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Re: Mathematica integration Vs Sympy

• To: mathgroup at smc.vnet.net
• Subject: [mg130564] Re: Mathematica integration Vs Sympy
• From: Brentt <brenttnewman at gmail.com>
• Date: Sun, 21 Apr 2013 05:16:26 -0400 (EDT)
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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 at 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*%281-a*%281-x%29%29%29%2Cx]
> ]
> into the online sympy [ http://live.sympy.org/ ] console
> the following:
>
> a = Symbol('a'); g = 1/(x*(1-a*(1-x))) ; 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
>
>

• Follow-Ups:
  • Re: Re: Mathematica integration Vs Sympy
    • From: Alex Krasnov <akrasnov@eecs.berkeley.edu>
  • Re: Mathematica integration Vs Sympy
    • From: Murray Eisenberg <murray@math.umass.edu>

• References:
  • Mathematica integration Vs Sympy
    • From: Sergio R <sergiorquestion@gmail.com>