Re: NIntegrate[UnitStep[...]PDF[...],{x,...}] hard to integrate

• To: mathgroup at smc.vnet.net
• Subject: [mg93410] Re: NIntegrate[UnitStep[...]PDF[...],{x,...}] hard to integrate
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Thu, 6 Nov 2008 04:07:49 -0500 (EST)

```On 11/5/08 at 4:57 AM, erwann.rogard at gmail.com (er) wrote:

>f[x_]=(0.014999775454701741*E^(5.264*x) +
>E^(2.632*x)*(-0.012692488700608462 - 0.14964297032258167*x))/
>(12.579644604014753 + 7.102251209677398*E^(2.632*x) + E^(5.264*x))

>g[t_]:=NIntegrate[ UnitStep[t-f[x]] PDF[
>NormalDistribution[0,1/2],x], {x,-Infinity,Infinity} ]

>g[0] runs for a very long time before forcing the kernel to quit,
>let alone FindRoot[g[t]==0.25,{t,-1,1}]

>Note that f[x]PDF[NormalDistribution[0,1/2],x] integrates to 0 and i
>virtually zero outside [-2,2]

>I use Mathematica 5.2/Ubuntu/4GB/IntelCoreDuo2

>any recommandation?

I wonder if you have defined g as you intended.

A plot of f[x] shows it everywhere positive except the interval
from 0 to ~.8. This can be refined by doing

In[8]:= FindRoot[f[x], {x, .8}]

Out[8]= {x->0.84719}

So, UnitStep[t-f[x]] will be 0 outside this interval when t = 0.
This means g[0] amounts to integrating the PDF of a normal
distribution from 0 to .84719. For any distribution the integral
of the PDF is the CDF. The median of NormalDistribution[0,a] is
0. So, g[0] must be

In[10]:= CDF[NormalDistribution[0, 1/2], .84719] - .5

Out[10]= 0.454903

The way you have set the problem up, you are asking Mathematica
to do a lot more work than is needed to solve the problem.

```

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