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Re: help in solving a double integration....

  • To: mathgroup at
  • Subject: [mg66087] Re: [mg66062] help in solving a double integration....
  • From: Daniel Lichtblau <danl at>
  • Date: Sat, 29 Apr 2006 03:40:27 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

ashesh wrote:
> Hi all,
> Would like to solve the following double integration.
> int_x_0^t int_y_0^t F(x) exp(-a|x-y|) G(y) dy dx
> F(x) = (3x/t);   0 < x <(t/3)
>        = (2t-3y)/t;  (t/3) < x < (2t/3)
>        = 0; otherwise
> G(y) = F(y)
> Where a, t are real numbers. And there is absolute value of x-y, i.e.,
> it is |x-y| in the exponent.
> I am not sure as to how these fucntions, F(x) and G(y) can be fed to
> mathematica, as they have different values in different ranges.
> Hope some one can help me solve this integration.

Define your function using Piecewise.

f[x_,t_] := Piecewise[{{3*x/t,0<x<t/3}, {2*t-3*y/t,t/3<x<2*t/3}}, 0]

I pass some assumptions to integrate regarding the parameters {a,t}. I 
think assumptions on both are needed in the current version of 
Mathematica though the development version does not seem to worry about a.

InputForm[ii = Integrate[f[x,t]*Exp[-a*Abs[x-y]]*f[y,t],
   {x,0,t}, {y,0,t}, Assumptions->{t>0,a>0}]]

(-81 + 324*E^((a*t)/3) - 243*E^((2*a*t)/3) + 54*a*t - 108*a*E^((a*t)/3)*t +
   108*a*E^((2*a*t)/3)*t - 108*a*t^2 + 216*a*E^((a*t)/3)*t^2 +
   72*a^2*E^((a*t)/3)*t^2 - 108*a*E^((2*a*t)/3)*t^2 -
   72*a^2*E^((2*a*t)/3)*t^2 - 144*a^2*E^((a*t)/3)*t^3 +
   144*a^2*E^((2*a*t)/3)*t^3 + 16*a^3*E^((2*a*t)/3)*t^3 +
   72*a^2*E^((a*t)/3)*t^4 - 72*a^2*E^((2*a*t)/3)*t^4 -
   36*a^3*E^((2*a*t)/3)*t^4 + 24*a^3*E^((2*a*t)/3)*t^5)/

Sanity check: evaluate numerically at some parameter values and compare 
to direct quadrature result for same.

In[22]:= ii /. {t->4.,a->2.}
Out[22]= 39.0948

In[23]:= With[{t=4,a=2},NIntegrate[f[x,t]*Exp[-2*Abs[x-y]]*f[y,t],
   {x,0,t}, {y,0,t}]]
Out[23]= 39.0948

Daniel Lichtblau
Wolfram Research

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