Sum of $n \sigma(n)$ What is known about the asymptotic behavior of
$$
-\frac{\pi^2}{18}x^3+\sum_{n\le x}n\sigma(n) ?
$$
It seems to be $O(x^{2+\varepsilon})$ but I cannot prove this.
 A: Look at the corresponding Dirichlet series and make use of Perron's formula to get the asymptotic.
$$f(n) = n \sigma(n) \implies F(s) = \sum_{n=1}^{\infty} \dfrac{n \sigma(n)}{n^s} = \sum_{n=1}^{\infty} \dfrac{\sigma(n)}{n^{s-1}} = \zeta(s-1)\zeta(s-2)$$
$F(s)$ has a pole at $s=2$ and $s=3$. Now making use of Perron's formula,
the leading order terms of $\displaystyle \sum_{n \leq x} n \sigma(n)$ are the leading order terms in the integral (keeping $c>3$) $$\dfrac1{2 \pi i}\displaystyle \int_{c-i T}^{c+i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}{s} ds$$
for a sufficiently large and appropriately chosen $T$.
For $c>3$, we have $$\dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s dx = \sum_{n=1}^{\infty} \dfrac{ n\sigma(n) }{2 \pi i} \int_{c-iT}^{c+iT}\left(\dfrac{x}{n} \right)^s \dfrac{ds}s$$
$$\sum_{n=1}^{\infty} \dfrac{ n\sigma(n) }{2 \pi i} \int_{c-iT}^{c+iT}\left(\dfrac{x}{n} \right)^s \dfrac{ds}s = \sum_{n \leq x} n \sigma(n) + \mathcal{O} \left( \sum_n \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \right)$$
Hence, we have $$\dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s dx = \sum_{n \leq x} n \sigma(n) + \mathcal{O} \left( \sum_n \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \right)$$
Now we look at the error term. We will use the following bounds for $\zeta$. $$\left \vert \zeta(1+k+it) \right \vert \ll \dfrac1{k}$$
$$\left \vert \zeta(k+it) \right \vert \ll (1+\vert t \vert)^{1-k} \log (1+ \vert t \rvert)$$ where $k > 0$.
First we will look at the error
$$ \sum_n \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right)$$
split this into two pieces. Say $\vert n - x \vert \leq 0.1x$ and $\vert n - x \vert > 0.1x$.
In the case, $\vert n - x \vert > 0.1x$, the error is $$ \sum_{n:\vert n - x \vert > 0.1x} \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^c}{T} \sum_n \dfrac{\sigma(n)}{n^{c-1}} = \dfrac{x^c}{T} \zeta(c-1) \zeta(c-2)$$
Choosing $c = 3 + \dfrac1{\log x}$.
$$ \sum_{n:\vert n - x \vert > 0.1x} \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^\left(3 + 1/\log x\right)}{T} \zeta(2) \zeta(1 + 1/ \log(x)) \ll \dfrac{x^3}{T} \log(x)$$
In the case, $\vert n - x \vert \leq 0.1x$, taking $x = \text{integer} + 1/2$, the error is $$ \sum_{n:\vert n - x \vert \leq 0.1x} \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^c}{T} \sum_{n:\vert n - x \vert \leq 0.1x} \left(\left( \dfrac1{n^{c-2 - \epsilon}}\right) \dfrac1{\log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^3}{T} \log(x)$$
Now we need to evaluate some asymptotic for $\displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds$. Let $d \in (2,3)$, then we have that
$$\displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds = \underbrace{\zeta(2) \dfrac{x^3}3}_{\text{Leading order term}}$$ $$+ \underbrace{\displaystyle \dfrac1{2 \pi i} \int_{d - i T}^{d + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds}_{\text{Vertical Integral}}$$ $$+ \underbrace{\displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{d - i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds + \displaystyle \dfrac1{2 \pi i} \int_{c + i T}^{d + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds}_{\text{Horizontal Integrals}}$$ Let $d = 2+k$, where $k>0$.
Bounds for the vertical integral:
$$\left \lvert \displaystyle \dfrac1{2 \pi i} \int_{d - i T}^{d + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds \right \rvert \leq \dfrac1{2 \pi} \int_{-T}^{T} \vert \zeta(1+k+it) \vert \vert \zeta(k+it) \vert \dfrac{x^{2+k}}{\sqrt{(2+k)^2 + t^2}} dt$$
$$ \ll \dfrac1{2 \pi k}\int_{-T}^{T} \dfrac{(1+\vert t \vert)^{1-k} x^{2+k}}{1 + \vert t \vert} \log(1 + \vert t \vert) dt \ll x^{2+k} T^{1-k} \log (T)$$
Bounds for the horizontal integral:
$$\left \vert \displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{d - i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds \right \vert \leq \dfrac1{2 \pi} \int_c^d \dfrac{\vert \zeta(\sigma -1 + iT) \vert \vert \zeta(\sigma -2 + iT) \vert x^{\sigma}}{\sqrt{T^2 + \sigma^2}} d \sigma \ll x^{2+k} T^{1-k} \log (T) + \dfrac{x^3 \log(x)}{T}$$
Hence, now we have $$\sum_{n \leq x} n \sigma(n) = \dfrac{\pi^2}{18}x^3 + \mathcal{O} \left(x^{2+k} T^{1-k} \log (T) + \dfrac{x^3 \log(x)}{T}\right)$$
Now the optimal choice for $k$ and $T$ is $\epsilon>0$ and $T = \sqrt{x}$ respectively. This gives us an error of $\mathcal{O}(x^{2.5 + \epsilon})$.
Hence, you have that $$\sum_{n \leq x} n \sigma(n) = \dfrac{\pi^2}{18}x^3 + \mathcal{O} \left(x^{2.5+\epsilon}\right)$$
You can also try to use elementary methods using partial summation and hyperbola method but I believe it is a bit too much work and using Perron's formula is simpler.
EDIT Deleted some incorrect stuff which I wrote initially.
A: See my blog post regarding the average of $\sigma(n)$.  This post is a two part series, part I looks at the upper bound, and Part II proves Pétermann's lower bound, which is significantly more difficult.  All results regarding $n\sigma(n)$ follow right away from partial summation.
In Part I, the hyperbola method is used to show that $$\sum_{n\leq x} \sigma(n) =\sum_{n\leq x}\sigma(n)=\frac{\pi^{2}}{12}x^2+O(x\log x),$$ which should be exactly what you are looking for.  From here, partial summation yields $$\sum_{n\leq x} n\sigma(n) =\frac{\pi^{2}}{18}x^3+O(x^2\log x).$$
I will post Part II soon which proves that the error term is not $o(x^2\log \log x),$ and oscillates from negative to positive with a magnitude of $x^2\log \log x$ infinitely often.
