$\sum_{p,m\geq 3}(-1)^{m(p-1)/2}e^{-p^my}\log p = O(y^{-1/3})$ Show that for sufficiently small $y$ we have $\sum_{p,m\geq 3}(-1)^{m(p-1)/2}e^{-p^my}\log p = O(y^{-1/3})$ where $m\geq 3$ represents all positive integers from $3$ onwards, while $p\geq 3$ represents all odd primes.
I was thinking of splitting according to the parity of $m$ and the remainder of $p$ mod $4$ and then using partial summation and prime number theorems. However, the part with $m$ even is
$$ \sum_{\substack{m\geq 4 \\ 2\mid m}}\sum_{p>2} e^{-p^my}\log p = \sum_{\substack{m\geq 4 \\ 2\mid m}}\int_0^{\infty}\left[\sum_{2<p\leq t} \log p\right](myt^{m-1}e^{-t^my})dt \leq \sum_{\substack{m\geq 4 \\ 2\mid m}}m\int_0^{\infty}yt^{m}e^{-t^my}dt  \\ = \sum_{\substack{m\geq 4 \\ 2\mid m}}\Gamma\bigg(1+\frac{1}{m}\bigg)y^{-\frac{1}{m}} \leq \sum_{\substack{m\geq 4 \\ 2\mid m}}y^{-\frac{1}{m}}$$
and the latter unfortunately is not $O(y^{-1/3})$ (when I plug in the ratio into Wolfram Alpha, it gives me that it diverges). For odd $m$ it is at least nice that you can cancel things from the main terms arising from the prime number theorem for arithmetic progressions and it might work fine.
Ideas how to fix the approach? Any help appreciated!
 A: Sum over $m$ first. Since $p^k \geqslant 1 + k(p-1)$ for all $k \in \mathbb{N}$ by Bernoulli's inequality we have
$$\sum_{m = 3}^{\infty} e^{-p^my} \leqslant \sum_{k = 0}^{\infty} e^{-p^3y - k(p-1)p^3y} = \frac{e^{-p^3y}}{1 - e^{-(p-1)p^3 y}} \tag{1}$$
for all $y > 0$ and all $p > 1$.
If we keep $y$ away from $0$, the denominator of the right hand side of $(1)$ is bounded away from $0$ and we can brutally estimate
\begin{align}
\sum_{p \geqslant 3} e^{-p^3y}\log p
&= \int_0^{\infty} e^{-t^3y}\,d\vartheta_3(t) \\
&= \int_0^{\infty} 3t^2y\vartheta_3(t) e^{-t^3y}\,dt \\
&= \int_0^{\infty} y\vartheta_3(\sqrt[3]{u})e^{-uy}\,du \\
&\leqslant 2\int_0^{\infty} y u^{1/3}e^{-uy}\,du \\
&= 2\Gamma \biggl(\frac{4}{3}\biggr)y^{-1/3}
\end{align}
where $\vartheta_3(t)$ is the sum of the logarithms of the odd primes $\leqslant t$ and we used the easy estimate $\vartheta_3(t) \leqslant \vartheta(t) \leqslant 2t$ for all $t > 0$.
However, we want to estimate things for small $y$, so we cannot keep $y$ away from $0$. Nevertheless, the above is useful for that too. First, if $p \equiv 3 \pmod{4}$, then we have an alternating sum, and
$$\Biggl\lvert \sum_{m = 3}^{\infty} (-1)^m e^{-p^my}\Biggr\rvert \leqslant e^{-p^3y}\,,$$
thus we have an estimate similar to $(1)$ with denominator bounded away from $0$ uniformly in $y$. Next, if [for $p \equiv 1 \pmod{4}$] on the left hand side of $(1)$ we let the sum start at $m = r \geqslant 3$ where $p^{r+1} > \frac{1}{y}$ we get an estimate
$$\sum_{m = r}^{\infty} e^{-p^my} \leqslant 2e^{-p^ry} \leqslant 2e^{-p^3y}\,.$$
We can then estimate as above and obtain
$$\Biggl\lvert \sum_{(p,m) \in A(y)} (-1)^{m(p-1)/2}e^{-p^my}\log p\Biggr\vert \leqslant 4\Gamma\biggl(\frac{4}{3}\biggr)y^{-1/3}\,,$$
where
$$A(y) = \{(p,m) : p \equiv 3 \pmod{4} \text{ or } p^{m+1}y \geqslant 1\}\,.$$
It remains to show
$$\sum_{\substack{p \equiv 1 \pmod{4} \\ m \geqslant 3 \\ p^{m+1}y < 1}} e^{-p^my}\log p \ll y^{-1/3} \tag{2}$$
for $y < 1/5$, say.
For each fixed $m$ to be considered in that we have
$$\sum_{p^{m+1} < y^{-1}} e^{-p^my}\log p \leqslant \sum_{p < y^{-1/(m+1)}} \log p \leqslant 2\cdot y^{-1/(m+1)} \leqslant 2y^{-1/4}\,,$$
and $5^{m+1} < y^{-1}$ implies $m+1 < \frac{\lvert \log y\rvert}{\log 5}$, thus the sum on the left hand side of $(2)$ is no larger than
$$\frac{2}{\log 5}y^{-1/4}\log \frac{1}{y}$$
which is of smaller order than $y^{-1/3}$.
