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newbie question about symbolic integration in sage

asked 2012-09-22 08:45:33 -0600

kayhan gravatar image

updated 2012-09-22 16:06:33 -0600

kcrisman gravatar image

Hi,

I am new to sage and my question can be naive, I apologize in advance. I would like to use sage for some symbolic integration operations to avoid human mistake. Here is my first test (which is going to important for my application).

Integration of Normal distribution from -infinity to +infinity. Obviously the result should be one and I am wondering why I don't get it:


x, sigma, mu = var('x','sigma','mu')

q = 1/((2*pi)^(1/2)*sigma)*exp(-(x - mu)^2/(2*sigma^2))

show(q)

import sympy

show((sympy.integrate(q, (x,-sympy.oo, sympy.oo))).simplify())

Here is the result:

Integral(2**(1/2)*exp(-mu**2/(2*sigma**2))*exp(-x**2/(2*sigma**2))*exp(mu*x/sigma**2)/(2*pi**(1/2)*sigma),(x,-oo,oo))

I am wondering why I don't get 1? Am I missing anything?

Thanks

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answered 2012-09-22 09:36:35 -0600

achrzesz gravatar image
sage: x, sigma, mu = var('x','sigma','mu')                   
sage: q = 1/((2*pi)^(1/2)*sigma)*exp(-(x - mu)^2/(2*sigma^2))
sage: assume(sigma>0)                                        
sage: integral(q,(x,-oo,oo))                                     
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Thank you :) It is true that sign of sigma matters and what you said works and it yields "1". But when I type: integral(-q*log(q),(x,-oo,oo)) to compute entropy. it is still confused about sign of mu ?! even after assuming mu>0 produces stack overflow!

kayhan gravatar imagekayhan ( 2012-09-22 13:25:18 -0600 )edit

Notice that @achrzesz used Maxima's integration (Sage's default integral mode) and not Sympy. Apparently sympy can't do the integral in question, perhaps because of the lack of an assumption framework, who knows...

kcrisman gravatar imagekcrisman ( 2012-09-22 16:09:54 -0600 )edit

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Asked: 2012-09-22 08:45:33 -0600

Seen: 276 times

Last updated: Sep 22 '12