Skip to main content

calculus and analysis - NIntegrate and Integrate giving different results for ill-behaved function


I'm trying to integrate the following function with Mathematica 8:


$$ I(a,b)= \int_0^1 \mathrm{d}x\int_0^1\mathrm{d}y \,\theta(1-x-y) \frac{1}{x a^2-y(1-y)b^2},$$ where $\theta$ is the Heaviside function. However, I find different results with Integrate or NIntegrate and I don't understand why.


More specifically, for a=100 and b=90:


NIntegrate[HeavisideTheta[1 - x - y]/(x a^2 - y (1 - y) b^2), {x, 0, 1}, {y, 0, 1}]


gives


0.0000927294,

while


Integrate[HeavisideTheta[1 - x - y]/( x a^2 - y (1 - y) b^2), {x, 0, 1}, {y, 0, 1}, PrincipalValue -> True]

gives


+0.0000600275+0.000314159 I.


What is the correct result? Why does Integrate give a complex result?



Answer



Reversing the order of integration produces a solution:


ans= Integrate[HeavisideTheta[1 - x - y]/(x 100^2 - y (1 - y) 90^2), {y, 0, 1}, {x, 0, 1}, 
PrincipalValue -> True]
(* Log[(100*10^(38/81))/(81*19^(19/81))]/10000 *)

N[ans]
(* 0.000060027526501455836 *)


Solutions of this sort are what I would expect based on outlining a pencil-and-paper derivation.


Attempting to solve the integral numerically produces error messages. For instance,


NIntegrate[HeavisideTheta[1 - x - y]/(x 100^2 - y (1 - y) 90^2), {y, 0, 1}, {x, 0, 1}, 
MinRecursion -> 30, MaxRecursion -> 60]

produces


NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>
NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 0.0000591404833446268` and 0.0003356281358114434` for the integral and error estimates. >>
(* 0.0000591404833446268 *)


Of course,NIntegrate has many options, and one or more of them may produce an acceptable answer.


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.