MathGroup Archive 2001

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

Search the Archive

Re: Integration by substitution

  • To: mathgroup at
  • Subject: [mg29865] Re: Integration by substitution
  • From: ggarrett7 at (G. A. Garrett)
  • Date: Sat, 14 Jul 2001 01:36:55 -0400 (EDT)
  • References: <9ig788$kh5$>
  • Sender: owner-wri-mathgroup at

For your concideration. Note that the following integral
does not evaluate:

  Mathematica 4.0 for Microsoft Windows
  Copyright 1988-1999 Wolfram Research, Inc.
   -- Terminal graphics initialized -- 

   In[1]:= Integrate[Sech[(x-x0)/k]^2,{x,-Infinity,Infinity}]

                          x - x0 2
   Out[1]= Integrate[Sech[------] , {x, -Infinity, Infinity}]

Even though the same function with just scaling or just displacement
does evaluate.

   In[2]:= Integrate[Sech[x/k]^2,{x,-Infinity,Infinity},Assumptions ->

   Out[2]= 2 k

   In[3]:= Integrate[Sech[x0-x]^2,{x,-Infinity,Infinity}]

   Out[3]= 2

Exploring further, we note that replacing the limits of integration
produce an answer:

   In[4]:= Integrate[Sech[(x-x0)/k]^2,{x,-b,b}]

                    -b - x0            b - x0
   Out[4]= -(k Tanh[-------]) + k Tanh[------]
                       k                 k

But taking the limit does not:

   In[5]:= Limit[Out[4],b->Infinity]

                          -b - x0            b - x0
   Out[5]= Limit[-(k Tanh[-------]) + k Tanh[------], b -> Infinity]
                             k                 k

yet the equation clearly has a solution given the proper constraint:

   In[6]:= Simplify[Out[4] /. b->Infinity,k>0]

   Out[6]= 2 k

But applying this contraint to the original integral has no affect.

   In[7]:= Integrate[Sech[(x-x0)/k]^2,{x,-Infinity,Infinity},Assumptions->

                          x - x0 2
   Out[7]= Integrate[Sech[------] , {x, -Infinity,
Infinity},Assumptions-> k>0]

Any ideas on what it is about this equation that produces this effect
so that one can keep an eye out for it in the future? Or perhaps there
is a rule I could define that would allow Integrate to "discover" the

Note that in my previous post, it might be more correct to state that
the limits of the integration should also be changed to +- Infinity/a
after the substitution, in which case Mathematica will choke and give
no answer.


  • Prev by Date: Re: Changing Pure Function in Module
  • Next by Date: Re: about ConstrainedMin
  • Previous by thread: Re: Integration by substitution
  • Next by thread: Can Anyone Help With This?