Listing out all the integers

$I = \displaystyle \int_{1}^{\infty} \frac{ \{x\}}{x^3} \ dx$

$I = \displaystyle \int_{1}^{\infty} \frac{x - \lfloor x \rfloor}{x^3} \ dx$

Here, I am going to split this into two integrals , $I_{1}$ & $I_{2}$ for the sake of convenience.

$I_{1} = \displaystyle \int_{1}^{\infty} \frac{1}{x^3} \ dx$

$I_{1} =\displaystyle \lim_{t\rightarrow \infty } \left. \frac{1}{x^2} \right |_{t}^{1} = 1 - \lim_{t\rightarrow \infty} \frac{1}{t}= 1$

Moving on to $I_{2}$

$I_{2} = \displaystyle \int_{1}^{\infty} \dfrac{\lfloor x \rfloor}{x^3} \ dx = \displaystyle \int_{1}^{2}\frac{1}{x^3} \ dx + \int_{2}^{3}\frac{2}{x^3} \ dx + \int_{3}^{4}\frac{3}{x^3} \ dx + \ldots$

$I_{2} = \displaystyle \frac{1}{2} \times [ \left. \dfrac{1}{x^2}\right| _{2}^{1} + \left. \dfrac{2}{x^2} \right| _{3}^{2} +\left. \dfrac{3}{x^2}\right | _{4}^{3} + \ldots ]$

$I_{2} =\displaystyle \frac{1}{2} \times [ \displaystyle 1 - \frac{1}{4} + \frac{2}{4} -\frac{2}{9} + \frac{3}{9} - \frac{3}{16} + \ldots ]$

$I_{2} = \displaystyle \frac{1}{2} \times [ \displaystyle 1 + \frac{1}{4} +\frac{1}{9}+ \frac{1}{16} \ldots ]$

$I_{2} = \displaystyle \frac{1}{2} \times \zeta (2) = \frac{1}{2} \times \frac{\pi^2}{6} = \frac{\pi^2}{12}$

$I = I_{1} - I_{2} = 1 - \displaystyle \frac{\pi^2}{12}$

Comparing, $A = 12$ .

Consider the two integrals $S_1 = \int_{x=1}^\infty \frac{x}{x^3}\:dx\ \ \ \ \ \text{and}\ \ \ \ \ S_2 = \int_{x=1}^\infty \frac{\{x\}}{x^3}\:dx.$ The difference between the integrands is $1/x^3$ for $1 \leq x < 2$ , $2/x^3$ for $2 \leq x < 3$ , and so on. We therefore write $S_1 - S_2 = \int_1^\infty \frac{1}{x^3}\:dx + \int_2^\infty \frac{1}{x^3}\:dx + \cdots.$ In general, $\int_n^\infty \frac{1}{x^3} = \left.-\frac{1}{2x^2}\right|_n^\infty = \frac{1}{2n^2}.$ Thus $S_1 - S_2 = \frac{1}{2\cdot 1^2} + \frac{1}{2\cdot 2^2} + \frac{1}{2\cdot 3^2} + \cdots = \tfrac12\sum_{n=1}^\infty \frac{1}{n^2}.$ The sum is well-known to be equal to $\pi^2/6$ . Thus $S_1 - S_2 = \frac{\pi^2}{12}.$ Integral $S_1$ is easily evaluates (it is equal to 1), so $S_2 = 1 - \frac{\pi^2}{12}.$ Thus the solution is $\boxed{12}$ .

The integral is equal to :

$\int_1^2 \! \dfrac{x-1}{x^3} \, \mathrm{d}x + \int_2^3 \! \dfrac{x-2}{x^3} \, \mathrm{d}x + \int_3^4 \! \dfrac{x-3}{x^3} \, \mathrm{d}x. + ... = \sum_{i = 1}^{\infty} \int_i^{i+1} \! \dfrac{x-i}{x^3} \mathrm{d}x = I.$

We get

$I =\displaystyle \sum_{i = 1}^{\infty} \left[ \left(\dfrac{-1}{x} \right)_{i}^{i+1} - \left(\dfrac{-i}{2x^2} \right)_{i}^{i+1}\right]$

$= \displaystyle \sum_{i = 1}^{\infty} \left[\dfrac{-1}{i+1} + \dfrac{1}{i} +\dfrac{i}{2(i+1)^2} - \dfrac{1}{2i}\right]$

$= \displaystyle \sum_{i = 1}^{\infty} \left[\dfrac{-2 - i}{2(i+1)^2} + \dfrac{1}{2i} \right]$

$= \displaystyle \sum_{i = 1}^{\infty} \left[\dfrac{-2 - i}{2(i+1)^2}\right] + \displaystyle \sum_{i = 1}^{\infty}\dfrac{1}{2i}$

$= \displaystyle \sum_{i = 1}^{\infty} \left[\dfrac{-2 - i}{2(i+1)^2}\right] + \displaystyle \sum_{i = 1}^{\infty}\dfrac{1}{2(i+1)} + \dfrac{1}{2}$

$= \displaystyle \sum_{i = 1}^{\infty} \left[\dfrac{-2 - i}{2(i+1)^2} + \dfrac{1}{2(i+1)} \right] + \dfrac{1}{2}$

$= \dfrac{1}{2} - \displaystyle \sum_{i = 1}^{\infty} \left[\dfrac{1}{2(i+1)^2} \right]$

$= \dfrac{1}{2} - \dfrac{1}{2} ( \zeta(2) - 1)$

$= 1 - \dfrac{\pi^2}{12} .$

Thus A = 12.

$\begin{aligned} I & = \int_1^\infty \frac{\{x\}}{x^3} \, dx \\ & = \sum_{k=1}^\infty \int_k^{k+1} \frac{x-k}{x^3} \, dx \\ & = \sum_{k=1}^\infty \int_k^{k+1} \left( \frac{1}{x^2} - \frac{k}{x^3} \right) \, dx \\ & = \sum_{k=1}^\infty \left[ -\frac{1}{x} + \frac{k}{2x^2} \right]_k^{k+1} \\ & = \sum_{k=1}^\infty \left(-\frac{1}{k+1} + \frac{k}{2(k+1)^2} + \frac{1}{k} - \frac{k}{2k^2} \right) \\ & = \sum_{k=1}^\infty \color{#3D99F6}{\frac{1}{2k(k+1)^2} \quad \quad \small \text{By partial fractions}} \\ & = \sum_{k=1}^\infty \color{#3D99F6}{\frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2}\right)} \\ & = \frac{1}{2} \left(1 - \zeta(2) + 1\right) \\ & = 1 - \frac{\pi^2}{12} \end{aligned}$

$\implies A = \boxed{12}$