       Re: Integral confusion

• To: mathgroup at smc.vnet.net
• Subject: [mg107301] Re: [mg107262] Integral confusion
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Mon, 8 Feb 2010 03:37:00 -0500 (EST)
• References: <201002071112.GAA25294@smc.vnet.net>

```A key point is that Log[ k g[x] ] == Log@k + Log@g[x], so the multipliers
2 and -2 inside the logs are simply (arbitrary) constants of integration.

Anyway, Mathematica's result has the intended derivative:

f[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify
D[f@x, x]

Log[-2 (1 + x)] - Log[2 (6 + x)]

1/(1 + x) - 1/(6 + x)

The question is only, "Where is it defined and continuous?"

The first term of the integrand, 1/(x+1), has a pole at -1. The second
term has a pole at -6.

Hence, plotting over the range between:

Plot[f@x, {x, -6, -1}, PlotRange -> All]

Plot[1/(x + 1) - 1/(x + 6), {x, -6, -1}]

The two plots seem consistent; I see no reason to question Mathematica's
result.

Near -6 (but greater), the second term dominates in both integrand and
integral. x+6 is small but positive, so 1/(6+x) is large but positive.

Near -1 (but smaller), the first term dominates. x+1 is small and
negative, so 1/(x+1) is large and negative, hence Log[-2 (1+x)] is large
and positive.

If we try to plot on other ranges, we'll have Complex results from Log...
but D and Integrate assume complex variables anyway.

Bobby

On Sun, 07 Feb 2010 05:12:45 -0600, Jon Joseph <josco.jon at gmail.com> wrote:

> 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.=
>

--
DrMajorBob at yahoo.com

```

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