Re: Definite Integration in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg74683] Re: Definite Integration in Mathematica*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sat, 31 Mar 2007 01:31:22 -0500 (EST)*References*: <etqo3f$10i$1@smc.vnet.net> <ett73k$h3g$1@smc.vnet.net> <euigeg$ov1$1@smc.vnet.net>

"dimitris" <dimmechan at yahoo.com> wrote: > > Back in December, in the thread which led to this one, I gave an > > antiderivative which is continuous along the whole real line: > > > > ArcTan[(4x + Sqrt[2(15 + Sqrt[241])])/(2 - Sqrt[2(-15 + Sqrt[241])])] + > > ArcTan[(4x - Sqrt[2(15 + Sqrt[241])])/(2 + Sqrt[2(-15 + Sqrt[241])])] > > Could we show me how you get this antiderivative for the integrand > > f[x_] = (x^2 + 2*x + 4)/(x^4 - 7*x^2 + 2*x + 17); Let's begin with Mathematica's own antiderivative: ArcTan[(1 + x)/(4 - x^2)] Recalling that tan(p + q) = (tan(p) + tan(q))/(1 - tan(p)tan(q)), let's rewrite (1 + x)/(4 - x^2) in the form ((a*x + b) + (c*x + d))/(1 - (a*x + b)*(c*x + d)) because that will allow us to break apart Mathematica's antiderivative into a sum of two arctangents having arguments which are _always real_. In[9]:= ArcTan[a*x + b] + ArcTan[c*x + d] /. FullSimplify[SolveAlways[ (1 + x)/(4 - x^2) == ((a*x + b) + (c*x + d))/(1 - (a*x + b)*(c*x + d)), x][[3]]] Out[9]= ArcTan[Root[-1 + 15*#1 - 58*#1^2 - 17*#1^3 + 3*#1^4 & , 1] + x*Root[4 - 8*#1 + 21*#1^2 - 17*#1^3 + 3*#1^4 & , 1]] + ArcTan[Root[-1 + 15*#1 - 58*#1^2 - 17*#1^3 + 3*#1^4 & , 2] + x*Root[4 - 8*#1 + 21*#1^2 - 17*#1^3 + 3*#1^4 & , 2]] Out[9] is equivalent to the continuous antiderivative which I'd given previously. The process I used to obtain it here could be automated. The only thing that probably seems mysterious above is why I chose the _third_ solution in particular. Explanation: SolveAlways returned six solutions. In the first two of those, a and c were zero, and so those two were not valid. In the last two of those, we got complex quantities; either of those could in fact have been used, but I preferred an expression where everything was real. The middle two solutions involved real quantities only. I could just as well have chosen to use the fourth solution, instead of the third; doing so would also have produced a continuous antiderivative equivalent to that which I'd given previously. David