Return to Answer

With some effort I can evaluate the integral in closed form: $$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$ Hence the desired equality $\frac{f(a)-I(a)}{g(a)-I(1-a)}=1$ reduces to a simple 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 $a=1/2$, $f(1/2)=g(1/2)=-2 -2 \zeta(1/2)$.

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

With some effort I can evaluate the integral in closed form: $$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$ Hence the desired equality $\frac{f(a)-I(a)}{g(a)-I(1-a)}=1$ reduces to a simple equation for $f(a)-g(a)$: $$f(a)-g(a)=\frac{\zeta (a)-a \zeta (1-a)-a\zeta (a)+2 a-1}{(a-1) a},\;\;0<a<1.$$

With some effort I can evaluate the integral in closed form: $$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$ Hence the desired equality $\frac{f(a)-I(a)}{g(a)-I(1-a)}=1$ reduces to a simple 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 $a=1/2$, $f(1/2)=g(1/2)=-2 -2 \zeta(1/2)$.

With some effort I can evaluate the integral in closed form: $$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$ Hence the desired equality reduces to a simple consistency equation for $f(a)$ and $g(a)$.

With some effort I can evaluate the integral in closed form: $$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$ Hence the desired equality $\frac{f(a)-I(a)}{g(a)-I(1-a)}=1$ reduces to a simple equation for $f(a)-g(a)$: $$f(a)-g(a)=\frac{\zeta (a)-a \zeta (1-a)-a\zeta (a)+2 a-1}{(a-1) a},\;\;0<a<1.$$