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