Piltz divisor problem

For an integer $k\geq 2$ and a positive integer $n$, let

$$d_{k}(n)=\#\{(n_{1},\dots ,n_{k})\in \mathbb{Z}_{\geq 1}^{k}\;:\;n_{1}n_{2}\cdots n_{k}=n\},$$

so that $d_{2}(n)=d(n)=\sigma_{0}(n)$ is the number of divisors of $n$, and for real $x\geq 1$ put

$$D_{k}(x)=\sum _{n\leq x}d_{k}(n).$$

For $\operatorname{Re}(s)>1$ one has $\sum _{n\geq 1}d_{k}(n)n^{-s}=\zeta(s)^{k}$, where $\zeta$ is the Riemann zeta function, and for $c>1$,

$$D_{k}(x)={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }\zeta ^{k}(w){\frac {x^{w}}{w}}\,dw .$$

Let $P_{k}$ be the polynomial of degree $k-1$ with real coefficients determined by

$$xP_{k}(\log x)=\operatorname*{Res}_{w=1}\;\zeta ^{k}(w){\frac {x^{w}}{w}},$$

equivalently the unique polynomial of degree $k-1$ for which $D_{k}(x)-xP_{k}(\log x)=o(x)$ as $x\to \infty$, and define the error term

$$\Delta _{k}(x)=D_{k}(x)-xP_{k}(\log x).$$

For $k=2$ this reads $D_{2}(x)=x\log x+(2\gamma -1)x+\Delta _{2}(x)$, with $\gamma$ the Euler–Mascheroni constant.

Define

$$\alpha _{k}=\inf \{\theta \geq 0\;:\;\Delta _{k}(x)=O(x^{\theta })\ \text{as }x\to \infty \},$$

so that $\alpha _{k}$ is the smallest number for which $\Delta _{k}(x)=O\!\left(x^{\alpha _{k}+\varepsilon }\right)$ holds for every $\varepsilon >0$.

Determine the exact value of $\alpha _{k}$ for every integer $k\geq 2$.