Matematické Fórum

Marian · 08. 11. 2017 14:58 — Editoval Marian (08. 11. 2017 15:00)

Calculate the definite integral

$\boldsymbol{I_k:=\int_{0}^{1}\{\frac{k}{\sqrt[k]{x}}\}\mathrm dx},\qquad k\ge 2,\quad k\in\mathbb N,$

where {z} denotes the fractional part of the real number z.

Also, evaluate the limit

$\boldsymbol{\lim_{k\to\infty}I_k}.$

Detailed solution required...

laszky · 08. 04. 2019 23:56 — Editoval laszky (09. 04. 2019 02:11)

↑ Marian:

Hi,

The inequality $p\leq \frac{k}{\sqrt[k]{x}} \leq p+1$ is equivalent to $\left(\frac{k}{p+1}\right)^k \leq x \leq \left(\frac{k}{p}\right)^k$ , hence

$\int_0^1 \{\frac{k}{\sqrt[k]{x}} \}\,\mathrm{d}x = \sum_{p=k}^{\infty}\int_{\left(\frac{k}{p+1}\right)^k}^{\left(\frac{k}{p}\right)^k} \frac{k}{\sqrt[k]{x}} -p\, \mathrm{d}x = \int_0^1\frac{k}{\sqrt[k]{x}}\,\mathrm{d}x - \sum_{p=k}^{\infty}p \left(\left(\frac{k}{p}\right)^k-\left(\frac{k}{p+1}\right)^k\right) =$
$= k\left[\frac{x^{\frac{k-1}{k}}}{\frac{k-1}{k}}\right]_0^1 - k^k \sum_{p=k}^{\infty}\left(\frac{1}{p^{k-1}} -\frac{1}{(p+1)^{k-1}}+ \frac{1}{(p+1)^k}\right) = \frac{k^2}{k-1} - k^k \left(\frac{1}{k^{k-1}} + \sum_{p=k}^{\infty}\frac{1}{(p+1)^k}\right) =$
$= \frac{k^2-k^2+k}{k-1} - k^k\sum_{p=k}^{\infty}\frac{1}{(p+1)^k} = \frac{k}{k-1}-k^k \zeta(k,k+1)$

with Hurwitz zeta function $\zeta(x,a)$ .

By computer calculations, the limit $\lim_{k\to\infty}I_k$ seems to be close to 0.418, but I did not yet verify it rigorously (if it is even possible).

Marian · 09. 04. 2019 14:12

It is possible to find the exact value of the limit investigated. It is not so complicated to find it. Try to rewrite the Hurwitz function and work with the infinite series that is multiplied by k^k.

Let yourself be surprised...

laszky · 10. 04. 2019 10:15

↑ Marian:

Thank you for that hint.

Since for the i-th term of the sum $\sum_{i=1}^{\infty}\left(\frac{k}{k+i}\right)^k$ there holds

$\lim_{k\to\infty}\left(\frac{k}{k+i}\right)^k = \lim_{k\to\infty}\left(1-\frac{i}{k+i}\right)^{k+i}\left(1-\frac{i}{k+i}\right)^{-i}=\mathrm{e}^{-i}$ , there is

$\lim_{k\to\infty} I_k\;=\;\lim_{k\to\infty} \frac{k}{k-1}-\sum_{p=k}^{\infty}\left(\frac{k}{p+1}\right)^k\;=\;\lim_{k\to\infty} \frac{k}{k-1}-\sum_{i=1}^{\infty}\left(\frac{k}{k+i}\right)^k\;=$
$=\;1-\sum_{i=1}^{\infty}\mathrm{e}^{-i}\;=\;1-\frac{\frac{1}{\mathrm{e}}}{1-\frac{1}{\mathrm{e}}}\;=\;\frac{\mathrm{e}-2}{\mathrm{e}-1}\;\approx\;0.418$

Marian · 10. 04. 2019 10:37

Now that's what I call fine job. Nice.

Matematické Fórum

#1 08. 11. 2017 14:58 — Editoval Marian (08. 11. 2017 15:00)

Fractional part integral

#2 08. 04. 2019 23:56 — Editoval laszky (09. 04. 2019 02:11)

Re: Fractional part integral

#3 09. 04. 2019 14:12

Re: Fractional part integral

#4 10. 04. 2019 10:15

Re: Fractional part integral

#5 10. 04. 2019 10:37

Re: Fractional part integral

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