Edit (before anyone fights through it): found the mistake.
Let $k\in \mathbb N$ and $x\in \mathbb R$. I can't figure out what is wrong with the following proof that \[ \sum _{n\leq x}(\log n)^k\log (x/n)\ll _kx(\log x)^{k/2+1}\] which is obviously nonsense since the LHS is \[ \approx x(\log x)^{k}.\] By one of the Perron formulas, the sum in question is \[ =\frac {(-1)^k}{2\pi i}\int _{(c)}\zeta ^{(k)}(s)\frac {x^sds}{s^2}\] for any $c>1$. If for some $c>1$ we could show the bound \[ \zeta ^{(k)}(s)\ll \log x\hspace {10mm}(1)\] for $\sigma =c$ and $t\ll 1$ then the part of the integral with $t\ll 1$ would be \[ \ll x^c\int _{\sigma =c\atop {|t|\ll 1}}\frac {|\zeta ^{(k)}(s)|ds}{|s|^2}\ll x^c\log x.\] Since for large $t$ we have $\zeta ^{(k)}(s)\ll (\log t)^{k+1}$ (I think? In any case, a $\log t$ power?), the part of the integral with $t\gg 1$ is trivially $\ll x^c$. So the whole integral is $\ll x^c\log x$ and on choosing \[ c=1+\frac {k\log \log x}{2\log x}\] we then get the claim, so our task is to establish $(1)$ for $\sigma =c$ and $|t|\ll 1$. So let's suppose $s$ satisfies these conditions and prove $(1)$.
Since \[ \left (\frac {d}{dU}\right )\left [\frac {(\log U)^k}{U^s}\right ]=\frac {k(\log U)^{k-1}}{U^{s+1}}-s\frac {(\log U)^k}{U^{s+1}}\ll \frac {(\log U)^k}{U^{\sigma +1}}\] we have with partial summation, for any $Y>X>0$, \[ \sum _{X<n\leq Y}\frac {(\log n)^k}{n^s}=\frac {(\log Y)^k}{Y^s}\sum _{X<n\leq Y}1-\int _X^Y\left (\frac {d}{dU}\right )\left [ \frac {(\log U)^k}{U^s}\right ]\left (\sum _{X<n\leq U}1\right )dU\] \[ \ll \frac {(\log Y)^k}{Y^{\sigma -1}}+\int _X^Y\frac {(\log U)^kdU}{U^{\sigma }}\] so that letting $Y\rightarrow \infty $ we have \[ \sum _{n>X}\frac {(\log n)^k}{n^s}\ll \int _X^\infty \frac {(\log U)^kdU}{U^{\sigma }}\] and the integral here is I believe \[ \sum _{r=0}^{k-1}\frac {(-1)^rk!}{(k-r)!(1-\sigma )^{r+1}}\left (\frac {(\log U)^{k-r}}{U^{\sigma -1}}\right )_X^\infty +\frac {(-1)^kk!}{(1-\sigma )^k}\int _X^\infty \frac {dU}{U^\sigma }.\] Take $X>x$. Since in particular $\sigma \geq 1+1/\log x\geq 1+1/\log X$, all the terms are \[ \ll \frac {(\log X)^{k+1}}{X^{\sigma -1}}\] so we've shown \[ \sum _{n>X}\frac {(\log n)^k}{n^s}\ll \frac {(\log X)^{k+1}}{X^{\sigma -1}}.\] On the other hand for $U$ large $(\log U)^k/U^{\sigma -1}$ is a decreasing function of $U$ (we worked out the derivative above) so \[ \sum _{n\leq X}\frac {(\log n)^k}{n^s}\ll (\log X)\max _{N\leq X}\left [ \sum _{N<n\leq 2N}\frac {(\log n)^k}{n^\sigma }\right ]\ll (\log X)\max _{N\leq X}\left [ \frac {(\log N)^k}{N^{\sigma -1}}\sum _{N<n\leq 2N}\frac {1}{n}\right ] \ll \log X\] so actually \[ \sum _{n=1}^\infty \frac {(\log n)^k}{n^s}\ll \log X\left (1+\frac {(\log X)^{k}}{X^{\sigma -1}}\right ).\] The LHS is $(-1)^k\zeta ^{(k)}(s)$ and the RHS is, on setting $X=x^2$, \[ \ll \log x\left (1+\frac {(\log x)^{k}}{x^{2(c-1)}}\right )=2\log x\] and we've shown $(1)$.