To address your last comment on evaluating infinite sums of powers of logarithms:
Letting
$$(-1)^n \zeta^{(n)}(s)=\sum_{k=1}^\infty \frac{(\log k)^n}{k^s}$$
the question can be recast as the evaluation of the quantity $(-1)^n \zeta^{(n)}(0)$ for various $n\in \mathbb N$.
Apostol and Choudhury had given a bunch of formulae for evaluating $\zeta^{(n)}(s)$; in particular, there is the formula
$$(-1)^n \zeta^{(n)}(0)=\frac{\Im((-\log\,2\pi-i\pi/2)^{n+1})}{\pi(n+1)}+\frac1{\pi}\sum_{j=1}^{n-1} a_j j!\binom{n}{j}\Im((-\log\,2\pi-i\pi/2)^{n-j})$$
where
$$a_j=c_{j+1}+\sum_{\ell=0}^j \frac{(-1)^\ell \gamma_\ell}{\ell!} c_{j-\ell}$$
$$c_k=-\frac{\gamma c_{k-1}}{k}-\frac1{k}\sum_{\ell=1}^{k-1} (-1)^\ell \zeta(\ell+1) c_{k-\ell-1},\qquad c_0=1$$
the $\gamma_n$ are the Stieltjes constants, and $\gamma=\gamma_0$ is the Euler-Mascheroni Constant.
Here are a few explicit evaluations of the formula:
$$\zeta^{\prime\prime}(0)=\frac{\gamma^2}{2}-\frac{\pi^2}{24}-\frac{(\log (2\pi))^2}{2}+\gamma_1$$
$$-\zeta^{\prime\prime\prime}(0)=-\gamma^3-\frac32 \gamma ^2 \log(2\pi)+\frac{\pi^2}{8}\log(2\pi)+\frac{(\log(2\pi))^3}{2}-3\gamma\gamma _1-3\gamma_1\log (2\pi)-\frac32\gamma_2+\zeta(3)$$
As can be surmised, the closed forms become rather unwieldy as $n$ gets large...
An extension of these results to noninteger powers would involve having to appropriately define the differintegral of the zeta function; going the series route would now involve terms containing the incomplete gamma function, but I've no knowledge of any closed forms for the resulting sum.
Thanks for bringing this to my attention. I checked the lecture tape, and found that what Feynman originally said was, "When I took a small fraction of 1024, as the fraction went to zero, what would the answer be here?" This was edited in 1961 by Robert Leighton into the form it now takes in The Feynman Lectures on Physics. I agree with you that the current version is a bit confusing, and will recommend that it be changed from, "When we take a small fraction Δ of 1024 as Δ approaches zero..." to, "When we take a small fraction Δ/1024 as Δ approaches zero..."
Mike Gottlieb [Editor, The Feynman Lectures on Physics New Millennium Edition]
Best Answer
From this we have:
$$\ln ^k(x)=\sum _{n=k}^{\infty } \frac{(-1)^n (1-x)^n k! S_n^{(k)}}{n!}$$
where: $ S_n^{(k)}$ is Stirling number of the first kind.