       Re: Integration Constants 2

• To: mathgroup at smc.vnet.net
• Subject: [mg5187] Re: Integration Constants 2
• From: Daniel Lichtblau <danl>
• Date: Sat, 9 Nov 1996 02:24:00 -0500
• Organization: wolfram.com
• Sender: owner-wri-mathgroup at wolfram.com

```Dennis Michael Winters wrote:
>
> I posted a message recently (Integration...Constants) that received
> the actual problem:
>
> In:=Integrate[(-2 x+ao)^-2 (-x+bo)^-1, x]
>
> Mathematica gives the following output:
>
> Out=-(1/((ao-2 bo)(ao- 2 x))) - (Log(-bo+x)/(ao^2 - 4 ao bo + 4 bo^2)) +
>
> Log(-ao + 2 x)/(ao^2 - 4 ao bo + 4 bo^2)
>
> This answer is incorrect.  If performed by hand, the answer is shown to
> have positive values for ao and bo.  This is confirmed by the fact that
> this integration is a classical result of chemistry.  Years of chemical
> reaction rate analysis show the values to be positive.  Again, this
> calculation was performed using the unix version of 2.2.3.  Any help from
> those who have already responded would be most appreciated.
>
> --
> ----------------------------------
> -Dennis M. Winters               -
> -P.O. BOX 21215                  -
> -EMORY UNIVERSITY                -
> -ATLANTA, GA 30322               -
> -dwint01 at emory.edu               -
> -or winters at euch6h.chem.emory.edu-
> ----------------------------------

Below I show the result from version 3.0, which is mathematically
equivalent to your result. I then show that the integrand and the
differentiated result differ by a constant (which is in fact zero).

Out//InputForm=
1/((ao - 2*bo)*(-ao + 2*x)) - Log[-bo + x]/(ao - 2*bo)^2 +
Log[-ao + 2*x]/(ao - 2*bo)^2

In:= ee = (-2 x+ao)^-2 (-x+bo)^-1;

In:= Integrate[ee, x] // InputForm
Out//InputForm=
1/((ao - 2*bo)*(-ao + 2*x)) - Log[-bo + x]/(ao - 2*bo)^2 +
Log[-ao + 2*x]/(ao - 2*bo)^2

In:= D[%, x] - ee // Together
Out= 0

I do not understand what you mean by the phrase, "answer ... ha[s]
positive values for ao and bo". The variables ao and bo are parameters.

Daniel Lichtblau
Wolfram Research
danl at wolfram.com

```

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