Justify an approximation of $\int_{0}^1\binom{ \left\{ 1/z\right\}}{z}dz$, where $ \left\{ z \right\}$ denotes the fractional part function

Question

I'm interested in to know an approximation for this integral

$$\int_{0}^1\binom{ \left\{ \frac{1}{z} \right\}}{z}dz,\tag{1}$$ that I've created this morning. Here $ \left\{ x \right\}$ denotes the fractional part function. And here is the plot, showed by Wolfram Alpha online calculator from which I've created the integral using

plot Binom[frac(1/x),x], from x=0 to 1

Then I know (if there are no mistakes in my change of variable) $$\int_{0}^1\binom{ \left\{ \frac{1}{z} \right\}}{z}dz=\sum_{k=1}^\infty\int_k^{k+1}\frac{\Gamma(-k+u+1)}{\Gamma(1+1/u)\Gamma(-k+u+1-1/u)}\frac{du}{u^2}.\tag{2}$$ And then my idea is try to perform an approximation of this gamma factors, integrate and take an approximation of the series.

Question. How do you calculate and justify an approximation of the integral $$\int_{0}^1\binom{ \left\{ \frac{1}{z} \right\}}{z}dz\,?$$ Many thanks.

Answer 1

It can be shown that the partial sums have an asymptotic expansion of the form

$$ \sum_{k=1}^n\int_k^{k+1}\cdots\,du \approx c_0 + \frac{c_1}{n} + \frac{c_2}{n^2} + \cdots, \tag{$*$} $$

where

$$ c_0 := \sum_{k=1}^\infty\int_k^{k+1}\cdots\,du. $$

This tells us that the convergence of the partial sums to $c_0$ is very slow; to naively compute $10$ digits of $c_0$ we might need to evaluate something like the $10^{10}\text{th}$ partial sum, which is just not feasible.

We can instead try some sort of convergence acceleration. Sequences with this kind of asymptotic expansion are often well-suited to Richardson extrapolation, and it ends up working pretty well here.

Using Mathematica we can calculate the first $50$ partial sums to roughly $100$ digits relatively quickly with the command

a[k_] := a[k] = NIntegrate[ Gamma[-k + u + 1]/(Gamma[1 + 1/u] Gamma[-k + u + 1 - 1/u] u^2), {u, k, k + 1}, WorkingPrecision -> 100]; Subscript[T, 0] = Table[Sum[a[k], {k, 1, n}], {n, 1, 50}];

Now $T_0$ is a list containing the numerical values of the first $50$ partial sums. We then iteratively compute lists $T_1, T_2, \ldots, T_{49}$, each of which is an accelerated version of the previous one, using the formula

$$ T_{k+1}(n) := \frac{x_n T_k(n+1) - x_{n+k+1}T_k(n)}{x_n - x_{n+k+1}} \qquad (1 \leq n \leq 50 - k - 1), $$

where $x_j$ is a sequence of constants. Because the asymptotic expansion $(*)$ involves only powers of $n$, we take $x_j = 1/j$.

The Mathematica code is

x[j_] := 1/j; For[k = 0, k <= Length[Subscript[T, 0]] - 2, k++, Subscript[T, k + 1] = Table[(x[j] Subscript[T, k][[j + 1]] - x[j + k + 1] Subscript[T, k][[j]])/(x[j] - x[j + k + 1]), {j, 1, Length[Subscript[T, 0]] - k - 1}] ]

The content of the last three lists is now

T_47 = {0.82738000080545448263010087550271301001436936357076159313066822525902388323, 0.82738000080545448263010087550271301001436952134826362854767855646955812112, 0.8273800008054544826301008755027130100143695297480234033866397050979721660} T_48 = {0.82738000080545448263010087550271301001436952463529492095219960503644425107, 0.8273800008054544826301008755027130100143695300980133940049297529574894179} T_49 = {0.8273800008054544826301008755027130100143695302094974444753936335273066662}

Based on these computations, I suggest that

$$ \int_{0}^1\binom{ \left\{ \frac{1}{z} \right\}}{z}dz \doteq 0.82738\ 00008\ 05454\ 48263\ 01008\ 75502\ 71301\ 00143\ 69530. $$

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I learned this version of the Richardson extrapolation from section 1.5 of the book Scientific Computation on Mathematical Problems and Conjectures by Richard S. Varga. https://epubs.siam.org/doi/book/10.1137/1.9781611970111