Use of $\sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{6} $ in evaluating $I$

Question

Let $$I=\int_1^{\infty} \frac{\{x\}}{x^3} dx,$$ where $\{x\}$ represents fractional part of $x$.

Now, using $\{x\}+\lfloor x \rfloor=x$, $I$ can be rewritten as $I=\int_1^{\infty} \frac{x-\lfloor x \rfloor}{x^3} dx$ and consecutively broken as $$I=\int_1^{\infty} \frac{1}{x^2} dx-\int_1^{\infty}\frac{\lfloor x \rfloor}{x^3} dx.$$

The first integral simplifies to 1 and my question is how do we evaluate the second integral? Clearly, $[x]$ is a discontinuous function and needs to be broken at integers so I tried rewriting it as $$I=\sum_{r=1}^{\infty} \int_{r}^{r+1} \frac{r}{x^3} dx$$ and it almost resembled the series; $\sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{6} $, can this series be used here, if so how?

Answer 1

We can write $\{x\}=x-\lfloor x\rfloor$, where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. Then we have \begin{align*} I&=\int_1^{\infty} \frac{\{x\}}{x^3} dx\\ &=\int_1^2 \frac{x-\lfloor x\rfloor}{x^3} dx+\int_2^3 \frac{x-\lfloor x\rfloor}{x^3} dx+\int_3^4 \frac{x-\lfloor x\rfloor}{x^3} dx+\cdots\\ &=\sum_{n=1}^{\infty}\int_n^{n+1} \frac{x-\lfloor x\rfloor}{x^3} dx\\ &=\sum_{n=1}^{\infty}\int_0^1 \frac{u}{(n+u)^3} du\qquad\qquad\text{(where }u=x-n\text{)}\\ &=\sum_{n=1}^{\infty}\left[-\frac{n+2u}{2(n+u)^2}\right]_{u=0}^{u=1}\\ &=\frac12\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}-\frac{1}{(n+1)^2}\right)\\ &=\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right)-\frac12\sum_{n=1}^{\infty}\frac{1}{(n+1)^2}\\ &=\frac{1}{2}-\frac12\left(\frac{\pi^2}{6}-1\right)\\ &=\boxed{1-\frac{\pi^2}{12}}. \end{align*}

Answer 2

Your way is fine indeed

$$\sum_{r=1}^{\infty} \int_{r}^{r+1} \frac{r}{x^3} dx=\sum_{r=1}^{\infty}\frac{2r+1}{2r(r+1)^2}=\frac12\sum_{r=1}^{\infty}\left(\frac1{r}-\frac1{r+1}\right)+\frac12 \sum_{r=1}^{\infty}\frac1{(r+1)^2}=\frac{\pi^2}{12}$$

using that by telescoping

$$\sum_{r=1}^{\infty}\left(\frac1{r}-\frac1{r+1}\right)=1$$

and

$$\sum_{r=1}^{\infty}\frac1{(r+1)^2}=-1+\sum_{r=1}^{\infty}\frac1{r^2}=-1+\frac{\pi^2}{6}$$

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As an alternative

$$I=\int_1^{\infty} \frac{\{x\}}{x^3} dx=\lim_{n\to \infty} \sum_{k=1}^n\int_0^1\frac{u}{(k+u)^3}du=\lim_{n\to \infty} \sum_{k=1}^n\frac1{2k(k+1)^2}$$

and

$$\sum_{k=1}^n\frac1{2k(k+1)^2}=\frac12\sum_{k=1}^n\left(\frac1{k}-\frac1{k+1}\right)-\frac12\sum_{k=1}^n\frac1{(k+1)^2}\to 1-\frac{\pi^2}{12}$$

Answer 3

I hope this offers some new insight: $$ \begin{align} \int_1^\infty\frac{\{x\}}{x^3}\,\mathrm{d}x &=\sum_{k=1}^\infty\int_0^1\frac{x}{(k+x)^3}\,\mathrm{d}x\tag{1a}\\ &=\sum_{k=1}^\infty\int_0^1\left(\frac1{(k+x)^2}-\frac{k}{(k+x)^3}\right)\mathrm{d}x\tag{1b}\\ &=\sum_{k=1}^\infty\left[\left(\frac1k-\frac1{k+1}\right)-\frac12\left(\frac{k}{k^2}-\frac{k}{(k+1)^2}\right)\right]\tag{1c}\\ &=\sum_{k=1}^\infty\left[\left(\frac1k-\frac1{k+1}\right)-\frac12\left(\frac{k-1}{k^2}-\frac{k}{(k+1)^2}+\frac1{k^2}\right)\right]\tag{1d}\\ &=\color{#C00}{\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)}-\frac12\color{#090}{\sum_{k=1}^\infty\left(\frac{k-1}{k^2}-\frac{k}{(k+1)^2}\right)}-\frac12\color{#F80}{\sum_{k=1}^\infty\frac1{k^2}}\tag{1e}\\ &=\color{#C00}{1}-\frac12\cdot\color{#090}{0}-\frac12\cdot\color{#F80}{\frac{\pi^2}6}\tag{1f}\\ &=1-\frac{\pi^2}{12}\tag{1g} \end{align} $$ Explanation:
$\text{(1a):}$ rewrite the integral as a sum of integrals over unit intervals
$\text{(1b):}$ $\frac{x}{(k+x)^3}=\frac1{(k+x)^2}-\frac{k}{(k+x)^3}$
$\text{(1c):}$ evaluate the integrals
$\text{(1d):}$ add and subtract $\frac1{k^2}$
$\text{(1f):}$ evaluate two telescoping sums and the Basel Identity
$\text{(1g):}$ simplify