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Re: Integration by substitution

  • To: mathgroup at
  • Subject: [mg29824] Re: [mg29800] Integration by substitution
  • From: "J Rockmann" <mtheory at>
  • Date: Thu, 12 Jul 2001 02:52:37 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

----- Original Message -----
From: "G. A. Garrett" <ggarrett7 at>
To: mathgroup at
Subject: [mg29824] [mg29800] 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--
If I'm not mistaken, Mathematica will consider "sp" as a single variable as
it would "x".  Therefore, the two statements are not equivalent to
Mathematica after any algebraic manipulation has been performed on your
naming of the variable "sp=sa".
I hope this helps,
Jonathan Rockmann
mtheory at

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