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


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 $a=1/2$, $f(1/2)=g(1/2)=-2 -2 \zeta(1/2)$.

  • $\begingroup$ Dear Carlo, thank you very much for your very different approach. It is very interesting. On the other hand, what do you think that can we apply Hopital rule, because of the result which we have obtained on develop.wolframcloud.com ? Also please remember, we know that the limit appearance of the left side is $0/0$ (under $u → ∞$) . $\endgroup$ – Testform Dec 13 '18 at 9:50
  • $\begingroup$ Then you just have a=1/2, with f(1/2)=g(1/2). No need for l’Hopital. $\endgroup$ – Carlo Beenakker Dec 13 '18 at 11:18
  • $\begingroup$ I know, thank you very much. However, the important thing for me here, Can I apply Hopital rule to get this 1/2. $\endgroup$ – Testform Dec 13 '18 at 12:13
  • $\begingroup$ Certainly, that gives a=1/2. $\endgroup$ – Carlo Beenakker Dec 13 '18 at 13:42
  • $\begingroup$ Dear Carlo, thank you for your kind support. $\endgroup$ – Testform Dec 13 '18 at 18:10

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