We have the following limit with and $a \in \mathbb{R}$ and $ u \in \mathbb{R}$ . And here, ${\lfloor x \rfloor}$ is floor function

$$\lim_{u \rightarrow \infty} \frac{f(a)-\int_1^u ( {x-\lfloor x \rfloor}) \cdot x^{-a-1} dx} {g(a)-\int_1^u ({x-\lfloor x \rfloor}) \cdot x^{a-2}dx} =1 $$

Meanwhile know the following derivatives results. It means $f(a)$ and $g(a)$ are functions of only $a$ (they are not any related with $u$).

$$\ \frac{d} {du}f(a) =0....and ....\frac{d} {du}g(a) =0. $$

Thus, if the limit appearance of the left side (under $u→∞$) is $0/0$ , can we find the $a$ value by applying Hospital rule ?

NOTE: Please notice we know that we can obtain the following result:

$$ \frac{{\frac {d} {du}}\left[\int_1^u {(x-\lfloor x \rfloor}) \cdot x^{-a-1} dx\right]} {{\frac {d} {du}}\left[\int_1^u {(x-\lfloor x \rfloor}) \cdot x^{a-2}dx\right]} =u^{1-2a} $$

Also we have obtained on https://develop.wolframcloud.com $$ {{\frac {d} {du}}\left[\int_1^u {(x-\lfloor x \rfloor}) \cdot x^{-a-1} dx\right]}={(u-\lfloor u \rfloor}) \cdot u^{-a-1}$$

We have the following limit with and $a \in \mathbb{R}$ and $ u \in \mathbb{R}$ . And here, ${\lfloor x \rfloor}$ is floor function

$$\lim_{u \rightarrow \infty} \frac{f(a)-\int_1^u ( {x-\lfloor x \rfloor}) \cdot x^{-a-1} dx} {g(a)-\int_1^u ({x-\lfloor x \rfloor}) \cdot x^{a-2}dx} =1 $$

Meanwhile know the following derivatives results. It means $f(a)$ and $g(a)$ are functions of only $a$ (they are not any related with $u$).

$$\ \frac{d} {du}f(a) =0....and ....\frac{d} {du}g(a) =0. $$

Thus, if the limit appearance of the left side (under $u→∞$) is $0/0$ , can we find the $a$ value by applying Hopital rule ?

NOTE: Please notice we know that we can obtain the following result:

$$ \frac{{\frac {d} {du}}\left[\int_1^u {(x-\lfloor x \rfloor}) \cdot x^{-a-1} dx\right]} {{\frac {d} {du}}\left[\int_1^u {(x-\lfloor x \rfloor}) \cdot x^{a-2}dx\right]} =u^{1-2a} $$

Also we have obtained on https://develop.wolframcloud.com $$ {{\frac {d} {du}}\left[\int_1^u {(x-\lfloor x \rfloor}) \cdot x^{-a-1} dx\right]}={(u-\lfloor u \rfloor}) \cdot u^{-a-1}$$