Is this estimate true? Can anyone give a proof of it?

$$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p)\qquad (p\text{ prime, } p\to\infty) $$
where $ (ab)_p\equiv ab\;(\operatorname{mod}p)$, $0<(ab)_p<p$.

Note: we have $$\lim_{p\rightarrow \infty}\frac1{p\ln^2p}{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}}=\frac{1}{2}.$$

Is this estimate true? Can anyone give a proof of it?

$$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p)\qquad (p\text{ prime, } p\to\infty) $$
where $ (ab)_p\equiv ab\;(\operatorname{mod}p)$, $0<(ab)_p<p$.

p is a prime $$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p) $$
$ (ab)_p\equiv ab(\mod p), 0<(ab)_p<p $

($\lim_{p\rightarrow \infty}\frac{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}}{p\ln^2p}=\frac{1}{2}$).

Is this estimate true? Can anyone give a proof of it?

$$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p)\qquad (p\text{ prime, } p\to\infty) $$
where $ (ab)_p\equiv ab\;(\operatorname{mod}p)$, $0<(ab)_p<p$.

Note: we have $$\lim_{p\rightarrow \infty}\frac1{p\ln^2p}{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}}=\frac{1}{2}.$$

p is a prime $$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p) $$
$ (ab)_p\equiv ab(\mod p), 0<(ab)_p<p $.

p is a prime $$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p) $$
$ (ab)_p\equiv ab(\mod p), 0<(ab)_p<p $

($\lim_{p\rightarrow \infty}\frac{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}}{p\ln^2p}=\frac{1}{2}$).