On a sum involving divisor function and the integral part function

For any real number t, let \(\lfloor t\rfloor \) $\lfloor t\rfloor$ be the largest integer not exceeding y. As usual, let d(n) be the divisor function. Recently, Ma and Sun obtain the following \(\begin{aligned}\sum _{n\le x}d\left( \left\lfloor \frac{x}{n}\right\rfloor \right) =\lambda x+O_{\varepsilon }\left( x^{11/23+\varepsilon }\right) .\end{aligned}\) $$\begin{array}{c}\hfill \sum _{n\le x}d\left(⌊\frac{x}{n}⌋\right)=\lambda x+O_{\epsilon }\left(x^{11/23+\epsilon }\right).\end{array}$$ where \(\lambda =\sum _{k=1}^{\infty }\frac{d(k)}{k(k+1)}\) $\lambda ={\sum }_{k=1}^{\infty }\frac{d(k)}{k(k+1)}$ is an absolute constant and \(\varepsilon \) $\epsilon$ is an arbitrarily small positive number. In this article, we give a slight generalization of their formula. Specifically, we prove that for any \(c>0\) $c>0$ , the asymptotic formula \(\begin{aligned}\sum _{n\le x^{1/c}}d\left( \left\lfloor \frac{x}{n^{c}}\right\rfloor \right) =d_{c}x^{1/c}+O_{\varepsilon ,c}\left( x^{\theta _{c}+\varepsilon }\right) \end{aligned}\) $$\begin{array}{c}\hfill \sum _{n\le x^{1/c}}d\left(⌊\frac{x}{n^c}⌋\right)=d_c{x}^{1/c}+O_{\epsilon ,c}\left(x^{{\theta }_c+\epsilon }\right)\end{array}$$ holds, where \(\theta _{c}<\frac{1}{c}\) ${\theta }_c<\frac{1}{c}$ and \(d_{c}=\sum _{k\ge 1}d(k)\left( \frac{1}{k^{1/c}}-\frac{1}{(k+1)^{1/c}}\right) \) $d_c={\sum }_{k\ge 1}d(k)\left(\frac{1}{k^{1/c}},-,\frac{1}{{(k+1)}^{1/c}}\right)$ is a constant.