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Topic: Mathematica integration Vs Sympy
Replies: 7   Last Post: Apr 23, 2013 9:38 PM

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Murray Eisenberg

Posts: 2,105
Registered: 12/6/04
Re: Mathematica integration Vs Sympy
Posted: Apr 22, 2013 3:13 AM
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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*%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?


---
Murray Eisenberg murray@math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2838 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305









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