An integral with the floor function

Relevant wiki: Integration of Piecewise Functions

Let $n$ be an integer . If $n \leq x < n+1$ , then $\lfloor x \rfloor = n$ . And we denote $I$ as the integral in question. We have

$\begin{aligned} I &=& \int_1^\infty \dfrac{\lfloor x \rfloor }{x^3} \, dx \\ &=& \int_1^2 \dfrac1{x^3} \, dx + \int_2^3 \dfrac1{x^3} \, dx + \int_3^4 \dfrac1{x^3} \, dx + \cdots + \int_n^{n+1} \dfrac1{x^3} \, dx + \cdots \\ &=& \sum_{m=1}^\infty \int_m^{m+1} \dfrac m{x^3} \, dx \\ &=& \sum_{m=1}^\infty m \cdot \left [-\dfrac1{2x^2} \right ]_m^{m+1} \\ &=& \sum_{m=1}^\infty \dfrac 12 \cdot \left [-\dfrac m{(m+1)^2} + \dfrac m{m^2} \right ] \\ &=& \dfrac12 \sum_{m=1}^\infty \cdot \left [\dfrac1m -\dfrac1{m+1} + \dfrac1{(m+1)^2} \right ] \\ &=& \dfrac12 \; \underbrace{ \sum_{m=1}^\infty \left( \dfrac1m - \dfrac1{m+1} \right)}_{ = 1 } + \dfrac12 \; \underbrace{ \sum_{m=1}^\infty \dfrac1{(m+1)^2} }_{= \pi^2 /6 - 1} = \dfrac{\pi^2}{12} . \\ \end{aligned}$

We can evaluate the two series in the penultimate step above by noticing that the first series is a telescoping series , and the second series can be expressed as $\zeta(2) - 1$ , where $\zeta(\cdot)$ denotes the Riemann Zeta function .

Now, we want to express $I = \dfrac{\pi^2}{12}$ in the form of $\dfrac 3a + \dfrac{\pi b}{12} + \dfrac{\pi^2 c}{24 } - 1$ .

Notice that $\dfrac{\pi^2}{12} = (\color{#20A900}1 - 1) + \dfrac{\pi \cdot \color{#20A900}0}{12} + \dfrac{\pi^2 \cdot \color{#20A900}2}{24}$ . So $\dfrac3a = 1, b = 0, c = 2 \Rightarrow c^{a+b} = 2^{0+3} = \boxed8$ .

$\begin{aligned} I & = \int_1^\infty \frac {\lfloor x \rfloor}{x^3} dx \\ & = \sum_{k=1}^\infty \int_k^{k+1} \frac k{x^3} dx \\ & = \sum_{k=1}^\infty - \frac k{2x^2} \bigg|_k^{k+1} \\ & = \frac 12 \sum_{k=1}^\infty \left(\frac k{k^2} - \frac k{(k+1)^2} \right) \\ & = \frac 12 \sum_{k=1}^\infty \left(\frac 1k - \frac {k+1-1}{(k+1)^2} \right) \\ & = \frac 12 \sum_{k=1}^\infty \left({\color{#3D99F6}\frac 1k - \frac 1{k+1}} + \frac 1{(k+1)^2} \right) \\ & = \frac 12 {\color{#3D99F6}\left(\frac 11\right)} + \frac 12 \sum_{k=1}^\infty \frac 1{(k+1)^2} \\ & = \frac 12 + {\color{#3D99F6}\frac 12 \sum_{k=1}^\infty \frac 1{k^2}} - \frac 12 \\ & = \color{#3D99F6} \frac {\pi^2}{12} \end{aligned}$

Therefore, $\dfrac 3{\color{#3D99F6}a} + \dfrac {\pi \color{#3D99F6}b}{12} + \dfrac {\pi^2\color{#3D99F6}c}{24} - 1 = \dfrac 3{\color{#3D99F6}3} + \dfrac {\pi\cdot \color{#3D99F6}0}{12} + \dfrac {\pi^2\cdot \color{#3D99F6}2}{24} - 1$ $\implies c^{a+b} = 2^{3+0} = \boxed{8}$ .

$\begin{aligned} I&=\int\limits_1^\infty\frac{\lfloor x \rfloor}{x^3}\,\mathrm dx=\sum\limits_{k=1}^\infty\int\limits_0^1 \frac{k}{(k+x)^3}\,\mathrm d x=-\sum\limits_{k=1}^\infty \frac k2\left[\frac{1}{(k+1)^2}-\frac{1}{k^2}\right]\\ &=-\sum\limits_{k=1}^\infty \frac k2\left[\frac{-2k-1}{(k+1)^2 k^2}\right]=\frac 12\sum\limits_{k=1}^\infty\left[\frac{2}{(k+1)^2}+\frac{1}{(k+1)^2 k}\right]\\ &=\frac{1}{2}\left[2\left(\pi^2/6-1\right)+2-\pi^2/6\right]=\pi^2/12\\ &=\frac 33+\frac{2\pi^2}{24}-1 \end{aligned}$

So $c^{a+b}=2^{3+0}=\boxed{8}$