number theory - Average order of $mathrm{rad}(n)$

Let $\mathrm{rad}(n)$ denote the radical of an integer $n$, which is the product of the distinct prime numbers dividing n. Or equivalently, $$\mathrm{rad}(n)=\prod_{\scriptstyle p\mid n\atop p\text{ prime}}p.$$ Assume $\mathrm{rad}(1)=1$, so that $\mathrm{rad}(n)$ is multiplicative.

I was wondering if one has a nice asymptotic formula for the sum $$\sum_{n\le x}\mathrm{rad}(n).$$
At first, I wanted to use the Wiener-Ikehara Theorem. Using Euler Product, we have
$$\begin{align} \begin{split} R(s)&=\sum_{n\ge 1}\frac{\mathrm{rad}(n)}{n^s}=\prod_{p}(1+\frac{p}{p^s}+\frac{p}{p^{2s}}+\cdots)\\ &=\prod_{p}(1+\frac{p}{p^s}\frac{1}{1-p^{-s}})\\ &=\zeta(s)\prod_{p}(1+\frac{p-1}{p^s}). \end{split} \end{align}$$
However, the Wiener-Ikehara Theorem seems not to work here because the product part diverges when $s=1.$

Comparing term-wisely with the product and using $1$$\frac{\zeta(s)^2}{\zeta(2s)}

Moreover, if multiplicative function $\mathrm{core}(n)$ is defined to map positive integers ''n'' to square-free numbers by reducing the
exponents in the prime power representation modulo 2, or in formula
: $$\mathrm{core}(p^e) = p^{e\mod 2}.$$ with $\mathrm{core}(1) =1 .$ Or equivalently, $$\mathrm{core}(p^{2k+1})=p,$$ $$\mathrm{core}(p^{2k})=1$$
Then we always have $$\mathrm{core}(n) \le \mathrm{rad}(n).$$
Since the Dirichlet generating function of $\mathrm{core}$ is
:

$$\begin{align} \begin{split} C(s)&=\sum_{n\ge 1}\frac{\mathrm{core}(n)}{n^s} =\prod_{p}(1+\frac{p}{p^s}+\frac{1}{p^{2s}}+\frac{p}{p^{3s}}+\frac{1}{p^{4s}}+\cdots)\\ &=\prod_{p}(1+\frac{\frac{p}{p^s}+\frac{1}{p^{2s}}}{1-\frac{1}{p^{2s}}})\\ &=\frac{\zeta(2s)\zeta(s-1)}{\zeta(2s-2)}. \end{split} \end{align}$$

$$\frac{\zeta(2s)\zeta(s-1)}{\zeta(2s-2)}
Here we may use the Wiener-Ikehara Theorem to derive asymptotic bounds.

This was as far as I could work out.
Are there any results considering the average order of $\mathrm{rad}(n)$ or analytic expression of its Dirichlet series? Thanks.

Answer

The general methodology used in this answer will handle your sum: Mean Value of a Multiplicative Function close to $n$ in Terms of the Zeta Function. (See also this MSE answer, and this answer on Math Overflow)

The main theorem proven in that answer is that if we let $g$ be defined so that $(1*g)(n)=\frac{f(n)}{n}$, then for any $0<\sigma<1$ we have that

$$\sum_{n\leq x}f(n)=\frac{x^2}{2}\sum_{d=1}^\infty\frac{g(d)}{d}+O\left(x^{1+\sigma}\sum_{d=1}^\infty \frac{|g(d)|}{d^\sigma}\right).$$

Note this result does not make sense unless $f$ is close to $n$ as a multiplicative function, as otherwise the series involving $g(n)$ will not converge.

In our case, $f(p^k)=p$ for all $k$, and so $f(p^k)/p^k = 1/p^{k-1}$. Convolving with the mobius function we find that $$g(p^{k})=\begin{cases} 0 & \text{if }k=1\\ \frac{1}{p^{k-1}}-\frac{1}{p^{k-2}} & \text{if }k\geq2 \end{cases}$$
Thus, since $$\sum_{n=1}^{\infty}\frac{g(n)}{n}=\prod_{p}\left(1+\frac{g(p)}{p}+\frac{g(p^{2})}{p^{2}}+\cdots\right) = \prod_{p}\left(1+\frac{\frac{1}{p}-1}{p^{2}}+\frac{\frac{1}{p}-1}{p^{4}}+\cdots\right)=\prod_{p}\left(1-\left(\frac{p-1}{p}\right)\left(\frac{1}{p^{2}-1}\right)\right)=\prod_{p}\left(1-\frac{1}{p(p+1)}\right),$$ by taking $\sigma = 1+\frac{1}{\log x}$ we obtain

$$\sum_{n\leq x}\text{rad}(n)=\frac{x^2}{2}\prod_p \left(1-\frac{1}{p(p+1)}\right)+O\left(x^{3/2}\log x\right).$$