Integral with fractional part

[Tolaso J Kos]

Let $n \in \mathbb{N}$ and let $\{ \cdot \}$ denote the fractional part. Prove that:

$$\int_0^1 x^n \left\{ \frac{1}{x} \right\}^n \, {\rm d}x = 1- \frac{1}{n+1} \sum_{k=2}^{n+1} \zeta(k)$$

where $\zeta$ is the Riemann zeta function.

[galactus]

Hey T. Heck, I think this one is more involved than that triple one I managed to get sometime back.

Anyway, I think I have it. Though, I may have to come back to finish posting.

$$\int_{0}^{1}x^{n}\left\{\frac{1}{x}\right\}^{n}dx$$

Let $x=1/u, \;\ dx=\frac{-1}{u^{2}}du$

$$\begin{align*}\int_{1}^{\infty}\frac{\left\{u\right\}^{n}}{u^{n+2}}du

=\sum_{k=1}^{\infty}\int_{k}^{k+1}\frac{\left\{u\right\}^{n}}{u^{n+2}}du

=\sum_{k=1}^{\infty}\int_{k}^{k+1}\frac{(u-k)^{n}}{u^{n+2}}du\end{align*}$$

Let $u=t+k, \;\ du=dt$

$$\begin{align}\sum_{k=1}^{\infty}\int_{0}^{1}\frac{t^{n}}{(t+k)^{n+2}}dt

=\int_{0}^{1}t^{n}\sum_{k=1}^{\infty}\frac{1}{(t+k)^{n+2}}dt

=\int_{0}^{1}t^{n}\cdot \frac{1}{(n+1)!}\int_{0}^{\infty}\frac{w^{n+1}e^{-wt}}{e^{w}-1}dwdt\end{align}$$

$$=\frac{1}{(n+1)!}\int_{0}^{\infty}\frac{w^{n+1}}{e^{w}-1}dw\int_{0}^{1}t^{n}e^{-tw}dt$$

Note that $$\int_{0}^{\infty}\frac{w^{n+1}}{e^{w}-1}dw=\Gamma(n+2)\zeta(n+2)$$

Now, gotta try and finish this up.

[r9m]

$$=\frac{1}{(n+1)!}\int_{0}^{\infty}\frac{w^{n+1}}{e^{w}-1}dw\int_{0}^{1}t^{n}e^{-tw}dt$$

We can rewrite the inner integral as: \begin{align*}\int_0^{1} t^ne^{-wt}\,dt &= (-1)^n\frac{d^n}{dw^n}\left(\int_0^1 e^{-wt}\,dt\right) \\&= (-1)^n \left(\frac{1-e^{-w}}{w}\right)^{(n)} \\&= (-1)^n\sum\limits_{j=0}^{n} \binom{n}{j}(1-e^{-w})^{(j)}\left(\frac{1}{w}\right)^{(n-j)}\\&= n!w^{-(n+1)}\left(1-e^{-w}\sum\limits_{j=0}^{n} \frac{w^{j}}{j!}\right)\\&= n!w^{-(n+1)}e^{-w}\sum\limits_{j=n+1}^{\infty} \frac{w^{j}}{j!}\end{align*}

where, the notation $f^{(n)}$ denotes the $n-$th derivative of the function $f$ and $f^{(0)} = f$.

Thus, \begin{align*}\frac{1}{(n+1)!}\int_0^{\infty}\frac{w^{n+1}}{e^{w} - 1}\int_0^{1} t^ne^{-wt}\,dt\,dw &= \frac{1}{n+1}\int_0^{\infty} \frac{e^{-2w}}{1-e^{-w}}\sum\limits_{j=n+1}^{\infty} \frac{w^{j}}{j!}\,dw\\&= \frac{1}{n+1}\sum\limits_{k=1}^{\infty}\sum\limits_{j=n}^{\infty}\frac{1}{(j+1)!}\int_0^{\infty}w^{j+1}e^{-(k+1)w}\,dw\\&= \frac{1}{n+1}\sum\limits_{k=1}^{\infty}\sum\limits_{j=n}^{\infty}\frac{1}{(k+1)^{j+2}}\\&= \frac{1}{n+1}\sum\limits_{k=1}^{\infty}\frac{1}{k(k+1)^{n+1}}\\&= \frac{1}{n+1}\sum\limits_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}-\sum\limits_{l=2}^{n+1}\frac{1}{(k+1)^l}\right)\\&= 1 - \frac{1}{n+1}\sum\limits_{l=2}^{n+1}\zeta(l)\end{align*}

[galactus]

Thanks for finishing, rd

[Tolaso J Kos]

Another way that is not that different from those presented above.

\begin{align*}
\int_{0}^{1}x^n \left \{ \frac{1}{x} \right \}^n \, dx &=\sum_{k=1}^{\infty}\left [ \int_{1/(k+1)}^{1/k} x^n \left \{ \frac{1}{x} \right \}^n \right ]\, dx \\
&= \sum_{k=1}^{\infty}\left [ \int_{1/(k+1)}^{1/k}x^n \left ( \frac{1}{x}-k \right )^n \right ]\,dx\\
&=\sum_{k=1}^{\infty}\left [ \int_{1/(k+1)}^{1/k}\left ( 1-kx \right )^n \right ]\, dx \\
&= \sum_{k=1}^{\infty}\frac{1}{k}\left ( \int_{1/(k+1)}^{1/k} \left ( 1-y \right )^n \, dy \right )\\
&= \frac{1}{n+1}\sum_{k=1}^{\infty}\frac{1}{k(k+1)^{n+1}}
\end{align*}

Let

$$F(n)= \sum_{k=2}^{\infty}\frac{1}{(k-1)k^{n+1}}$$

Hence:

\begin{align*}
F(n) &=\sum_{k=2}^{\infty} \frac{1}{k^n} \left ( \frac{1}{k-1} - \frac{1}{k} \right )\\
&=\sum_{k=2}^{\infty}\frac{1}{(k-1)k^n} - \sum_{k=2}^{\infty}\frac{1}{k^{n+1}} \\
&=F(n-1) -\zeta(n+1) +1 \\
&=\cdots \\
&=F(0) - \left ( \zeta(n+1) + \zeta(n)+\cdots + \zeta(2) \right )+n
\end{align*}

Hence

$$F(n)=n+1 - \sum_{k=2}^{n+1}\zeta(k)$$

The result follows.

[galactus]

That was a fun one. Kudoes fellas