Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: How to make results from Integrate and NIntegrate agree

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66642] Re: [mg66610] How to make results from Integrate and NIntegrate agree
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 25 May 2006 02:58:21 -0400 (EDT)
  • References: <200605240701.DAA08462@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 24 May 2006, at 16:01, riazzi at hotmail.com wrote:

> Hello,
>
> I need to evaluate the integral of a discontinuous function for which
> the results from Integrate and NIntegrate do not agree.   A more  
> simple
> function which demonstrates the problem I face is the following, which
> is a simplified form (for nonnegative, real arguments) of the
> FractionalPart function from functions.wolfram.com.  I'm using this
> form because Integrate returns my integral unevaluated when I use the
> built in FractionalPart[].
>
> In[1]=
> FrPart[x_]:=(1/2)-ArcTan[Cot[Pi*x]]/Pi
>
>
> So here's the example function to be integrated, showing it's
> discontinuities at {1/n : n a Natural number}:
>
> In[2]=
> Plot[FrPart[1/x], {x,1/6,1},PlotPoints->100];
>
>
> If I integrate this numerically, I get what appears to be the right
> answer, and the plot looks like what I would expect to see:
>
> In[3]=
> nintgrl[x_] := If[x = (1/6), 0, NIntegrate[FrPart[1/u], {u, (1/6),
> x},
>     MaxRecursion -> 30, WorkingPrecision -> 30, PrecisionGoal -> 25]]
> Plot[nintgrl[x], {x, 1/6, 1}, PlotStyle -> {Red}];
> nintgrl[1]
>
> Out[3]= <<Plot omitted>>
> 0.3417594692
>
> If I evaluate the integral piecewise, I obtain an answer that agrees
> with NIntegrate's:
>
> In[4]=
> area = Sum[Integrate[1/u -n, {u,1/(n+1),1/n}],{n,1,5}]
> N[area]
>
> Out[4]= -(29/20) + Log[6/5] + Log[5/4] + Log[4/3] + Log[3/2] + Log[2]
> 0.341759
>
>
> If I evaluate the integral with Integrate, here is what I get:
>
> In[5]=
> FullSimplify[Integrate[FrPart[1/x], x]]
>
> Out[5]= (x/2) - (x * ArcTan[Cot[Pi/x]])/Pi + Log[x] - 1
>
> (NOTE: I am aware that having Integrate evaluate the definite integral
> Integrate[FrPart[1/u], {u,1/6,1}] returns the correct answer in this
> case.  For my more complicated function over a different range it does
> not.)
>
>
> Here's how the Integrate result looks, with a constant of integration
> of 5/6 + Log[6] added to make the plot start at zero like the
> NIntegrate result:
>
> In[6]=
> Plot[(x/2) - (x * ArcTan[Cot[Pi/x]])/Pi + Log[x] - 1 + 5/6 + Log[6],
> {x, 1/6, 1}, PlotStyle -> Blue];
>
>
> Looking at limits as x approaches (1/n) from either direction, I see
> that the magnitude of the jump at (1/n) equals (1/n).  If I remove the
> jumps (by subtracting off their contribution via the function adj[x]),
> the NIntegrate and Integrate plots overlap perfectly:
>
> In[7]=
> adj[x_] := Sum[(1/n)*UnitStep[x - 1/n], {n, 2,  5}];
> adjPlot = Plot[(x/2) - (x * ArcTan[Cot[Pi/x]])/Pi + Log[x] - 1 + 5/6 +
> Log[6] - adj[x], {x, 1/6, 1}, PlotStyle -> Blue];
>
>
> But I don't want to subtract off the contribution of the jumps, nor do
> I want to sum up a (for me infinite) collection of piecewise  
> terms.  So
> what I'm looking for is a way to make Integrate produce a closed-form
> expression for my integrand that does not produce these
> discontinuities, plus some insight as to why they appear.  I'd guess
> the latter has to do with branch cuts in the complex plane for the
> functions involved, and where the solution might have to do with
> picking a different contour or somesuch, but that's getting into areas
> I've long forgotten or never learned.
>
> Thanks in advance,
>     Will R.
>

I suggest downloading from MathSource Maxim Rytin's excellent package  
"Piecewise".

http://library.wolfram.com/infocenter/MathSource/5117/


It can deal correctly with FractionalPart as well as with your  
function FrPart:

<< Piecewise`


Simplify[PiecewiseIntegrate[FractionalPart[1/x], {x, 1/6, 1}]]


-(29/20) + Log[6]

N[%]

0.341759


FrPart[x_] := 1/2 - ArcTan[Cot[Pi*x]]/Pi


PiecewiseIntegrate[FrPart[1/x], {x, 1/6, 1}]

-(29/20) + Log[6]

The package also contains a numerical integrator NPiecewiseIntegrate:


NPiecewiseIntegrate[FractionalPart[1/x], {x, 1/6, 1}]


0.3417594692280566

I am not sure if PiecewiseIntegrate can manage a general formula for

PiecewiseIntegrate[FractionalPart[1/x], {x, a, b}]

It seems to me a lot to ask and as it appears to be taking a long  
time, which I could not to spare today I just aborted.


Andrzej Kozlowski

Tokyo, Japan



  • Prev by Date: Re: How to make results from Integrate and NIntegrate agree
  • Next by Date: Differential equations
  • Previous by thread: How to make results from Integrate and NIntegrate agree
  • Next by thread: Re: How to make results from Integrate and NIntegrate agree