Can we obtain the following result for ... $f(x)={x-\lfloor x \rfloor}$ ... ? It is very simple but important question. Here ${\lfloor x \rfloor}$ is floor function and $a \in \mathbb{R}$ . Thank you for your kind comment.
$$ \frac{{\frac {d} {du}}\left[\int_1^u f(x) \cdot x^{-a-1} dx\right]} {{\frac {d} {du}}\left[\int_1^u f(x) \cdot x^{a-2}dx\right]} =u^{1-2a} $$
HINT
Use the Fundamental Theorem of Calculus. For any (constant) $a \in \mathbb{R}$: $$ \frac{d}{dz} \int_a^z f(u)du = f(z). $$ That should simplify your problem greatly.
UPDATE
In your case $f(x) = x - \lfloor x \rfloor$, so it has a countable number of discontinuities, which does not prevent the Riemann integral from existing. You have to define, for example, $$ \int_1^x g(f(u)) du := \int_{\lfloor x \rfloor}^x g(f(u))du + \sum_{k=2}^{\lfloor x \rfloor} \int_1^k g(f(u)) du. $$ This way, all the integrals on the RHS are well-defined and the LHS can be defined in a meaningful way.