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Re: Integration by substitution


Dear Mr Rockmann,
It is true that "sp" and "tp" are single variables, but under the substitution
sp->s/a and tp->t/a, the 2 expressions

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

and

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

are mathematically equivalent. This can be checked by substituting and taking
into account that ds = a dsp and that the integration limits do not change.

J Rockmann wrote:

> ----- Original Message -----
> From: "G. A. Garrett" <ggarrett7 at netscape.net>
To: mathgroup at smc.vnet.net
> Subject: [mg29843]  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 msn.com

--

Ignacio Rodriguez Ramirez de Arellano
Unidad de RMN
Universidad Complutense
Paseo Juan XXIII, 1
Madrid 28040, Spain

Tel. 34-91-394-3288
Fax  34-91-394-3245
e-mail: ignacio at sgirmn.pluri.ucm.es





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