Evaluate$\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx where [x] denotes greatest integer function and $0<\sigma<1$ Evaluate$\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx where [x] denotes greatest integer function and $0<\sigma<1$.
My try:-  1-(x-[x])$\leq 1 \Rightarrow$ $\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx $\leq$ $\int_{1}^{\infty}$ $\frac {1}{x^{2-\sigma}}$dx= $\frac{1}{1-\sigma}$
$\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx= $\int_{1}^{2}$ $\frac{1-(x-1)}{x^{2-\sigma}}$dx+$\int_{2}^{3}$ $\frac{1-(x-2)}{x^{2-\sigma}}$dx+...
But on integration I am not getting a finite value.
 A: Trying to continue along your way.
You wrote
$$\int_1^\infty(1-x+\lfloor x\rfloor )\, x^{\sigma -2}\,dx=\sum_{n=1}^\infty \int_n^{n+1}(n+1-x)\,x^{\sigma -2}\,dx=\sum_{n=1}^\infty I_n$$
$$I_n=\int_n^{n+1}(n+1-x)\,x^{\sigma -2}\,dx=\frac{ (n+\sigma )n^{\sigma }-n (n+1)^{\sigma }}{n (1-\sigma) \sigma }=\frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma } } {(1-\sigma)\, \sigma  }$$ and here the problem starts to be difficult if you are not familiar with the zeta function.
Hoping that you are, the result should be
$$\frac{1+\sigma\,  \zeta (1-\sigma ) } {(1-\sigma) \,\sigma  }$$
A: This is not a complete answer, but hopefully will provide some insight:
Let $f(x) = \frac{1-(x-\lfloor x\rfloor)}{x^{2-\sigma}}$, then as $0\leqslant x-\lfloor x\rfloor<1$ and $x^{2-\sigma}>0$, $f$ is nonnegative on $[1,\infty)$. Hence by Tonelli's theorem,
$$
\int_1^\infty f(x)\ \mathsf dx = \int_1^\infty \sum_{n=1}^\infty f_n(x)\ \mathsf dx = \sum_{n=1}^\infty \int_n^{n+1} f_n(x)\ \mathsf dx, 
$$
where $f_n(x) = f(x)\cdot\mathsf 1_{[n,n+1)}$. By induction (the tricky part), we can show that
$$
\int_n^{n+1} f_n(x)\ \mathsf dx = \frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma }}{\sigma(1-\sigma)}.
$$
Hence
$$
\int_1^\infty f(x)\ \mathsf dx = \frac{1+\sigma\,  \zeta (1-\sigma ) } {(1-\sigma) \,\sigma  },
$$
where
$$
\zeta(s) := \sum_{n=1}^\infty \frac1{n^s}
$$
is the Riemann zeta function.
