       Re: Re: Integral confusion

• To: mathgroup at smc.vnet.net
• Subject: [mg107314] Re: [mg107297] Re: Integral confusion
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Tue, 9 Feb 2010 02:43:40 -0500 (EST)
• References: <hkm7d8\$os0\$1@smc.vnet.net> <201002080836.DAA29853@smc.vnet.net>

```> But Mathematica does (normally?) return results for indefinite integrals
> without an explicit constant of integration.

What's "explicit" or weird about the constants of integration in this
problem?

I don't know Integrate's steps in obtaining the result, but I'm sure it

Bobby

On Mon, 08 Feb 2010 02:36:16 -0600, Nasser M. Abbasi <nma at 12000.org> wrote:

>
> "Jon Joseph" <josco.jon at gmail.com> wrote in message
> news:hkm7d8\$os0\$1 at smc.vnet.net...
>> All: Is this integral wrong? If not could someone explain the minus sign
>> inside the log?
>>
>> Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify
>>
>> log(-2 (x + 1)) - log(2 (x + 6))
>>
>> Thanks, Jon.=
>>
>
> Well, lets see:
>
> log(-2 (x + 1)) - log(2 (x + 6))
>           = log(-2)+log(1+x) -log(2)-log(x+6)
>           = log(-1)+log(2)+log(1+x)-log(2)-log(x+6)
>           = log(-1)+ log(1+x) - log(x+6)
>
> but log(-1) = Sqrt[-1]*Pi
>
> so result is
>
>          Sqrt[-1]*Pi + log(1+x) - log(x+6)
>
> But we all know that the result should be
>
>          log(1+x) - log(x+6)
>
> So, an extra term, Sqrt[-1]*Pi term pops up. But this term is a
> constant, so
> its derivative is zero, i.e. a constant of integration.
>
> Since
>
> D[log(-2 (x + 1)) - log(2 (x + 6)),x] will give back
>
>                1/(x + 1) - 1/(x + 6)
>
> So, in theory, the answer given by Mathematica is NOT wrong.
>
> But Mathematica does (normally?) return results for indefinite integrals
> without an explicit constant of integration. So I am not sure why it
> does in
> this case, and if it it does, why did not pick this constant? Why not
> C
> as it does for DSolve[]?
>
> So, if I have to guess, I'd say this result is at least very weired, but
> mathematically it is not wrong?
>
> --Nasser
>
>
>

--
DrMajorBob at yahoo.com

```

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