Just compute the definite integral, and evaluate at the limits manually.
That is, use integrate(1/(1-z), z).

However you're going to have problems because as Chris notes, log(1 - z) =
log(-1) + log(z - 1) is not always valid (I think it may be true if z < 1,
if I remember the rule correctly).  The issue is that for real values, the
integral of 1/x should be log(abs(x)) (this isn't true in the complex case,
which is why SymPy doesn't return that value).

I believe the result you get with the I*pi is mathematically correct. That
is, if you plug in any real value for z, you'll get the same result as from
-log(abs(z - 1)).

Aaron Meurer

On Fri, Feb 13, 2015 at 8:15 PM, Chris Smith <[email protected]> wrote:

> You can do a little manual simplification like this:
>
> >>> integrate(1/(1-z),(z,0,z))
> -log(z - 1) + I*pi
> >>> exp(_)
> -1/(z - 1)
> >>> 1/(-1/_/-1)  # the badger-face inversion :-)
> 1/(-z + 1)
> >>> log(_)
> log(1/(-z + 1))
>
> But that can't simplify to -log(1-z) because that is not valid unless 1-z
> is positive.
>
> >>> expand_log(log(1/Dummy(positive=True)))
> -log(_Dummy_85)
> >>> expand_log(log(1/Dummy(nonnegative=True)))
> log(1/_Dummy_86)
> >>> expand_log(log(1/Dummy(negative=True)))
> log(1/_Dummy_107)
>
>
> On Friday, February 13, 2015 at 6:37:56 PM UTC-6, Ian Bell wrote:
>>
>> There was an embedded image, but clearly that didn't work.
>>
>> >> integrate(1/(1-z),(z,0,z))
>>
>> was the command.  I used a picture to show that you get an imaginary term
>> as part of the return value
>>
>> On Friday, February 13, 2015 at 4:11:54 PM UTC-7, Aaron Meurer wrote:
>>>
>>>
>>>
>>> On Fri, Feb 13, 2015 at 3:42 PM, Ian Bell <[email protected]> wrote:
>>>
>>>> I am trying to do an integration like this:
>>>>
>>>
>>> Was there supposed to be something here? It's just showing up as empty
>>> space for me.
>>>
>>> Aaron Meurer
>>>
>>>
>>>>
>>>>
>>>>
>>>> Ultimately the result I should be able to get is -log(1-z)
>>>>
>>>> Manual integration shows that you can integrate(1/(1-z),z) ->
>>>> -log(z-1), evaluation at the limits yields -log(z-1) - (-log(-1)) which you
>>>> can simplify to -log(1-z).  How can I tell sympy to delay evaluation until
>>>> after it has done the simplification?  I guess that it first does log(-1),
>>>> which it isn't happy about...
>>>>
>>>> Ian
>>>>
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