Re: Integration by substitution

• To: mathgroup at smc.vnet.net
• Subject: [mg29865] Re: Integration by substitution
• From: ggarrett7 at netscape.net (G. A. Garrett)
• Date: Sat, 14 Jul 2001 01:36:55 -0400 (EDT)
• References: <9ig788\$kh5\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

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

Mathematica 4.0 for Microsoft Windows
-- Terminal graphics initialized --

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

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

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

In[2]:= Integrate[Sech[x/k]^2,{x,-Infinity,Infinity},Assumptions ->
k>0]

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
does

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->
k>0]

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

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
solution?

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