Looking at the attached worksheet,
View 565_int vs sum for mrb.mw on MapleNet or Download 565_int vs sum for mrb.mw
View file details, it appears that the absolute value, minus 1/2, of the integral of (-1)^x*x^(1/x) from 1 to infinity would equal the partial sum of (-1)^x*x^(1/x) from 1 to where the upper summation is even and growing without bound [0]. Is anyone interested in improving or disproving this conjecture?
[0] Google "MRB Constant"
Comments
Undefined
The function f(x) = (-1)^x * x^(1/x) = exp(I*Pi*x + ln(x)/x) is not integrable on [1,infinity): note that f(x) ~ exp(I*Pi*x) as x -> infinity. What does have a limit as N -> infinity (for integers N) is
int(f(x), x = 1 .. 2*N) . Similarly, sum(f(n),n=1..2*N) has a limit as N -> infinity.
Point taken
For x>0and f(x) = (-1)^x * x^(1/x) , the next document, View 565_part 2 sum vs int.mw on MapleNet or Download 565_part 2 sum vs int.mw
View file details , shows the difference of and the ratio of abs(int(f(x), x = 1 .. 2*N))-1/2 and sum(f(n),n=1..2*N) as integer N -> infinity. The difference seems to go to 0 and the ratio seems to go to 1. However, it would be nice to have a proof of some sort to show that they both converge upon the same number. I have given a simple geometric example of the summing of f in the following: http://marvinrayburns.com/what_is_mrb.mht. Perhaps a similar example of the integrating of f could help make the proof -- that is if it is true.
not true
My conjecture is not true. See the following link for some bounds of the value.
http://marvinrayburns.com/latest.html
Some research done
It appears that RICHARD J. MATHAR has written a paper on this subject of this subject of Int(f(x)) where f(x) = (-1)^x * x^(1/x) = exp(I*Pi*x + ln(x)/x). arxiv1.library.cornell.edu/PS_cache/arxiv/pdf/0912/0912.3844v1.pdf