On the average order of $2^{\Omega(n)}$. Let $\Omega(n)$ denote the number of prime factors of $n$ counted with multiplicity (e.g. $\Omega(9)=2$) and define
$$
M(x)=\sum_{n\le x}2^{\Omega(n)},\quad F(s)=\sum_{n\ge1}{2^{\Omega(n)}\over n^s}.
$$
It follows from Euler product formula that $F(s)$ can be rewritten as
$$
F(s)\zeta^2(s){(1-2^{-s})^2\over1-2^{1-s}}\underbrace{\prod_{p>2}\left(1-{2\over p^s}\right)^{-1}\left(1-{1\over p^s}\right)^2}_{H(s)},
$$
where $H(s)$ is a product that converges absolutely when $\Re(s)>\frac12$. Since it is well known that there exists $c>0$ such that in the region $|t|\ge2,\sigma\ge1-c/\log|t|$ we have $\zeta(\sigma+it)=O(\log|t|)$, I decide to estimate $M(x)$ via contour integration.
Applying Perron's formula, I deduce that for $2\le T\le x,\varepsilon>0,\delta=c(\log T)^{-1},a=1+(\log x)^{-1}$ there is
$$
M(x)={1\over2\pi i}\int_{a-iT}^{a+iT}F(s){x^s\over s}\mathrm ds+O\left(x^{1+\varepsilon}\over T\right).
$$
To continue, we move the path of integration to $\Re(s)=1-\delta,|t|\le T$ so that
$$
M(x)={1\over2\pi i}\oint\limits_{|s-1|=(\log x)^{-1}}F(s){x^s\over s}\mathrm ds+O(x^{1-\delta}\log^3T)+O\left(x^{1+\varepsilon}\over T\right).
$$
For the remaining residue integral, using the fact that
$$
F(s)={1\over(s-1)^3}{H(1)\over4\log2}[1+O(|s-1)]
$$
when $s\to1$, we get
$$
M(x)={H(1)\over8\log2}x\log^2x+O(x\log x)+O(x^{1-\delta}\log^3T)+O\left(x^{1+\varepsilon}\over T\right).
$$
However, I am not able to choose an appropriate $T$ so that O terms would grow slower than $x\log^2x$.
I wonder how I can proceed to find the remainder of $M(x)$, and elementary methods to this problem is also welcomed!
 A: As we can see from the Dirichlet series. It is natural for us to isolate the even prime factor of $n$:
$$
M(x)=\sum_{0\le k\le\log_2x}2^k\sum_{\substack{m\le x/2^k\\m\equiv1(2)}}2^{\Omega(m)}.
$$
For the inner sum, we observe its Dirichlet series has the following property:
$$
\sum_{\substack{m\ge1\\m\equiv1(2)}}{2^{\Omega(m)}\over m^s}=\prod_{p>2}\left(1-{2\over p^s}\right)^{-1}=\zeta^2(s)(1-2^{-s})^2H(s),
$$
where $H(s)$ converges absolutely for $\Re(s)>1/2$. This indicates that if $h(n)$ is characterized by
$$
H(s)(1-2^{-s})^2=\sum_{n\ge1}{h(n)\over n^s},
$$
then we see that if $\tau(n)$ denotes the divisor function then
$$
2^{\Omega(m)}=\sum_{n=ab}\tau(a)h(b),
$$
so we have
\begin{aligned}
\sum_{\substack{m\le y\\m\equiv1(2)}}2^{\Omega(m)}
&=\sum_{b\le y}h(b)\sum_{a\le y/b}\tau(a)
=\sum_{b\le y}h(b)\left[\frac yb\log\frac yb+O\left(\frac yb\right)\right] \\
&=y\log y\sum_{b\le y}{h(b)\over b}+O(y)={H(1)\over4}y\log y+O(y).
\end{aligned}
Plugging this result back into $M(x)$, we obtain
\begin{aligned}
M(x)
&={H(1)\over4}x\sum_{0\le k\le\log_2x}\log{x\over2^k}+O(x\log x) \\
&={H(1)\over4}x\sum_{0\le k\le\log_2x}(\log x-k\log2)+O(x\log x) \\
&={H(1)\over8\log2}x\log^2x+O(x\log x).
\end{aligned}
