Re: [sympy] ntegrate does not know the sign of abs()

[Aaron Meurer]

It's definitely a bug. integrate should catch NotImplementedError from limit() and return an unevaluated Integral. 

Also, not knowing the sign of abs is definitely a bug as well. Please open an issue for this at http://code.google.com/p/sympy/issues/list. 

Aaron Meurer

On Jun 18, 2012, at 12:53 PM, Barak Yair Reif <barak...@gmail.com> wrote:

I tried to look for this bug and haven't found it (maybe my lacking search skills), still I felt uncomfortable to open an issue. 

Here is the code:

x=Symbol("x")

f1=2*E**(-2*x)

F1=integrate(f1,(x,0,oo))

print F1 #Works great:

a=Symbol("a")

f2=abs(a)*E**(-abs(a)*x)

F2=integrate(f2,(x,0,oo)) #fails

and the output is:

Traceback (most recent call last):

  File "C:/Users/Barak/Desktop/python/temp", line 18, in <module>

    F2=integrate(f2,(x,0,oo))

  File "C:\Python27\lib\site-packages\sympy\utilities\decorator.py", line 24, in threaded_func

    return func(expr, *args, **kwargs)

  File "C:\Python27\lib\site-packages\sympy\integrals\integrals.py", line 847, in integrate

    return integral.doit(deep = False)

  File "C:\Python27\lib\site-packages\sympy\integrals\integrals.py", line 393, in doit

    function = antideriv._eval_interval(x, a, b)

  File "C:\Python27\lib\site-packages\sympy\core\expr.py", line 229, in _eval_interval

    B = limit(self, x, b)

  File "C:\Python27\lib\site-packages\sympy\series\limits.py", line 116, in limit

    return i*limit(d, z, z0, dir)

  File "C:\Python27\lib\site-packages\sympy\series\limits.py", line 192, in limit

    r = gruntz(e, z, z0, dir)

  File "C:\Python27\lib\site-packages\sympy\series\gruntz.py", line 678, in gruntz

    r = limitinf(e, z)

  File "C:\Python27\lib\site-packages\sympy\core\cache.py", line 101, in wrapper

    func_cache_it_cache[k] = r = func(*args, **kw_args)

  File "C:\Python27\lib\site-packages\sympy\series\gruntz.py", line 480, in limitinf

    c0, e0 = mrv_leadterm(e, x)

  File "C:\Python27\lib\site-packages\sympy\core\cache.py", line 101, in wrapper

    func_cache_it_cache[k] = r = func(*args, **kw_args)

  File "C:\Python27\lib\site-packages\sympy\series\gruntz.py", line 562, in mrv_leadterm

    f, logw = rewrite(exps, Omega, x, w)

  File "C:\Python27\lib\site-packages\sympy\series\gruntz.py", line 627, in rewrite

    raise NotImplementedError('Result depends on the sign of %s' % sig)

NotImplementedError: Result depends on the sign of -sign(Abs(a))

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[Aaron Meurer]

It looks like in the git master the integral works:

In [1]: var('a')
Out[1]: a

In [2]: integrate(abs(a)*exp(-abs(a)*x), (x, 0, oo))
Out[2]:
⎧ π
⎪ 1 for │periodic_argument(│a│, ∞)│ < ─
⎪ 2
⎪
⎪∞
⎨⌠
⎪⎮ -x⋅│a│
⎪⎮ ℯ ⋅│a│ dx otherwise
⎪⌡
⎪0
⎩

But the limit bug still exists (it's simply bypassed by the new
integration algorithm), so if you could still open an issue for that,
that would be great.

Aaron Meurer

[Tom Bachmann]

> But the limit bug still exists (it's simply bypassed by the new
> integration algorithm)

Are you sure of that? I vaguely remember fixing an issue related to
limits and integral evaluation. Not sure, though.

[Barak Yair Reif]

Opened: http://code.google.com/p/sympy/issues/detail?id=3298 

> sympy+unsubscribe@googlegroups.com.

[Aaron Meurer]

I meant the other bug:

In [5]: limit(integrate(abs(a)*exp(-abs(a)*x), x), x, oo)
---------------------------------------------------------------------------
NotImplementedError Traceback (most recent call last)
<ipython-input-5-bb74dd8ee8de> in <module>()
----> 1 limit(integrate(abs(a)*exp(-abs(a)*x), x), x, oo)

/home/asmeurer/Dropbox/sympy/sympy/series/limits.pyc in limit(e, z, z0, dir)
134 i, d = e.as_independent(z)
135 if i is not S.One and i.is_bounded:
--> 136 return i*limit(d, z, z0, dir)
137 else:
138 i, d = S.One, e

/home/asmeurer/Dropbox/sympy/sympy/series/limits.pyc in limit(e, z, z0, dir)
260
261 try:
--> 262 r = gruntz(e, z, z0, dir)
263 if r is S.NaN:
264 raise PoleError()

/home/asmeurer/Dropbox/sympy/sympy/series/gruntz.pyc in gruntz(e, z, z0, dir)
676 r = None
677 if z0 == oo:
--> 678 r = limitinf(e, z)
679 elif z0 == -oo:
680 r = limitinf(e.subs(z, -z), z)

/home/asmeurer/Dropbox/sympy/sympy/core/cache.pyc in wrapper(*args, **kw_args)
89 except KeyError:
90 pass
---> 91 func_cache_it_cache[k] = r = func(*args, **kw_args)
92 return r
93 return wrapper

/home/asmeurer/Dropbox/sympy/sympy/series/gruntz.pyc in limitinf(e, x)
470 e = e.subs(x, p)
471 x = p
--> 472 c0, e0 = mrv_leadterm(e, x)
473 sig = sign(e0, x)
474 if sig == 1:

/home/asmeurer/Dropbox/sympy/sympy/core/cache.pyc in wrapper(*args, **kw_args)
89 except KeyError:
90 pass
---> 91 func_cache_it_cache[k] = r = func(*args, **kw_args)
92 return r
93 return wrapper

/home/asmeurer/Dropbox/sympy/sympy/series/gruntz.pyc in mrv_leadterm(e, x)
548 #
549 w = Dummy("w", real=True, positive=True, bounded=True)
--> 550 f, logw = rewrite(exps, Omega, x, w)
551 series = calculate_series(f, w, logx=logw)
552 series = series.subs(log(w), logw) # this should not be necessary

/home/asmeurer/Dropbox/sympy/sympy/series/gruntz.pyc in rewrite(e,
Omega, x, wsym)
616 wsym = 1/wsym #if g goes to oo, substitute 1/w
617 elif sig != -1:
--> 618 raise NotImplementedError('Result depends on the sign
of %s' % sig)
619 #O2 is a list, which results by rewriting each item in
Omega using "w"
620 O2 = []

NotImplementedError: Result depends on the sign of -sign(Abs(a))

Aaron Meurer

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[Tom Bachmann]

ah yes, the sign of -abs(a) not being known.

[Chris Smith]

Since the sign can be 0 or 1 maybe that's why it's confused. If you
give `a` a sign it works:

>>> var('a',positive=True)
a
>>> f2=abs(a)*E**(-abs(a)*x)
>>> integrate(f2,(x,0,oo))

1
>>> limit(integrate(abs(a)*exp(-abs(a)*x), x), x, oo)

0

/c

[Chris Smith]

One way to help integrate out would be to replace all Abs(foo) with
positive dummies, do the integration, then restore the original Abs
instances.

[Aaron Meurer]

On Jun 19, 2012, at 8:38 PM, Chris Smith <smi...@gmail.com> wrote:

> Since the sign can be 0 or 1 maybe that's why it's confused. If you
> give `a` a sign it works:

Ah, so this is actually not a bug, because if a is zero, then the
values of both the integral and the limit at infinity are different.
Not sure why I didn't notice that before...

Moral of the story: if you want something to be positive, define it
using Symbol(positive=True). abs() will give you nonnegative, which
is a little different...

Aaron Meurer

>
>>>> var('a',positive=True)
> a
>>>> f2=abs(a)*E**(-abs(a)*x)
>>>> integrate(f2,(x,0,oo))
> 1
>>>> limit(integrate(abs(a)*exp(-abs(a)*x), x), x, oo)
> 0
>
> /c
>