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MathGroup Archive 2006

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Re: Evaluating integrals

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66213] Re: Evaluating integrals
  • From: "Chris Chiasson" <chris.chiasson at gmail.com>
  • Date: Thu, 4 May 2006 05:20:51 -0400 (EDT)
  • References: <e39jvr$clu$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Your hand answer suggests the result is independant of mu c, and I was
wondering if it was.

In[1]:=
<< "Statistics`NormalDistribution`"
In[2]:=
g[c_] = PDF[NormalDistribution[\[Micro]c, sc], c]
Out[2]=
1/(E^((c - \[Micro]c)^2/(2*sc^2))*(Sqrt[2*Pi]*sc))
In[3]:=
pi = Exp[(-r)*b*c]
Out[3]=
E^((-b)*c*r)
In[4]:=
Integrate[pi*g[c], c]
Out[4]=
(1/2)*E^((1/2)*b*r*(b*r*sc^2 - 2*\[Micro]c))*Erf[(c + b*r*sc^2 -
\[Micro]c)/(Sqrt[2]*sc)]
In[5]:=
Block[{b = 5, r = 1, c = 3, sc = 2}, %]
Out[5]=
(1/2)*E^((5/2)*(20 - 2*\[Micro]c))*Erf[(23 - \[Micro]c)/(2*Sqrt[2])]
In[6]:=
Plot[%, {\[Micro]c, 0, 1}, PlotRange -> All]
In[7]:=
N[Block[{b = 5, r = 1, c = 3, sc = 2}, Exp[(r^2*b^2*sc^2)/2]]]
Out[7]=
5.184705528587072*^21


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