We have that
$$\limsup_{n\to\infty}\frac{\sigma(n)}{n\log(\log(n))}=e^\gamma$$
$$\limsup_{n\to\infty}\log(\tau(n))\frac{\log(\log(n))}{\log(n)}=\log(2).$$
I'm trying to find a similar result for $\sigma_k$ for $0<k<1$. The two results above correspond to the ends of the interval $\tau=\sigma_0$ and $\sigma=\sigma_1$. For an $n=p_1^{a_1}\dots p_r^{a_r}$,
$$\sigma_k(n)=\prod_{h=1}^r\frac{p_h^{k(a_h+1)}-1}{p_h^k-1}$$
$$\frac{\sigma_k(n)}{n^k}=\prod_{h=1}^r\frac{1-p_h^{-k(a_h+1)}}{1-p_h^{-k}}.$$
Each term in the product is a decreasing function of $p_h$, so for this quantity to be large we want the $p_h$ to be small, so assume they're the primes in increasing order $p_1=2$, etc. By choosing all of the $a_h$ to be large the numerators all approach 1, but the behavior of the denominators is more vexing. We have
$$\frac{\sigma_k(n)}{n^k}<\prod_{h=1}^r\frac 1{1-p_h^{-k}}$$
so define
$$f_k(x)=\prod_{p\leq x}\frac 1{1-p^{-k}}.$$
It seems reasonable to conjecture that $\frac{\sigma_k(n)}{n^k}$ is close to $f_k(p_r)$, that is, that the the above inequality is tight. This is why I want an asymptotic form for $f_k$.
I did some numerical computation for the $k=\frac 12$ case. Plotting $\log(f_{1/2})$ on a log-log plot got me a surprisingly linear-looking relationship, which corresponds to
$$f_{1/2}(x)\approx e^{x^\alpha}$$
where $\alpha=0.3751988607236585$. Other than guessing this is three eights, I don't know.
I tried using the Euler product
$$\sum_{n=1}^x\frac 1{n^k}<f_k(x)$$
but it appears this inequality is far from tight, as the left-hand side is asymptotically polynomial.