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


Mathematica certainly seems to be assuming a != 1, since of course the displayed output is undefined when a = 1.
However:
1/(x*(1  a*(1  x))) /. a > 1 1/x^2 Integrate[%, x] 1/x
which is correct. And:
Limit[(Log[1 + a (1 + x)]  Log[x])/(1 + a), a > 1] (1 + x)/x
which differs from 1/x by a constant.
The real question should be, I think: why does Mathematica generate conditions on parameters for some integrals but not others?
On Apr 21, 2013, at 5:16 AM, Brentt <brenttnewman@gmail.com> 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?
 Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 5491020 (H) University of Massachusetts 413 5452838 (W) 710 North Pleasant Street fax 413 5451801 Amherst, MA 010039305



