numerically.) So that is only part of the story.
The value at R=1 is the midpoint of the upper and lower limits, I haven't decided whether that is meaningful yet.
Хьюстон, это Саравак, от подписания.
More or less, that's it. And the answer is amazing, at least I thought so--it's all laid out in post #671 above.
Actually, that is none of the story. The angle theta is measured from the center of the shell, so the distance is always positive, no negative square root involved.I am rather bothered by this - at first, I felt like the phenomenon was driven by the fact that the integrand is the square root of a rational function, but we need to take the negative square root over part of the range of integration for R<1. But then the R=1 case should extend by continuity from the R>1 case, and yet it clearly doesn't. (I just verified the 2/R^2 formula numerically.) So that is only part of the story.
As R approaches 1 from above (that is, a test particle moving towards the shell), the small area directly below has a greater and greater influence, even as the effective area lessens because the outlying parts are more closer to perpendicular to the approach. At R=1, the effect is like a flat plane, and the gravity is 1 due to that plane effect--and only at theta equal zero. The effect of the entire rest of the shell only adds up to 1, that is the integral that you've calculated. But, at R=1, the gravity has to be 2/R^{2}, Newton's law, and together they do add up to 2.The value at R=1 is the midpoint of the upper and lower limits, I haven't decided whether that is meaningful yet.
On the "other side" of that single point, that effect completely reverses, and cancels out the entire effect of the rest of the shell. The integrands vary smoothly from there. So, the "solution" does involve a delta function.
ETA: You can play with this in wolframalpha, substituting various values for R. Sometimes, wolframalpha complains though, saying I must be trying to find the cosine of a cosine--maybe that's a problem with one of their servers, I don't know. When it works, it includes a nice graph of the integrand.
Integrate[ (sine(theta) * (R-cosine(theta)))/(1 + R^2 - 2*R*cosine(theta))^(3/2), {theta,0, \pi}]
The integrand can become negative for small values of theta, when R<1. This is what I mean - if you square it, the integrand is a rational function, so I thought that maybe the integral of the square root of the square would not have a discontinuity. But that would seem not to be the case.
Хьюстон, это Саравак, от подписания.
The integrand can be visualized as the gravitational effect of a spherical shell of radius 1 on a test particle at R. For R<1, the negative part of the integrand represents the part of the shell pulling the particle "up", away from the center of the shell.
Of course, Newton's laws have the gravity equal to 2/R^{2} down to the surface, at which point it becomes zero. But that "Newtonian" explanation would include an infinitely narrow, infinite spike of total area 1.
The indefinite integral isn't so hard, integrate by parts, then integrate again.
I think some additional insight can be had from it.
Хьюстон, это Саравак, от подписания.
I think you missed a couple thetas.
ETA: oops, indefinite integral. Sorry. So, that's it?
EETA: So evaluating that at pi and zero gives the 2/R^2 for R>1 and 0 for R<1, as it should, and your idea that the difference is from the positive/negative square root is correct. The evaluation boils down to 1/R^2 +/- 1/R^2, except at R=1
I used a substitution u = 1+R^2-2*R*cos(theta) and it also worked out.
EEETA: And technically it is undefined at R=1
Last edited by grapes; 2012-May-04 at 01:09 AM. Reason: Add babbling
Physically, the integral comes from the gravity on a rest point at A due to a thin hollow shell of radius r. *
*
Which, according to the shell theorem, is equal to zero for A < r, but GM/A^2 otherwise, which is
*
Yes, I derived this, when I wasn't quite sure about the sine in the numerator. I got what you got (to within a multiplicative constant). Even the constant can be taken care of by having the shell have the right Dirac density
I was surprised about how easy the indefinite integral was to find - I figured it might be one of these where we can find the definite integral explicitly, but not the indefinite. But it wasn't.
Хьюстон, это Саравак, от подписания.
How is it undefined at R=1? I can see that it is undefined at but in order to take the definite integral we have to take the limit letting approach zero, which after using L'Hopital gets us 1 for the definite integral. (this was the way i calculated the definite integral at R=1 the first time, integrate to the indefinite integral and then take the appropriate limit)
The integral is definitely defined in the sense of Lebesgue for R=1. I think grapes is probably referring to some of the later specific expressions (e.g., my indefinite integral has a 0/0 problem for R=1 if you just plug theta=0 into it to find the definite integral).
Хьюстон, это Саравак, от подписания.
Yes, I meant that the definite integral is undefined at R=1, because of its value at theta equal zero. We can define a value, in various ways and one way is to take the limit.
As Coelacanth pointed out, any value we set the undefined portion of the integrand to, results in the same Lebesgue value for the integral, which is that limit. But we have another option.
I think you could assign any finite value you like, and it would even remain Riemann-integrable
The limit at R=1 and theta=0 rather depends on the direction in the R-theta plane from which you approach, which seems to be the fundamental phenomenon driving the whole thing here
Хьюстон, это Саравак, от подписания.
So, if I'm up, what is the value of the following integral?
Хьюстон, это Саравак, от подписания.
Sure, the measure of the set over which we change the value is zero (it's an endpoint) so it shouldn't change the Lebesgue integral. We could even redefine the integrand to have a different value at every rational number (or whatever subset of with measure zero we choose) and it wouldn't change the Lebesgue integral. Incidentally, because we're only using an endpoint, this means the Lebesgue integral must equal the improper Riemann integral. Neither does the improper Riemann integral depend on the value we choose at (the limit of a function at a point doesn't depend on the value of the function at that point).
Which is?But we have another option.
All standard approaches seem to give the same answer, in fact i think there's a theorem that the Lebesgue integral must equal the improper Riemann integral in this case, so i'm not sure what the added benefit is to going the Lebesgue route rather than just the standard route of dealing with improper Riemann integrals.