Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

need help with integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51997] need help with integration
  • From: "symbio" <symbio at has.com>
  • Date: Sat, 6 Nov 2004 02:08:53 -0500 (EST)
  • Reply-to: "symbio" <symbio at sha.com>
  • Sender: owner-wri-mathgroup at wolfram.com

 Below I define two functions x[t] and h[t], then in eq1 I integrate
the integrand ( x[tau] h[t-tau] ) from -inf to +inf and get one set of
results and plots, then I integrate the same integrand as before but this
time in two steps, once from -inf to 0 and once from 0 to +inf, but the
results and plots from the integration performed in two steps are NOT the
same as the results and plots from integration in one step.  Is this a bug
and if so what's the work around?

Try this in the input cell

In[19]:=
x[t_] = (3Exp[-.5t]UnitStep[t]) + DiracDelta[t + 3]
h[t_] = UnitStep[t] - UnitStep[t - 2]
eq1 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], -\[Infinity], \[Infinity]}]
Plot[eq1, {t, -10, 10}, PlotRange -> All]
eq2 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], 0, \[Infinity]}]
Plot[eq2, {t, -10, 10}, PlotRange -> All]
eq3 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], -\[Infinity], 0}]
Plot[eq3, {t, -10, 0}, PlotRange -> All]

You will get the following output expressions (I couldn't paste the plots
here, but if you run the above input cells you should get the plots too,
then it will be more obvious what the problem is)

Out[21]=
\!\(1\/2\ \((1 + \(6.`\ \@\((\(-2\) + t)\)\^2\)\/\(\(\(2.`\)\(\
\[InvisibleSpace]\)\) - 1.`\ t\) + \(6.`\ \@t\^2\)\/t + \@\((3 +
t)\)\^2\/\(3 \
+ t\) - \(1 + t + \@\((1 + t)\)\^2\)\/\(1 + t\))\)\)



Out[23]=
\!\(\((\(-6.`\) +
16.30969097075427`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[\
\(-2\) + t] + \((\(\(6.`\)\(\[InvisibleSpace]\)\) -
6.`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[t]\)



Out[25]=
\!\(1\/2\ \((\(-\(\(1 +
t\)\/\@\((1 + t)\)\^2\)\) + \(3 + t\)\/\@\((3 + t)\)\^2)\)\)


  • Prev by Date: Re: Garbage collection problem
  • Next by Date: Re: Re: Garbage collection problem
  • Previous by thread: Re: Re: Inverse of "PowerExpand"
  • Next by thread: Re: need help with integration