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Re: specifying the integration interval using a function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg95835] Re: specifying the integration interval using a function
  • From: dh <dh at metrohm.com>
  • Date: Wed, 28 Jan 2009 06:32:15 -0500 (EST)
  • References: <glmt0r$mqd$1@smc.vnet.net>


Hi,

NIntegrate has the Attribute: HoldAll. It does NOT evaluate its 

arguments. Therefore, to force the argument to be evaluated, you have 

wrap it into Evaluate:

NIntegrate[f[y], Evaluate@Flatten[{y, s[x, D]}]]



hope this helps, Daniel



pfb wrote:

> Hi everybody,

> 

>  is it possible to specify the integration interval using a function?

> My problem is as follows:

> 

> I have some function f[x] I want to integrate. Actually I want to

> obtain a sort of running average, i.e. a

> function F[x,D] given by the integral of f[x] over the interval [x-D, x

> +D].

> So far, it's easy. I can do that with the following function

> 

> F[x_,D_]:= NIntegrate[f[y],{y,x-D,x+D}]

> 

> However, the function f may have some (integrable) singularities in

> the integration interval.

> I know that NIntegrate finds it helpful if one tells it the locations

> of the singularities.

> So I thought: easy! I just need a function s[x,D] whose output is  {x-

> D, s1,s2,s3, x+D}.,

> where s1, s2, .. are the singularities of f in the interval.

> 

> I have such a function, but I'm not able to feed it into NIntegrate.

> I have tried

> 

> F[x_,D_]:= NIntegrate[f[y],Flatten[{y,s[x,D]}]]

> 

> but mathematica complains that Flatten[{y,s[x,D]}] is not a correct

> integration range specification, despite

> its evaluation (in a separate cell) gives what I'd expect, i.e. {y,x-

> D,s1,s2,s3,x+D}.

> 

> I also tried something like

> 

> r[y_,x_,D_]:=Flatten[{y,s[x,D]}]

> 

> which again gives {y,x-D,s1,s2,s3,x+D}, and then tried

> 

> 

> F[x_,D_]:= NIntegrate[f[y],r[y,x,D]]

> 

> Mathematica complains also in this case: r[y,x,D] is not a correct

> integration range specification.

> 

> In both case it seems that the function providing the integration

> range is not evaluated.

> Has this anything to do with delayed set (:=)?

> 

> Is there another way of dealing with the intermediate points in an

> integration interval?

> 

> Thanks a lot

> 

> F

> 

> 




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