Delaying evaluation until after simplification

[Ian Bell]

I am trying to do an integration like this:



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

[Aaron Meurer]

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

 

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[Ian Bell]

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

[Chris Smith]

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)

[Aaron Meurer]

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

To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/95372b64-3f66-484b-b8c2-1efff0d3a87f%40googlegroups.com.

[Chris Smith]

You can also just use the "manual" hint:

>>> integrate(1/((1-z)),z,manual=True)

-log(-z + 1)