# Revision history [back]

another try:

put the functions in a python dictionary with the respective intervals as keys.

run through the sorted keys and add up the integrals for each fitting interval

from scipy.constants import h, c, k

a=11717

f1(x) = a*(x**-0.53)
f2(x) = a*(x**-0.75)

T = {(1,10000):f1,(10000,20000):f2, (20000,50000):f1}

# Blackbody Planky function
def B(Lambda,Temp):
return 2*h*c**2/(Lambda**5 *(exp(h*c/(Lambda*k*Temp))-1))

a1 = 300; a2 = 22000  # interval

def flux2(Lambda):
sum = 0
L = copy(T.keys())
L.sort()
for k in L:
if k[1] <= a1: continue
if k[0] <= a1 and a2 <= k[1]:
print 'one: key=',k, 'interval from', a1, 'to', a2
return numerical_integral(2*pi*x*B( Lambda,T[k]),a1,a2)[0]
if k[0] <= a1:
print 'two: key=',k, 'interval from', a1, 'to', k[1]
sum += numerical_integral(2*pi*x*B( Lambda,T[k]),a1,k[1])[0]
continue
if a1 <= k[0] and k[1] <= a2:
print 'three: key=',k, 'interval from', k[0], 'to', k[1]
sum += numerical_integral(2*pi*x*B( Lambda,T[k]),k[0],k[1])[0]
continue
if k[0] <= a2 <= k[1]:
print 'four: key=',k, 'interval from', k[0], 'to', a2
sum += numerical_integral(2*pi*x*B( Lambda,T[k]),k[0],a2)[0]
if a1 >= k[1]:
pass
return sum

print 2.5*log(flux2(9000*10**-10))