Can we find the $a$ value?

Question

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

Answer 1

The large-$u$ limit of the integral can be evaluated in terms of a zeta function: $$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$ Hence the desired equality $\frac{f(a)-I(a)}{g(a)-I(1-a)}=1$ reduces to an equation for $f(a)-g(a)$: $$f(a)-g(a)=\frac{\zeta (a)-a \zeta (1-a)-a\zeta (a)+2 a-1}{(a-1) a},\;\;0<a<1.$$ If you know the numerator and denominator both vanish separately in the large-$u$ limit you need $a=1/2$, $f(1/2)=g(1/2)=-2 -2 \zeta(1/2)$.