Using partial summation to bound $\sum n^{-s} \log^\ell n$ I am having trouble obtaining the following estimate in a book:
$$
\sum_{n > X} n^{-s} (\log n)^l = \int_{X}^{\infty} t^{-s} (\log t)^l \ dt + O(|s|(\log X)^{l+1}/X).
$$
Here $s$ is a complex number with real part $>1$.
I would appreciate any suggestions! (I was starting to suspect if there was a typo or not, but I am not sure yet...)
 A: Here is one way of using summation by parts to  better understand this sum. 
I don't know what variant of summation by parts you prefer, but I prefer to base exposition on Riemann-Stieltjes integration, where this is just integration by parts. Then
$$ \sum_{n > X} \frac{\log^\ell n}{n^s} =  \int_X^\infty \frac{\log^\ell t}{t^s} d\lfloor t \rfloor = \int_X^\infty \frac{\log^\ell t}{t^s}dt - \int_X^\infty \frac{\log^\ell t}{t^s} d\{t\},$$
where $\lfloor \cdot \rfloor$ and $\{ \cdot \}$ denote the floor and fractional parts of integers. The first term on the right is the integral term you are looking for. So we only have to bound the remaining term.
Let us integrate by parts.
$$ -\int_X^\infty \frac{\log^\ell t}{t^s} d\{ t \} = -\frac{\log^\ell t}{t^s}\{t\} \bigg|_X^\infty + \int_X^\infty \frac{\ell\log^{\ell - 1} t - s \log^\ell t}{t^{s+1}}\{t\}dt.$$
The first term is $0$, since $\{X\} = 0$ (assuming you choose $X$ to be an integer --- otherwise its generically small) and $s > 1$ means the infinite limit is $0$.
As long as $X$ is sufficiently large (that is, $\log X > k/s$), the numerator term $s\log^\ell t$ dominates. Also, we can bound the contribution of $\{ t \}$ above by $1$. So we need to bound
$$ s\int_X^\infty \frac{\log^\ell t}{t^{s+1}} dt. \tag{1}$$
There are many ways of bounding this integral, and again I'm not sure which most closely falls into the methodology of your book. So I'll just do what feels most natural, which is to transform this into an incomplete gamma function.
Perform the substitution $\log t \mapsto t$ to transform this integral into
$$ s \int_{\log X}^\infty e^{-s t} t^{\ell + 1} \frac{dt}{t}.$$
This isn't quite in gamma form, so perform the substitution $t \mapsto t/s$ to get the integral
$$ s^{-\ell}\int_{s \log X}^\infty e^{-t} t^{\ell + 1} \frac{dt}{t} = s^{-\ell} \Gamma(s \log X, \ell + 1),$$
the incomplete gamma function. Asymptotic bounds for the incomplete gamma function are well-known and frequently used in cases like this. Applying the first term of the asymptotic expansion yields the bound
$$ s \frac{\log^\ell X}{X^s},$$
which is a bit smaller than the bound you asked for.
Notice that this is somewhat akin to saying that we remove a factor of $t^{-1}$ from $(1)$ (for convergence, sort of), and then say that the leftover initial mass approximates the size of the entire integral. I haven't seen this before, but it does agree with intuition.
I do not see a way to bound $(1)$ in an easy way to produce the exact upper bound your book asks for. For $X$ large enough, you might notice it is incredibly easy to beat your book's estimate by saying that $\log^\ell X < X^\epsilon$ for $X$ large enough. So then the integral in $(1)$ is bounded above by
$$ \int_X^\infty t^{-1-s + \epsilon} dt \sim X^{-s + \epsilon} \ll \frac{\log^\ell X}{X}.$$
Good luck.
