MathGroup Archive 2001

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

Search the Archive

Integration by substitution


I was wondering if someone could explain to me why the first input
below gives no answer while the second input does. It never crossed my
mind the Mathematica wouldn't have seen the simple substitution of
sp=s/a that produced an answer. I am running ver. 4.0 in a Windows
environment. I am basically looking for a characteristic that I can be
on the look out for in the future.

Integrate[Sech[s/a]^2 Sech[(t - s)/a]^2, {s, -Infinity, Infinity}]

Integrate[a Sech[sp]^2 Sech[(tp - sp)]^2, {sp, -Infinity, Infinity}]

Gregory
--Posted email rarely checked--


  • Prev by Date: boolean operations for polygons
  • Next by Date: Re: Saving the whole workspace...
  • Previous by thread: Re: boolean operations for polygons
  • Next by thread: Re: Integration by substitution