Fractional part integration

$\begin{aligned} I & = \int_1^\infty \frac{\{x\}}{x^n} dx \quad \quad \small \color{#3D99F6}{\text{Let } \{x\} = t \text{ and } x = k + t \text{, where } k =1,2,3... \space \Rightarrow dx = dt} \\ & = \sum_{k=1}^\infty \int_0^1 \frac{t}{(k+t)^n} dt \\ & = \sum_{k=1}^\infty \int_0^1 \left(\frac{1}{(k+t)^{n-1}} - \frac{k}{(k+t)^n} \right) dt \\ & = \sum_{k=1}^\infty \left[\frac{k}{(n-1)(k+t)^{n-1}} - \frac{1}{(n-2)(k+t)^{n-2}} \right]_0^1 \\ & = \sum_{k=1}^\infty \left[-\frac{k+(n-1)t}{(n-1)(n-2)(k+t)^{n-1}} \right]_0^1 \\ & = \frac{1}{(n-1)(n-2)} \sum_{k=1}^\infty \left( \frac{k}{k^{n-1}}-\frac{k+n-1}{(k+1)^{n-1}} \right) \\ & = \frac{1}{(n-1)(n-2)} \sum_{k=1}^\infty \left( \frac{k}{k^{n-1}}-\frac{k+1}{(k+1)^{n-1}} - \frac{n-2}{(k+1)^{n-1}} \right) \\ & = \frac{1}{(n-1)(n-2)} \left(1 - \sum_{k=1}^\infty \frac{n-2}{(k+1)^{n-1}} \right) \\ & = \frac{1}{(n-1)(n-2)} - \frac{\zeta(n-1) - 1}{n-1} \\ & = \boxed{\dfrac{n-1}{(n-1)(n-2)}-\dfrac{\zeta(n-1)}{n-1}} \quad n \ge 4 \end{aligned}$

For $n = 5$ , $I = \dfrac{4}{(3)(4)}-\dfrac{\zeta(4)}{4} = \dfrac{1}{3} - \dfrac{\pi^4}{360}$

$\Rightarrow A+B+C = 3+4+360 = \boxed{367}$

On taking fractional part of $x$ as $x \ - \lfloor x \rfloor$

Summation becomes:-

$\displaystyle \int_{1}^{\infty} \frac{1}{x^4} \ dx \ - \int_{1}^{\infty} \frac{\lfloor x \rfloor}{x^5} \ dx$

$= \frac{1}{3} \ - \ ( \int_{1}^{2} \frac{1}{x^5} \ dx \ + \int_{2}^{3} \frac{2}{x^5} \ dx \ + \int_{3}^{4} \frac{3}{x^5} \ dx \ ....)$

$= \frac{1}{3} \ - \frac{1}{4} \ ( \frac{1}{1^4} \ - \frac{1}{2^4} \ + \frac{2}{2^4} \ - \frac{2}{3^4} \ .... )$

$= \frac{1}{3} \ - \frac{1}{4} \ ( \frac{1}{1^4} \ + \frac{1}{2^4} \ + \frac{1}{3^4} \ .....)$

$= \frac{1}{3} \ - \frac{1}{4} \times \zeta(4)$

$= \frac{1}{3} \ - \frac{\pi^4}{360}$

Hence $A \ + \ B \ + \ C \ = \boxed{367}$

Generalised Form :-

$\displaystyle \int_1^\infty \dfrac{\{x\}}{x^n} \, \mathrm{d}x \ = \frac{1}{n-2} \ - \frac{1}{n-1} \times \zeta(n-1)$