Integration of fractional part function

Question

Does the integral $\displaystyle\int_{0}^{1}\{\frac{1}{x^n}\}dx$ exist? where $n\in \mathbb{Z}^+$ and $n\geq 2$, $\{x\} $is the fractional part. As is known, when $n=1$, the integral equals $1-\gamma$. here .what about the case $n\geq 2?$ I am wondering whether the following is right:

\begin{align} &\quad\ \int^1_0\left\{\frac{1}{x^n}\right\}\mathrm dx\\ &\overset{x\to\frac1 x}{=}\int^\infty_1\frac{\{x^n\}}{x^2}\mathrm dx\\ &\overset{x^n\to x}{=}\frac1n\int^\infty_1\frac{\{x\}}{x^{\frac1n+1}}\mathrm dx\\ &=\frac1n\sum_{k\geq 1}\int^{k+1}_k\frac{\{x\}}{x^{\frac1n+1}}\mathrm dx\\ &=\frac1n\sum_{k\geq 1}\int^{k+1}_k\frac{x-k}{x^{\frac 1n+1}}\mathrm dx\\ &\overset{x-k\to x}{=}\frac1n\sum_{k\geq 1}\int^1_0\frac{x\mathrm dx}{(x+k)^{\frac1n+1}}\\ &=\frac1{n-1}\sum_{k\geq 1}\left[\frac{kn+1}{(1+k)^{\frac1n}}-\frac{n}{k^{\frac1n-1}}\right]\\ &=\frac1{n-1}\lim_{N\to \infty}\sum^N_{k=1}\left[ \frac{n}{(1+k)^{\frac1n-1}}+\frac{1}{(1+k)^{\frac1n}}-\frac{n}{k^{\frac1n-1}}\right]\\ &=\frac1{n-1}\lim_{N\to \infty} n(N+1)^{1-\frac1n}-n+ \frac1{n-1}\lim_{N\to \infty}\sum^N_{k=1}\left[ \frac{1}{(1+k)^{\frac1n}}\right]\\ &=+\infty \end{align} Accurately， are the last two steps correct? Any help is appreciated, thanks!

Answer 1

Your last two steps appear to be incorrect to me. Here's what I am getting.

$$ \int_0^1 \left\{ \frac{1}{x^n} \right\} dx $$ Let $x = \dfrac{1}{u}$ $$ = \int_1^\infty \frac{\left\{ u^n \right\}}{u^2} du $$ Let $t = u^n$ $$ = \frac{1}{n} \int_1^\infty \frac{\left\{ t \right\}}{t^{1 + 1/n}} dt $$

$$ = \frac{1}{n} \sum_{k=1}^\infty \int_k^{k+1} \frac{t - k}{t^{1 + 1/n}} dt $$

$$ = \frac{1}{n} \sum_{k=1}^\infty \int_0^1 \frac{x}{(x + k)^{1 + 1/n}} dx $$

$$ \int_0^1 \frac{x}{(x + k)^{1 + 1/n}} dx = \frac{n}{n - 1} \left[ (1 + k)^{1 - 1/n} - k^{1 - 1/n} \right] + n k (1 + k)^{-1/n} - n k^{1 - 1/n} $$

$$ \int_0^1 \left\{ \frac{1}{x^n} \right\} dx = \frac{1}{n} \sum_{k=1}^\infty \left( \frac{n}{n - 1} \left[ (1 + k)^{1 - 1/n} - k^{1 - 1/n} \right] + n k (1 + k)^{-1/n} - n k^{1 - 1/n} \right) $$

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Testing convergence For large $k$, we can approximate: $(x + k)^{1 + 1/n} \approx k^{1 + 1/n}$ (since $x \in [0,1]$), so:

$$ \int_0^1 \frac{x}{(x + k)^{1 + 1/n}} dx \approx \int_0^1 \frac{x}{k^{1 + 1/n}} dx = \frac{1}{k^{1 + 1/n}} \int_0^1 x \, dx = \frac{1}{2 k^{1 + 1/n}}. $$

Since $n \geq 2$, $1/n \leq 1/2$, so $1 + 1/n \geq 3/2 > 1$. The series $\sum_{k=1}^\infty \frac{1}{k^{1 + 1/n}}$ converges (p-series with $p = 1 + 1/n > 1$). More precisely:

$$ \frac{x}{(x + k)^{1 + 1/n}} \leq \frac{x}{k^{1 + 1/n}}, $$

since $(x + k)^{1 + 1/n} \geq k^{1 + 1/n}$, so:

$$ \int_0^1 \frac{x}{(x + k)^{1 + 1/n}} dx \leq \frac{1}{2 k^{1 + 1/n}}, $$

and $\frac{1}{n} \sum_{k=1}^\infty \int_0^1 \frac{x}{(x + k)^{1 + 1/n}} dx < \infty$. Thus, the integral converges.

Answer 2

Thanks to Random. The integral really converges. By Euler-Maclaurin formula: (here) $$\sum_{i=1}^n\frac1{i^\beta}=\frac{n^{1-\beta}}{1-\beta}+\zeta(\beta)+\frac1{2n^\beta}-\frac\beta{12n^{\beta+1}}+\mathcal O\left(\frac1{n^{\beta+3}}\right),$$ it follows that $$\displaystyle \zeta(s)=\lim _{k \rightarrow \infty} (\sum_{n=1}^{k-1} \frac{1}{n^{s}}-\frac{k^{1-s}}{1-s}).$$

where $\zeta(s)$ is the zeta function, $Re(s)>0.$ Specifically，

\begin{align} &\quad\ \int^1_0\left\{\frac{1}{x^n}\right\}\mathrm dx\\ &\overset{x\to\frac1 x}{=}\int^\infty_1\frac{\{x^n\}}{x^2}\mathrm dx\\ &\overset{x^n\to x}{=}\frac1n\int^\infty_1\frac{\{x\}}{x^{\frac1n+1}}\mathrm dx\\ &=\frac1n\sum_{k\geq 1}\int^{k+1}_k\frac{\{x\}}{x^{\frac1n+1}}\mathrm dx\\ &=\frac1n\sum_{k\geq 1}\int^{k+1}_k\frac{x-k}{x^{\frac 1n+1}}\mathrm dx\\ &\overset{x-k\to x}{=}\frac1n\sum_{k\geq 1}\int^1_0\frac{x\mathrm dx}{(x+k)^{\frac1n+1}}\\ &=\frac1{n-1}\sum_{k\geq 1}\left[\frac{kn+1}{(1+k)^{\frac1n}}-\frac{n}{k^{\frac1n-1}}\right]\\ &=\frac1{n-1}\lim_{N\to \infty}\sum^N_{k=1}\left[ \frac{n}{(1+k)^{\frac1n-1}}-\frac{n-1}{(1+k)^{\frac1n}}-\frac{n}{k^{\frac1n-1}}\right]\\ &=\frac1{n-1}\lim_{N\to \infty} \left[n(N+1)^{1-\frac1n}-n\right]-\lim_{N\to \infty}\sum^N_{k=1}\left[ \frac{1}{(1+k)^{\frac1n}}\right]\\ &=\frac{n}{1-n}+\frac{n}{n-1}\lim_{N\to \infty} (N+1)^{1-\frac1n}-(H(\frac 1n)-1)\\ &=\frac{1}{1-n}+\frac{n}{n-1}\lim_{N\to \infty} (N+1)^{1-\frac1n}-H(\frac 1n)\\ &=\frac{1}{1-n}-\zeta(\frac 1n) \end{align}

where $\displaystyle H(s)=\lim_{N\to \infty}\sum^N_{k=1} \frac{1}{k^{s}}=\sum^{\infty}_{k=1} \frac{1}{k^{s}}$.

Comment: Meanwhile, the argument above is valid for any positive real number $n$ except $n=1$. When $n\to 1$, $\displaystyle \frac{1}{1-n}-\zeta(\frac 1n)\to 1-\gamma$.