Hi,
There are a couple of things to say here. First I think you have
confused when you give boundaries via x or via z. After substitution,
the integrand becomes sqrt(z)/sqrt(1-z**2), so let's discuss this. Then
we have:
In [26]: i = Integral(sqrt(z)/sqrt(1-z**2), (z, 0.3, 0.4))
In [29]: i.doit().subs(exp_polar, exp).n()
Out[29]: 0.0631741978430877
In [30]: i.n()
Out[30]: 0.0631741978430876
So this seems to work.
Some notes. First we need to replace exp_polar by exp here. This is
permitted, since |z| < 1. Arguably this is a shortcoming of evalf of
hyper. In fact the meijerg() evalf can evaluate for arbitrary polar
numbers, hijacking some obscure mpmath functionality. It is possible to
implement similar functionality for hyper(), but that's more work. At
the very least it should work for |z| < 1...
Since the integral can be done in terms of elliptic integrals this
hypergeometric function should have a closed form, but I cannot find it
documented anywhere.
Tom
On 07.07.2012 03:06, pallab wrote:
The answer seems to be wrong.
I do the following to numerically integral evaluation:
import numpy as np
import scipy as sp
import scipy.integrate
f=lambda x:np.sqrt(np.sin(x))
sp.integrate.quad(f,0.6,0.7)
get the following value:
0.07776347731181982
which matches with mathematica.
whereas
fI=sm.lambdify(z,sm.integrate(dydz(z)*tointegrate.subs(subdict)).subs(sm.exp_polar(2*sm.I*sm.pi),1.0))
and
fI(0.6)-fI(0.7)
gives:
mpf('-0.10638922100995812')
On Friday, July 6, 2012 12:00:19 PM UTC-4, pallab wrote:
It seems sympy can not integrate sqrt(sin(x)).
I did the following:
import sympy as sm
from sympy.abc import x,y,z
tointegrate=sm.sqrt(sm.sin(y))
sm.integrate(tointegrate)
output is : Integral(sqrt(sin(y)), y)
After a simple change of variable the integral is doable:
def ytoz(z):
return(sm.asin(z))
def dydz(z):
return(sm.diff(ytoz(z),z))
subdict={y:cvytoz(z)}
sm.integrate(dydz(z)*tointegrate.subs(subdict))
output is : z**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,),
z**2*exp_polar(2*I*pi))/(2*gamma(7/4))
best,
Pallab
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