
Re: Re: Mathematica integration Vs Sympy
Posted:
Apr 22, 2013 3:13 AM


In this case, the results are valid for a=1 in the sense of a limit, as Limit[u, a > 1] and limit(u, a, 1) demonstrate. This is not always the case. Example:
In: f = Integrate[x^n, x] Out: x^(1 + n)/(1 + n)
In: Limit[f, n > 1, Direction > 1] Out: Infinity
In: Limit[f, n > 1, Direction > 1] Out: Infinity
Alex
On Sun, 21 Apr 2013, Brentt wrote:
> > The result: > > In[0]: Integrate[1/(x*(1  a*(1  x))), x] > Out[0]: (Log[1 + a (1 + x)]  Log[x])/(1 + a) > > Seems to be true for all complex a and x . Why do you think it assumes a>1? > > > > > > > > On Sat, Apr 20, 2013 at 2:42 AM, Sergio R <sergiorquestion@gmail.com> wrote: > >> Hello all, >> >> Just for fun a put an integral I was doing via mathematica >> WolframAlpha >> [ >> http://www.wolframalpha.com/input/?i=Integrate[1%2F%28x*%281a*%281x%29%29%29%2Cx] >> ] >> into the online sympy [ http://live.sympy.org/ ] console >> the following: >> >> a = Symbol('a'); g = 1/(x*(1a*(1x))) ; u=simplify(integrate(g,x)) >> >> Then, to display the result, at the sympy ">>>" prompt, type u >> and hit return. >> >> To my surprise, sympy seems to give the right result without any >> assumption, while mathematica's result seems to assume a>1, which is >> not specified. Also for this case (a>1) sympy gives an extra constant >> which is not present in the mathematica result. >> >> Is there a way to make mathematica to output a general result like >> sympy >> in this case? >> >> Sergio >> >> > >

