Integral from Unsolvable Equation

[Mute]

diagram, shmiagram. What is the mathematical justification? To go from dr to dv, given r=u+v, u an unknown function of v, you need: (du/dv+1) dv. I don't see how any diagram changes this.

Come on now, don't be patronizing. Just because we're not quite following haruspex's diagrammatical arguments doesn't mean they're wrong. That said, haruspex, I do feel that if you are going to be making a diagrammatical argument it helps if you supply the picture you have in mind, rather than trying to describe the picture you have in your mind in detail. A picture of what you are thinking of will come across much easier than a description of the picture you are thinking of. (The typos are also hurting the presentation of your argument).

I'm very short on time these couple of days which is why I haven't tried to fully work through it myself. I took a short break and tried working through haruspex's argument, and I can almost see it - the slicing in the u-v plane is the last thing I need to work out, but I really need to go finish some other things now.

I also tried working through it quickly mathematically, but I think I have a sign error somewhere. Perhaps overlooked the absolute value signs haruspex mentions. Basically, using the fact that u(v) is an even function of v and so u'(v) is an odd function of v, one can perhaps show that the u'(v)+1 factor cancels out.

 

[PAllen]

Come on now, don't be patronizing. Just because we're not quite following haruspex's diagrammatical arguments doesn't mean they're wrong. That said, haruspex, I do feel that if you are going to be making a diagrammatical argument it helps if you supply the picture you have in mind, rather than trying to describe the picture you have in your mind in detail. A picture of what you are thinking of will come across much easier than a description of the picture you are thinking of. (The typos are also hurting the presentation of your argument).

I'm very short on time these couple of days which is why I haven't tried to fully work through it myself. I took a short break and tried working through haruspex's argument, and I can almost see it - the slicing in the u-v plane is the last thing I need to work out, but I really need to go finish some other things now.

I also tried working through it quickly mathematically, but I think I have a sign error somewhere. Perhaps overlooked the absolute value signs haruspex mentions. Basically, using the fact that u(v) is an even function of v and so u'(v) is an odd function of v, one can perhaps show that the u'(v)+1 factor cancels out.

Until the last post, haruspex didn't even mention that he was treating a single integral as a double integral, and insisted 'it was just substitution'. The words used in his last post, for the first time, would have at least given a clue what was in mind, rather than mis-using established terminology. Substitution, integration by parts, double integral are not normally interchanged. Especially confusing is that the r,s substitution was just that - an ordinary substitution for single integrals.

To me, there was not only refusal to provide explanatory detail, there was misuse of terminology.

Seeing that a couple of us thought dr = (u'+1)dv was required, some real attempt to explain would be helpful, rather than ignoring this request.

 

[haruspex]

Seeing that a couple of us thought dr = (u'+1)dv was required, some real attempt to explain would be helpful, rather than ignoring this request.

Would it have been less confusing if I'd said 'change of co-ordinates'?
How do you substitute for dxdy when changing to polar? You don't write dx = cos(θ)dr - r sin(θ)dθ etc. You use the Jacobian. For rs to uv we get drds = 2dudv.

 

[haruspex]

Diagram attached.

 

[PAllen]

Would it have been less confusing if I'd said 'change of co-ordinates'?
How do you substitute for dxdy when changing to polar? You don't write dx = cos(θ)dr - r sin(θ)dθ etc. You use the Jacobian. For rs to uv we get drds = 2dudv.

Well for starters, I don't see how, for most of your posts, or starting from the original problem statement, anyone would be thinking about dxdy, drds, or dudv. You, yourself wrote:

∫sdr

and ended up with simple integral on dv. No mention of double integration until the very end of this discussion.So, yes, a great deal of your thought process was not presented, and words and expressions you did use pointed in a different direction.

My integral in post #22 must be correct (at least for the simple case of a=1,b=2), and it remains an interesting task to understand its equivalence to yours, on the hypothesis that yours is correct.

 

[PAllen]

How do you substitute for dxdy when changing to polar? You don't write dx = cos(θ)dr - r sin(θ)dθ etc. You use the Jacobian. For rs to uv we get drds = 2dudv.

But if you have y=f(x), and a single integral on dx, and you introduce r,θ with polar definition, with x,y, and θ taken to be implicit functions of r, and you want to write an integral dr that yields the same answer as the original, that is exactly what you do.

 

[PAllen]

With the help of the diagram, combined with the observation that justifies haruspex's symmetry argument:

x(y) = y(x) [edit: meaning x for some y=k equals y for some x=k]

from the original implicit function defining equation,
and realizing that we are going from drds to dudv (rather than from dr to dv),
I agree with the gist of haruspex argument. I haven't checked every detail, but the overall approach seems valid.

Note: the symmetry argument resolves the discrepancy on limits of integration. It remains interesting to understand how the u'+1 factor from another valid argument can end up being ignored. Interesting, but I don't have time to play with this anymore for a while.

 

[PAllen]

The key difference in understanding was:

- treating u as just an auxiliary function to the process of changing variable of integration from r to v. Single integral variable change.

- versus, treating the integration after r,s were introduced as a trivial double integral, then changing coordinates (finally, using a symmetry argument to integrate over half the region).


These are completely different things, both are valid. As an exercise, I applied both to a definite integral of s=2r from zero to 1. You will see that the mechanics are completely different even for this trivial case, though the answers come out the same. The first just ends up being a convoluted way to change variable v=-r/2. The latter integrates a region bounded by various lines in the uv plane (no symmetry argument for this case of s=2r).

 

[JJacquelin]

Hi !

The parametric form allows to write explicitly the integral.
In attachment, the solving process involves a special function, namely the dilogarithm function, which is a particular case of polylogarithm.
It should be possible to proceed exactly on the same manner without dilogarithm. All could be done with the related integral instead of. The advantage of the use of dilogarithm is that the asymptotic formula is directly available. Using the related integral supposes to calulate the corresponding asymptotic development, what still increases the boring work.

 

[PAllen]

Hi !

The parametric form allows to write explicitly the integral.
In attachment, the solving process involves a special function, namely the dilogarithm function, which is a particular case of polylogarithm.
It should be possible to proceed exactly on the same manner without dilogarithm. All could be done with the related integral instead of. The advantage of the use of dilogarithm is that the asymptotic formula is directly available. Using the related integral supposes to calulate the corresponding asymptotic development, what still increases the boring work.

yes, the dilogarithm is very useful for this. I used it for the special case I solved at the beginning of the thread, to establish the general features.

 

[PAllen]

With the help of the diagram, combined with the observation that justifies haruspex's symmetry argument:

x(y) = y(x) [edit: meaning x for some y=k equals y for some x=k]

from the original implicit function defining equation,
and realizing that we are going from drds to dudv (rather than from dr to dv),
I agree with the gist of haruspex argument. I haven't checked every detail, but the overall approach seems valid.

Note: the symmetry argument resolves the discrepancy on limits of integration. It remains interesting to understand how the u'+1 factor from another valid argument can end up being ignored. Interesting, but I don't have time to play with this anymore for a while.

I can't specifically explain how the integral form with u'+1 (see post #22, and following) is equivalent to the identical looking one without this (and just a Jacobian factor which is constant 2 (or -2, depending on conventions) in this case, instead). The only explanation is that both are derived with valid methods (change of variable, single integral; change of coordinates, double integral).

I have verified the following:

- with the u'+1 factor, the integrand is not symmetric about v=0 (without the factor, the integrand [really, area of integration for double integral] is symmetric about v=0, corresponding to the r=s line, as Haruspex has pointed out). Thus, explicitly doing -∞ to +∞ is required if the u'+1 factor is used.

- I cannot solve the integral with the u'+1 factor in closed form, even with special functions (someone else may be more clever). However, numerically integrating for typical cases establishes that it is correct .