Is this function, defined by a limit, differentiable or at least continuous on [0, 1]?

Question

This question was inspired by the one at http://answers.yahoo.com/question/index;_ylt=ApfyiAkLxa3nBMvRv5dBST3sy6IX?qid=20070816104652AAunj7P

For x >=0, let f(x) = lim [1/1^x ....+1/n^x - Integral (1 to oo) dt/t^x]. Then f(x) = lim [1/1^x ....+1/n^x - (n^(1-x) -1)/(1 -x)], if x<>1, and f(1) = lim [1/1 ....+1/n - ln(n)] , so that f(1) is the Euler/Mascheroni constant. Since the function u --> 1/u^x is strictly decreasing for x>0, the limit of the sequence always exist and f is well defined. For x >1, 1- x <0, both the series and the integral converge and we have f(x) =lim [1/1^x ....+1/n^x - 1/(x -1)] = Z(x) - 1/(x -1), where Z is the Riemann Zeta function. We know Z is analytic on the complex plane, so has derivatives of all orders on (1, oo). Therefore, , f has derivatives of all orders on this set and f'(x) = Z'(x) + 1/(x-1)^2. But I couldn't conclude if f is differentiable, or at least continuous, on [0,1].

Can we come to any conclusion about the signa of f', so that we can find out where f is increasing or decreasing?

We know Z'(x) <0 for x >1, because Z is strictly decreasing on (1, oo)

I tried to represent f' as the limit of a series of functions g_n(x) = 1/1^x ...+1/n^x - Integral 1^oo dt/t^x. We see g_n is differentiable on [0, 1), and g'_n converges to some function h. If this convergence were uniform, then we'd have f' = h, but I couldn't prove this is uniform

Oh a mistake, Z is analytic on the part of Argand Gauss plane where the real part is >1. But this is enough for our purposes;

Answer 1

What you have is really two limits. An indefinite integral is itself a limit. So clear up some notation:

f(x) = lim [ Σ (1/n^x) - ∫ (dn / n^x ) ]

We know f(x) exists (see the other question).

First of all, it's definitely continuous almost everywhere, because it is of bounded variation. We know it has BV by Jordan's lemma (any diff of monotone functions has BV).

Is it continuous? Let's see:

Take lim to mean the lim in f(x) and LIM to mean lim (x→a).

LIM | f(x) - f(a) |

= LIM [ lim [ Σ (1/n^x) - ∫ (dn / n^x ) ]
- lim [ Σ (1/n^a) - ∫ (dn / n^a ) ] ]

= LIM lim [ Σ (1/n^x - 1/n^a) - ∫ (dn / n^x - dn / n^a ) ]

= lim LIM [ Σ (1/n^x - 1/n^a) - ∫ (dn / n^x - dn / n^a ) ]

= lim 0 = 0

The nice things here are:
1) you can subtract those sums/integrals because they all have positive terms (ie converge absolutely).
2) there's never any need to worry about the fact that the sums/limits on their own do not converge as N increases.
3) we can swap the limits because we know that either exists if we know convergence on N→∞ is uniform.

Is it? Well, let's think about that without muddling through any algebra. Just look at a picture:
http://ltcconline.net/greenl/courses/107/Series/inttes1.gif

Regardless of the exact shape, what we are summing is clearly bounded by the area in those little purple boxes. This area itself is bounded by the little purple boxes for the specific case x=0 - This means x=0 is the bounding case (gives maximal ε > sum of the boxes you didn't count yet, between N and ∞) for uniform convergence. I know that's not totally rigorous, but it's true and I think I've explained it well enough for Yahoo! Answers.

So f(x) is continuous. It is also differentiable, but there's so much computation going on, I won't even try to type it out here! It involve basically reducing the integral to a limit/sum you can manipulate. It's not so bad if you presume that the harmonic number function is differentiable - the integral part of f(x) is differentiable too of course. This has already been studied extensively, though. See:
http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant#Generalizations

These are called Euler's generalized constants.

Here's a plot of the function f(x)
http://math.colgate.edu/~kellen/interspace/fx.gif