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Re: Integration by substitution

    Regarding the 2nd integral.  Which of these two did you have in mind?
Integrate[a Sech[sp]^2 Sech[(tp - sp)]^2, {sp, -Infinity, Infinity}]
(8*a*E^(2*tp)*(2 - 2*E^(2*tp) + (1 + E^(2*tp))*Log[E^(2*tp)]))/(-1 +
sp = s/a;
Integrate[a Sech[sp]^2 Sech[(tp - sp)]^2, {sp, -Infinity, Infinity}]
    ":", "\<\"Invalid integration variable or limit(s) in \\!\\({s\\/a, \
\\*InterpretationBox[\\(-\[Infinity]\\), DirectedInfinity[-1]], \
\\*InterpretationBox[\\\"\[Infinity]\\\", DirectedInfinity[1]]}\\)
Integrate[a*Sech[s/a]^2*Sech[s/a - tp]^2, {s/a, -Infinity, Infinity}]

i.e. Mathematica *does* see that sp=s/a when you tell it that,& it makes the
substitution, but evidently it won't integrate over a function (expression)
like s/a, so it tells you that.  Probably, you will have to tell Mathematica
to expand the differential in the integral.  I don't know how to do this,
-mark harder

-----Original Message-----
From: G. A. Garrett <ggarrett7 at>
To: mathgroup at
Subject: [mg29852] [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}]
>--Posted email rarely checked--

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