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.

  • 1
    $\begingroup$ The integrals won't be a problem, the integrands are sufficiently well-behaved. Approximating the series will be the hard part, because convergence is slow (like $\sum k^{-2}$, obviously. So you'll have to come up with some convergence acceleration. What do you need that value for to justify such an effort? $\endgroup$
    – user436658
    Jan 18, 2018 at 13:18
  • $\begingroup$ This morning I was playing with the mentioned CAS when I've defined the integral, I did not know this integral, and I thought it was interesting to ask here for it @ProfessorVector $\endgroup$
    – user243301
    Jan 18, 2018 at 13:23
  • $\begingroup$ Why not use some kind of quadrature rule to approximate the integral? $\endgroup$ Jan 18, 2018 at 22:58
  • $\begingroup$ I did not think about it, because my belief was that through my strategy some approximation was possible. Since the binomials are defined in terms of gamma factors, maybe is possible to do an approximation of these ins the integrand. Many thanks @AntonioVargas . $\endgroup$
    – user243301
    Jan 19, 2018 at 12:03
  • $\begingroup$ And I am trying to define different integrals involving these binomials and the floor function. I am saying this, if you want to study/explore it in your home, because I don't know if these integrals were in the liteature. I know that Ovidiu Furdui wrote an ocean of formulas involving the fractional part function, and I wanted to emulate him now with binomials, but it seems that my integrals do not make nice pieces of math @AntonioVargas $\endgroup$
    – user243301
    Jan 19, 2018 at 18:33

1 Answer 1


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

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


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

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

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

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

  • 1
    $\begingroup$ The inverse symbolic calculator doesn't find a match for this number. $\endgroup$ Mar 15, 2018 at 23:40
  • $\begingroup$ Many thanks, I am going to study it. $\endgroup$
    – user243301
    Mar 16, 2018 at 9:25

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