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 <ian.h...@gmail.com> 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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