Both results are in fact the same. The first one is probably slightly more correct. You tend see these polar_lift
functions whenever SymPy tries to do something like square rooting something when it does not know the signs of the things inside (after integrating)
A polar_lift
does not appear below, but this basic Gaussian example shows that SymPy tries to be as general as possible:
from sympy import *
x = Symbol("x", real=True)
y = Symbol("y", real=True)
s = Symbol("s", real=True) # , positive=True
gaussian = exp(-((x-y)**2)/(2*(s**2)))
nfactor = simplify(integrate(gaussian, (x,-oo,oo)))
print(nfactor)
You need s
to be declared as positive: s = Symbol("s", real=True, positive=True)
. A similar thing happens with these kinds of polar_lift(b - 2*y)
functions in your example. It also happens with the question I reference below.
I have no idea why, but N(polar_lift(x))
for any float
or int
x
gives x
again; yet, SymPy does not simplify nicely with symbolic x
. Turns out if you keep on simplifying, you get nicer and nicer looking answers. I couldn't find anything about polar_lift related to pure math so I don't know what it actually does.
Remember for the simple example above how it gave a piece-wise? Same thing here. So we just take the first piece since the second piece is an un-evaluated integral.
In the code below, I use this question to remove the piece-wise function and then I simplify twice. And finally, I manually remove the polar_lift
.
import sympy as sp
x, y = sp.symbols("x y", real=True)
b, u, l, t = sp.symbols("b u l t ", real=True, positive=True)
Fortet = sp.integrate(sp.exp(-l * t) * (sp.sqrt(2 * sp.pi * t)) ** (-1) *
sp.exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, sp.oo), conds='none')
incorrect = Fortet.simplify().simplify()
correct = eval(str(incorrect).replace("polar_lift", ""))
correct = correct.factor()
print(correct)
The result is:
exp(b*u)*exp(-u*y)*exp(-sqrt(2*l + u**2)*sqrt(b**2 - 2*b*y + y**2))/sqrt(2*l + u**2)
That is close enough to your expression. I couldn't make SymPy simplify the sqrt(b**2 - 2*b*y + y**2)
to Abs(b-y)
no matter how hard I tried.
Note that either SymPy is still wrong or you are wrong since the powers in the numerator are opposite. So I checked on the Desmos for a numeric answer (top one is yours):
