The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Inactive » comp.soft-sys.math.mathematica

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 60
Registered: 9/24/11
Re: Mathematica integration Vs Sympy
Posted: Apr 21, 2013 5:18 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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 <> wrote:

> Hello all,
> Just for fun a put an integral I was doing via mathematica
> WolframAlpha
> [
> ]
> into the online sympy [ ] 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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.