Skip to main content

calculus and analysis - Problem with Integrate with Piecewise and PrincipalValue


Bug introduced in 8.0.4 or earlier and persisting through 11.3



In the course of developing an alternative solution for question 127301,


With,


$Version
(* "11.0.0 for Microsoft Windows (64-bit) (July 28, 2016)" *)

I attempted to perform the integral,


um = -(2/3) - 2/(3 (-1 + u)) - (2 u)/3 + u^2/3;
up = -(10/3) - 2/(3 (-1 + u)) + (8 u)/3 - u^2/3;
sv = Piecewise[{{um, u <= 1}, {up, u > 1}}];
Integrate[sv, {u, 0, 2}, PrincipalValue -> True]


but received the error message,



Integrate: Integral of ... does not converge on {0,2}.



Separating the term, -(2/(3 (-1 + u))), does not help.


sv1 = Piecewise[{{um + 2/(3 (-1 + u)), u <= 1}, {up + 2/(3 (-1 + u)), u > 1}}];
Integrate[sv1 - 2/(3 (-1 + u)), {u, 0, 2}, PrincipalValue -> True]

yielding the same error message. Yet,



Integrate[sv1, {u, 0, 2}] - 
Integrate[2/(3 (-1 + u)), {u, 0, 2}, PrincipalValue -> True]
(* -1 *)

does work. (The second integral equals 0, incidentally.) Is this a bug, or am I missing something? Thanks.


(Note that 10.4.1 produces the same results.)


Addendum: Workaround


Slightly shifting the Piecewise boundary at u = 1 so that the singular point lies within one or the other segment gives an accurate result. For instance, redefining sv as


sv = Piecewise[{{um, u <= 1 + 10^-10}, {up, u > 1 + 10^-10}}];


allows sv to be integrated by Integrate.


Integrate[sv, {u, 0, 2}, PrincipalValue -> True] // FullSimplify
(* -(4500000000000000000044999999999/4500000000000000000000000000000) *)

which is 1. to 20 significant figures. That this occurs is consistent with the suggestion by MichaelE2 that Integrate integrates each segment of Piecewise independently and, therefore, cannot handle singularities at the boundary between two segments. Nonetheless, I believe that it should be able to. Failing that, the documentation should describe this limitation.



Answer



I think this is a bug, because if we transform the Piecewise function into a combination of UnitStep (which is mathematically equivalent to the original function of course), Integrate integrates without difficulty:


um = -(2/3) - 2/(3 (-1 + u)) - (2 u)/3 + u^2/3;
up = -(10/3) - 2/(3 (-1 + u)) + (8 u)/3 - u^2/3;
sv = Simplify`PWToUnitStep@Piecewise[{{um, u <= 1}, {up, u > 1}}];

Integrate[sv, {u, 0, 2}, PrincipalValue -> True]
(* -1 *)

Tested on v9.0.1 and v11.2.


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.