Computing definite integrals in python

2
votes

I'm trying to write a loop that calculates the value of a definite integral at each step. The function bigF is very complicated. To put it in simple terms, it integrates a bunch of terms with respect to s, from s=tn-(n/2) to s=tn+(n/2). After the integration, bigF still has a variable t. So you can say bigF(t) = integral(f(s,t)), where f(s,t) is the big mess of terms after integrate.integ. In the last line, I want to evaluate bigF(t) at t=tn after bigF computes the integral of f(s,t)

After running, I get the error global name 's' is not defined. But s was meant to be just a dummy variable in the integration, since I am computing a convolution. What do I need to do?

import numpy as np  
import scipy.integrate as integ 
import math 

nt=5001#; %since (50-0)/.01 = 5000
dt = .01#; % =H
H=.01

theta_n = np.ones(nt)
theta_n[1]=0#; %theta_o
omega_n = np.ones(nt)
omega_n[1]=-0.4# %omega_o
epsilon=10^(-6)
eta = epsilon*10
t_o=0

def bigF(t, n):
    return integrate.integ((422.11/eta)*math.exp((5*(4*((eta*t-s-tn)^2)/eta^2)-1)^(-1))*omega, s,tn-(n/2),tn+(n/2))

for n in range(1,4999)
    tn=t_o+n*dt;
    theta_n[n+1] = theta_n[n] + H*bigF(tn, n);
1
pythonintegrationintegral
^ in Python means XOR, not "power". This is a serious bug in several places in your code. 1^1 is zero, much to your surprise. - John Zwinck
I'm guessing that you meant to pass a function to integ, not a computed value. So you should probably put lambda s: in front of your argument to integ. - happydave

1 Answers

0
votes

If you're doing a convolution, sounds like you want numpy.convolve.