Sum of logarithm reciprocals

Question

I am looking for an exact formula for the sum : $$\sum_{1<n\leq x}\frac{1}{\log n}$$ My attempt : Using the Perron's formula for partial Dirichlet series, we have : $$\sum_{1<n\leq x}\frac{1}{n^{s}}=\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}(\zeta(s+\omega)-1)\frac{x^{\omega}}{\omega}d\omega$$ $$=\frac{1}{2\pi i }\int_{c^{'}-i\infty}^{c^{'}+i\infty}(\zeta(z)-1)\frac{x^{z-s}}{z-s}dz$$ Now, we have : $$\sum_{1<n\leq x}\frac{1}{\log n}=\int_{0}^{\infty}\sum_{1<n\leq x}\frac{1}{n^{s}}ds=\frac{1}{2\pi i }\int_{c^{'}-i\infty}^{c^{'}+i\infty}\left(\zeta(z)-1\right)\text{Ei}(z\log x)dz$$ $\text{Ei}(\cdot)$ being the exponential integral function. By the residue theorem, the smooth part of this sum is given by : $$\text{Ei}(\log x)=\text{li}(x)$$ $\text{li}(\cdot)$ being the logarithmic integral function. My question is about the oscillatory part. How does one find an exact formula for this oscillatory part ?