Integral from Unsolvable Equation

[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 .