Integral confusion
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From: Bob Hanlon on 8 Feb 2010 03:34 f1[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify log(-2 (x+1))-log(2 (x+6)) f1'[x] 1/(x + 1) - 1/(x + 6) It means that for the result to be real that x must be less than -1. Reduce[-2 (x + 1) > 0] x < -1 Looking at both Logs Reduce[{-2 (x + 1) > 0, 2 (x + 6) > 0}] -6 < x < -1 Plot[f1[x], {x, -6, -1}] If you don't like the Log form, use FullSimplify f2[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // FullSimplify -2*ArcTanh[(2*x)/5 + 7/5] f2'[x] // Simplify // Apart 1/(x + 1) - 1/(x + 6) f1[x] == f2[x] // FullSimplify True Bob Hanlon ---- Jon Joseph <josco.jon(a)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.=
From: Simon on 8 Feb 2010 03:34 On Feb 7, 7:15 pm, Jon Joseph <josco....(a)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.= The integral is correct (just take its derivative and see). >From the look of the unsimplified output In[1]:= Integrate[1/(x + 1) - 1/(x + 6), x] Out[1]= 5 (1/5 Log[-2 (1 + x)] - 1/5 Log[2 (6 + x)]) I think that Mathematica is doing a Together before the integral. If you integrate each term separately, you get what you'd expect In[2]:= Integrate[{1/(x + 1), -1/(x + 6)}, x] Out[2]= {Log[1 + x], -Log[6 + x]} But of course, the different branches of Log only differ by a constant -- which is exactly what indefinite integrals don't care about. If x > 1, then you can simply factor out the minus sign using In[3]:= Log[-1] Out[3]= I \[Pi]
From: Nasser M. Abbasi on 8 Feb 2010 03:36 "Jon Joseph" <josco.jon(a)gmail.com> wrote in message news:hkm7d8$os0$1(a)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[1] 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
From: Jon Joseph on 8 Feb 2010 03:36 Thanks Bob and David. I completely understand your responses and I figured that was what Mathematica was doing. However, if I asked my students to integrate 1/(x+1) - 1/(x+6) and they came back with the answer Mathematica found I would, at the very least, call that student over for a discussion. Log arguments in absolute value are certainly more "traditional" . I appreciate Mathematica's complexity but it is unfortunate that a user has to piece together the underlying logic on such a simple problem. Jon On Feb 7, 2010, at 7:38 AM, Bob Hanlon wrote: > > f1[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify > > log(-2 (x+1))-log(2 (x+6)) > > f1'[x] > > 1/(x + 1) - 1/(x + 6) > > It means that for the result to be real that x must be less than -1. > > Reduce[-2 (x + 1) > 0] > > x < -1 > > Looking at both Logs > > Reduce[{-2 (x + 1) > 0, 2 (x + 6) > 0}] > > -6 < x < -1 > > Plot[f1[x], {x, -6, -1}] > > If you don't like the Log form, use FullSimplify > > f2[x_] = Integrate[1/(x + 1) - 1/(x + 6), x] // FullSimplify > > -2*ArcTanh[(2*x)/5 + 7/5] > > f2'[x] // Simplify // Apart > > 1/(x + 1) - 1/(x + 6) > > f1[x] == f2[x] // FullSimplify > > True > > > Bob Hanlon > > ---- Jon Joseph <josco.jon(a)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.= > >
From: DrMajorBob on 8 Feb 2010 03:36
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(a)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(a)yahoo.com |