使用Sympy进行Python中的符号集成

Question

I want to integrate exp(-(x^2 + y^2)) in python using sympy library. I could find the integral of exp(-(x^2))

>>> B1 = sympy.exp(-alpha1 * (r1_x**2))
>>> p = integrate(B1,r1_x)
>>> p
pi**(1/2)*erf(alpha1**(1/2)*r1_x)/(2*alpha1**(1/2))

But when I want to try integrate exp(-(x^2 + y^2))

>>> B1 = sympy.exp(-alpha1 * (r1_x**2 + r1_y**2))
>>> p = integrate(B1,r1_x)
>>> p
Integral(exp(-alpha1*(r1_x**2 + r1_y**2)), r1_x)

There is no output and python can't take the integral!

Answer 1

(I am the lead developer of SymPy)

DSM is correct that you can get this to work by calling expand, and that there is no general way to do this (because in general, integrals don't have closed forms).

I just wanted to point out that if SymPy cannot do an integral that does have a closed form, we consider this a bug, and you should feel free to report it at http://code.google.com/p/sympy/issues .

Answer 2

sympy doesn't always recognize every form, and so sometimes you have to give it a little help:

>>> import sympy
>>> alpha1, r1_x, r1_y = sympy.var("alpha1 r1_x r1_y")
>>> B1 = sympy.exp(-alpha1 * (r1_x**2 + r1_y**2))
>>> B1.integrate(r1_x)
Integral(exp(-alpha1*(r1_x**2 + r1_y**2)), r1_x)
>>> B1.expand(alpha1)
exp(-alpha1*r1_x**2)*exp(-alpha1*r1_y**2)
>>> B1.expand(alpha1).integrate(r1_x)
sqrt(pi)*exp(-alpha1*r1_y**2)*erf(sqrt(alpha1)*r1_x)/(2*sqrt(alpha1))