Calculating a difficult sum

CRGreathouse · 2009-08-25, 03:12

What are good ways to numerically calculate integrals? Background (skip, skim, or read depending on your patience):

I was trying to estimate the value of a difficult arithmetic sum recently and I'm not quite sure how to go about it.

The summands are ~ 1/xlog^2 x and so clearly converge, but a finite sum as far as I was able to manage (a few million terms, though I could reach a billion with a bit more time) aren't enough to get a value I can trust to even two decimal places, and it's surely meaningless beyond four.

The obvious next step was to take the sum as far as I cared to go and use an integral for the rest. Fortunately the terms are just well-behaved enough that I have viable upper and lower bounds. With some effort I managed to generate three integrals that give upper and lower bounds (the lower bound is two piecewise integrals).

So I typed all this up in my programming language of choice and wrote a wrapper function that let me set precision and how far to set the sum/integral cutoff. It was only then that I realized that the integral was apparently rather numerically unstable -- even though the bounding functions were smooth in the domain of consideration. Every time I raised the precision I got a new answer, to the limit of my patience.

My best though so far was to subtract the main term 1/xlog^2 x out of the integral, since it has closed-form integral -1/log x. But this hasn't helped much so far.

I'm just looking for general strategies, either better ways to calculate the integrals or a different approach to the sum. Because the sum includes primes it's a pain to work with numerically... but Pierre Dusart's bounds are tight enough at large numbers that I think I can make the integral approach work. But first I have to know that what I think is the integral is, in fact, the integral!

flouran · 2009-08-25, 04:50

Why not evaluate the integrals formulaically using an implementation of Risch's algorithm (although Cherry's algorithmic extension to Rich's result seems more practical; but, as you know, that's not saying much

), and then evaluate the resulting formula numerically at the points you want?

If I interpret Cherry's paper correctly (which you gave me 4 months ago), he does implement his algorithm in Macsyma.

Is Mathematica able to compute this integral (I know, it's a dumb question; but it's worth a shot)?

CRGreathouse · 2009-08-25, 05:05

Neither Mathematica nor Maxima can evaluate the integral in closed form. That's not surprising: most integrals can't be, and I wouldn't have posted if it could have been.

wblipp · 2009-08-25, 14:11

Sometimes a change of variable in the integral helps. The integral convergence may be the same problem as the summation convergence - the integrand decreases so slowly that you need to keep extending the range of integration. The idea is to get a new integration variable that rapidly covers ranges of the old integration variable. Perhaps z = log x.

William

2009-08-25, 03:12	#1
CRGreathouse Aug 2006 5,987 Posts	Calculating a difficult sum What are good ways to numerically calculate integrals? Background (skip, skim, or read depending on your patience): I was trying to estimate the value of a difficult arithmetic sum recently and I'm not quite sure how to go about it. The summands are ~ 1/xlog^2 x and so clearly converge, but a finite sum as far as I was able to manage (a few million terms, though I could reach a billion with a bit more time) aren't enough to get a value I can trust to even two decimal places, and it's surely meaningless beyond four. The obvious next step was to take the sum as far as I cared to go and use an integral for the rest. Fortunately the terms are just well-behaved enough that I have viable upper and lower bounds. With some effort I managed to generate three integrals that give upper and lower bounds (the lower bound is two piecewise integrals). So I typed all this up in my programming language of choice and wrote a wrapper function that let me set precision and how far to set the sum/integral cutoff. It was only then that I realized that the integral was apparently rather numerically unstable -- even though the bounding functions were smooth in the domain of consideration. Every time I raised the precision I got a new answer, to the limit of my patience. My best though so far was to subtract the main term 1/xlog^2 x out of the integral, since it has closed-form integral -1/log x. But this hasn't helped much so far. I'm just looking for general strategies, either better ways to calculate the integrals or a different approach to the sum. Because the sum includes primes it's a pain to work with numerically... but Pierre Dusart's bounds are tight enough at large numbers that I think I can make the integral approach work. But first I have to know that what I think is the integral is, in fact, the integral! Last fiddled with by CRGreathouse on 2009-08-25 at 03:12

2009-08-25, 04:50	#2
flouran Dec 2008 7²×17 Posts	Why not evaluate the integrals formulaically using an implementation of Risch's algorithm (although Cherry's algorithmic extension to Rich's result seems more practical; but, as you know, that's not saying much ), and then evaluate the resulting formula numerically at the points you want? If I interpret Cherry's paper correctly (which you gave me 4 months ago), he does implement his algorithm in Macsyma. Is Mathematica able to compute this integral (I know, it's a dumb question; but it's worth a shot)? Last fiddled with by flouran on 2009-08-25 at 04:52

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
mprime seems difficult/buggy	dchmelik	Software	3	2017-10-25 06:33
A rather difficult lottery	Oddball	Lounge	2	2010-05-06 02:18
Why is RH so difficult to prove?	Damian	Math	31	2008-10-03 02:11
Difficult Differential	Unregistered	Homework Help	9	2008-10-01 21:24
Geometry puzzle(difficult)	hyh1048576	Puzzles	6	2003-07-27 06:46

2009-08-25, 05:05	#3
CRGreathouse Aug 2006 5987₁₀ Posts	Neither Mathematica nor Maxima can evaluate the integral in closed form. That's not surprising: most integrals can't be, and I wouldn't have posted if it could have been.

2009-08-25, 14:11	#4
wblipp "William" May 2003 Near Grandkid 3×7×113 Posts	Sometimes a change of variable in the integral helps. The integral convergence may be the same problem as the summation convergence - the integrand decreases so slowly that you need to keep extending the range of integration. The idea is to get a new integration variable that rapidly covers ranges of the old integration variable. Perhaps z = log x. William