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 <email@example.com> wrote:
> The result: > > In: Integrate[1/(x*(1 - a*(1 - x))), x] > Out: (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 <firstname.lastname@example.org> 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?
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