Two Integrals
January 11, 2011
The exponential integral appears frequently in the study of physics, and the related logarithmic integral appears both in physics and in number theory. With the Euler-Mascheroni constant γ = 0.5772156649015328606065, formulas for computing the exponential and logarithmic integral are:
Since there is a singularity at Li(1) = −∞, the logarithmic integral is often given in an offset form with the integral starting at 2 instead of 0; the two forms of the logarithmic integral are related by Lioffset(x) = Li(x) – Li(2) = Li(x) – 1.04516378011749278. It is this form that we are most interested in, because the offset logarithmic integral is a good approximation of the prime counting function π(x), which computes the number of primes less than or equal to x:
| x | 106 | 1021 |
| Lioffset(x) | 78627 | 21127269486616126182 |
| π(x) | 78498 | 21127269486018731928 |
If you read the mathematical literature, you should be aware that there is some notational confusion about the two forms of the logarithmic integral: some authors use Li for the logarithmic integral and li for its offset variant, other authors turn that convention around, and still other authors use either notation in either (or both!) contexts. The good news is that in most cases it doesn’t matter which variant you choose.
Your task is to write functions that compute the exponential integral and the two forms of the logarithmic integral. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
[…] today’s Programming Praxis exercise, our task is to write functions to calculate the exponential and […]
My Haskell solution (see http://bonsaicode.wordpress.com/2011/01/11/programming-praxis-two-integrals/ for a version with comments):
ei :: Double -> Double ei x = 0.5772156649015328606065 + log x + sum (takeWhile (> 1e-17) [x**k / k / product [1..k] | k <- [1..]]) li :: Double -> Double li = ei . log liOffset :: Double -> Double liOffset x = li x - li 2My Python solution. This blog and my free time studies are drawing me more and
more towards Scheme and Haskell, but since there are two great solutions in
those languages already I felt I should offer a solution in a different
language. I’ve moved towards the newer “format” instead of the older printf
style string formatting.
#!/usr/bin/env python from __future__ import division import math GAMMA = 0.5772156649015328606065 def exp_int(x): s = GAMMA + math.log(x) term, k, f = x, 1, 1 while term > 1e-17: s += term k += 1 f *= k term = pow(x, k) / (k * f) return s def log_int(x): return exp_int(math.log(x)) def offset_log_int(x): return log_int(x) - 1.04516378011749278 if __name__ == "__main__": print "Li_offset(1e6) = {0:d}".format(int(round(offset_log_int(1e6)))) print "Li_offset(1e21) = {0:d}".format(int(round(offset_log_int(1e21)))) # Output: # Li_offset(1e6) = 78627 # Li_offset(1e21) = 21127269486616088576this is the best blog on the whole internet. thank you!
Why 1e-17 is chosen as the stopping point in all of your code?