Re: Simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg119260] Re: Simple integral
- From: Stefan Salanski <wutchamacallit27 at gmail.com>
- Date: Fri, 27 May 2011 06:13:26 -0400 (EDT)
- References: <irlel4$eo0$1@smc.vnet.net>
Your hand solution is simply the real part of Mathematica's solution. (note that the real part of the Log of a negative number, is the log of the absolute value.) The imaginary part of the solution comes from integrating over the singularities of the original function. Your function lends itself to computation with a contour integral, and so it can be seen that when one of the singularities is within your bounds of integration, there is a pole in the complex plane and so the residue at that point must be taken into account. See http://mathworld.wolfram.com/Contour.html . Since these poles are the only source of the imaginary piece, a plot of Im[ ] shows that its a step function. Plot[{Re[fn[x] /. {b -> 3, c -> 2}], Im[fn[x] /. {b -> 3, c -> 2}]}, {x, -10, 10}] So in practice, if you wanted to integrate from x1 to x2, where neither singularity is in (x1,x2), Mathematica's answer gives you exactly what your hand solution does (100% real). The only difference is if you try to integrate with one or both of the singularities inside (x1, x2) Mathematica also includes the imaginary part from the contour integral (which is not an artifact or anything, its supposed to be there). When you did it by hand, did you use complex analysis? -Stefan S On May 26, 7:47 am, Mariano Pierantozzi <mariano.pieranto... at gmail.com> wrote: > I am grateful to all who answered me. > All of you have suggest to me to use the command > Simplify[% // TrigToExp, b^2 - 4 c > 0] > In this way i've this solution: > (Log[-b + Sqrt[b^2 - 4*c] - 2*x] - Log[b + Sqrt[b^2 - 4*c] + 2*x])/ > Sqrt[b^2 - 4*c]. > But this is not the solution of the departure integral. > If "whit my hand" calculate the solution i've this result: > 1(1/Sqrt[b^2 - > 4 c]) (Log[(2 x + > b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))]) > And i thing it's not the same, because i've to do the limit x-->Infininit= y > and in this case the mathematica solution is, once again, in complex fiel= d. > Can you help me again? > Thank you again! > Mariano Pierantozzi > PhD Student > Energy Engineering > > 2011/5/23 Mariano Pierantozzi <mariano.pieranto... at gmail.com> > > > > > > > > > Hi, > > I've got some problem studing this simple integral: > > Integrate[1/(x^2 + b x + c), x]. > > The Mathematica solution is: > > (2 ArcTan[(b + 2 x)/Sqrt[-b^2 + 4 c]])/Sqrt[-b^2 + 4 c] > > > The problem is that my secon order polinomial have two real solutions, = so > > my > > delta (-b^2 + 4 c) is greater than zero. In this case the denominator o= f > > the > > solution does not exist or exist in complex field, but my x is a volume= ... > > I try in this way > > Integrate[1/(x^2 + b x + c), x, Assumptions -> {-b^2 + 4 c < 0}], but > > nothing! > > I can't trasform my arcotangent in two log. > > In summary I would like to have such a solution: > > Integrate[1/(x^2 + 5 x + 6), x] > > Log[2 + x] - Log[3 + x] > > but in general form. > > Sorry for my english! > > Mariano Pierantozzi > > PhD Student > > Energy Engineering