TITLE: The sum of fractional powers $\sum\limits_{k=1}^x k^t$.
QUESTION [8 upvotes]: This post is a continuation of Generalization of the Bernoulli polynomials ( in relation to the Index ) , the definition of the Bernoulli polynomial $B_t(x)$ with $|x|<1$ has an extension through $B_t(x+1)=B_t(x)+t x^{t-1}$. 
Two equivalent definitions for $B_t(x)$ with $|x|<1$: 
$$B_t(x):=-t\zeta(1-t,x)$$
or 
\begin{align*}  
B_t(x+1):=&-\frac{2\Gamma(1+t)}{(2\pi)^t}\cos \left( \frac{\pi t}{2} \right)  \sum_{k=0}^\infty (-1)^k \frac{(2\pi x)^{2k}}{(2k)!}\zeta(t-2k) \\
&-\frac{2\Gamma(1+t)}{(2\pi)^t}\sin \left( \frac{\pi t}{2} \right)  \sum_{k=0}^\infty (-1)^k \frac{(2\pi x)^{2k+1}}{(2k+1)!}\zeta(t-1-2k) 
\end{align*}
with $-t\in\mathbb{R}\setminus\mathbb{N}$. 
With https://www.researchgate.net/publication/238803313_Bernoulli_numbers_and_polynomials_of_arbitrary_complex_indices , page 86, Theorem 5, using equation (11) with the lower limit of $1$ instead of $0$ ($k=1$ instead of $k=0$)  the formula for the sum of fractional powers is 
$$S_x(t):=\sum\limits_{k=1}^x k^t =\frac{B_{t+1}(x+1)-B_{t+1}(1)}{t+1}$$ with $x\in\mathbb{N}_0$ and $t\in\mathbb{R}_0^+$ (general: $t$ can be complex but I don’t need this possibility here).  
The right side may be differentiated by $x$ and therefore one can write $$\frac{\partial}{\partial x} S_x(t)=B_t(x+1)$$ On the other hand differentiated by $t$ and the definition with $M_x(t):=\prod\limits_{k=1}^x k^{k^t} $ it's $$\ln M_x(t)=\frac{\partial}{\partial t}S_x(t)=\frac{\partial}{\partial t}\frac{B_{t+1}(x+1)-B_{t+1}(1)}{t+1}$$ 
Together one gets (by exchanging the derivatives, which is possible here) $$\frac{\partial}{\partial t}B_t(x+1)=\frac{\partial}{\partial x}\ln M_x(t)$$
Note: 
Perhaps this equation becomes a bit clearer if one looks at $$\frac{\partial}{\partial t}\Delta B_t(x)=\frac{\partial}{\partial x}\Delta \ln M_{x-1}(t)$$ with $\Delta B_t(x):=B_t(x+1)-B_t(x)=tx^{t-1}$ and $\Delta \ln M_x(t):=\ln M_{x+1}(t)-\ln M_x(t)=(x+1)^t\ln(x+1)$.
The problem now is: 
I need a formula for $\ln M_x(t)$ or $M_x(t)$, independend of $B_t(x)$ (otherwise it's a trivial identity), where $x$ and $t$ are variable. It could be a series of (more or less known) functions of $x$ (or perhaps $x$ and $t$) which becomes a sum/term for $t\in\mathbb{N}$ - similar to $B_t(x)$.  
Alternative: To proof that the two definitions above for $B_t(x)$ are indeed equivalent (a link to the literatur is enough).   
Note: 
The Euler-MacLaurin-formula can perhaps give a formula for $\ln M_x(t)$. Does someone know a link, where this is computed ?
Addition:
Maybe http://ac.els-cdn.com/S0377042798001927/1-s2.0-S0377042798001927-main.pdf?_tid=36ead884-7132-11e6-ac53-00000aab0f6b&acdnat=1472837296_60501a990f4d37792d48c76ad38c7e4b , page 198, equation (21), can help. (I will see.)

An application example with $\ln M_x(1)$:
The fourier series of $B_t(x)$ is
$$
\Re \left( \sum\limits_{k=1}^{\infty}{\frac{e^{i2\pi kx}}{\left( ik \right) ^t}} \right) =\frac{\left( 2\pi \right) ^t}{2\Gamma \left( 1+t \right)}B_t\left( x \right) 
$$ 
for $|x|<1$ and $t>0$.
It is known, that $\frac{d}{dx}\ln M_x(1)=-\ln\sqrt{2\pi}+\frac{1}{2}+x+\ln\Gamma(1+x)$.
Using
$$\frac{\partial}{\partial t}B_t(x)|_{t=1}=\frac{d}{dx}\ln M_{x-1}(1)$$ 
and derivating the fourier series of $B_t(x)$ (above) by $t$ and having regard to $(\ln\Gamma(1+t))'|_{t=1}=1-\gamma$ one gets
$$\sum_{k=1}^{\infty}{\frac{\ln k}{k}}\sin \left( 2\pi kx \right) =\frac{\pi}{2}\left( \ln \frac{\Gamma \left( x \right)}{\Gamma \left( 1-x \right)}-\left( 1-2x \right) \left( \gamma +\ln \left( 2\pi \right) \right) \right) 
$$
which can be seen in http://reader.digitale-sammlungen.de/en/fs1/object/display/bsb10525489_00011.html?zoom=1.0 (on the top of page 4) and in http://arxiv.org/pdf/1309.3824.pdf (page 30, formula 65.) 

A  second application example where I use $\frac{d}{dx}\ln M_x(m+1)|_{x=0}$ with $m\in\mathbb{N}_0$: 
Adamchik had computed $$\zeta’(-m)=\frac{B_{m+1}H_m}{m+1}-A_m$$ where $B_n$ are the Bernoulli-numbers, $H_n$ are the harmonic numbers and $A_n$ are the generalized Glaisher-Kinkelin constants. See e.g. 
http://www.sciencedirect.com/science/article/pii/S0377042798001927  (Article; last page, equation (24)) .
Dissolving the equation (5.4) on page 36 of
https://www.fernuni-hagen.de/analysis/docs/bachelorarbeit_aschauer.pdf
for $\ln M_x(k)$, using $\frac{B_{k+1}(x+1+w_2)- B_{k+1}(1+w_2)}{k+1}$ instead of $\sum\limits_{j=1}^x (w_2+j)^k$ and setting $(w_1;w_2):=(1;0)$ results in  
\begin{align*}  
\ln M_x(m)&=H_m\frac{B_{m+1}(x+1)- B_{m+1}(1)}{m+1}+\ln Q_m(x)+ \\
&+\sum_{k=0}^{m-1}\binom{m}{k}(-x)^{m-k}\sum_{v=0}^k \binom{k}{v}x^{k-v}(\ln A_v -\ln Q_v(x))
\end{align*} 
The definition of $Q_m(x)$ is (4.2) on page 13, it’s something like a modified Multiple-Gamma-Function. $\frac{d}{dx}\ln M_x(m)$ can be computed by using the differentiation rule (4.4) for the equation above.
Now one gets with $B_t(1)=-t\zeta(1-t)$ and $\frac{d}{dt}B_t(1)|_{t=m}=\frac{d}{dx}\ln M_x(m)|_{x=0}$ the equation chain 
$$\frac{B_{m+1}(1)}{m+1}+(m+1)\zeta’(-m)= \zeta(-m)+(m+1)\zeta’(-m)=(-t\zeta(1-t))’$$
$$=\frac{d}{dt}B_t(1)|_{t=m+1}=\frac{d}{dx}\ln M_x(m+1)|_{x=0}=H_{m+1}B_{m+1}(1)-(m+1)\ln A_m$$ and this result dissolved for $\zeta’(-m)$ and took into account that $H_{m+1}-\frac{1}{m+1}=H_m$ and $H_m B_{m+1}(1)=H_m B_{m+1}$ for $m\in\mathbb{N}_0$ one gets Adamchik’s result.  

Most simple solution for proofing $\displaystyle \frac{\partial}{\partial t}B_t(x+1)=\frac{\partial}{\partial x}\ln M_x(t)$ 
by using the 2nd development  of G Cab with the 
Hurwitz Zeta function: 
$\zeta(a,b):= \sum\limits_{k=0}^\infty (b+k)^{-a}$
$\displaystyle \frac{B_{t+1}(x+1)-B_{t+1}(1)}{t+1}=S_x(t)=\zeta(-t,1)-\zeta(-t,x+1)$ 
and therefore
$\displaystyle \frac{\partial}{\partial t}S_x(t)=\ln M_x(t)=\sum\limits_{k=0}^\infty (k+1)^t\ln(k+1) - \sum\limits_{k=0}^\infty (k+x+1)^t\ln (k+x+1)$
$\displaystyle \frac{\partial}{\partial x}S_x(t)= B_t(x+1)=-t\zeta(1-t,x+1)\,$ 
(as mentioned by gammatester, first link above)
\begin{align*}  
\frac{\partial}{\partial t}B_t(x+1)&= \frac{\partial}{\partial t}\frac{\partial}{\partial x}(\zeta(-t,1)-\zeta(-t,x+1)) \\
&=\frac{\partial}{\partial x}\frac{\partial}{\partial t}(\zeta(-t,1)-\zeta(-t,x+1))=\frac{\partial}{\partial x}\ln M_x(t) 
\end{align*}
Note:
Substituting $B_t(x)$ and $\ln M_x(t)$ by other formulas are leading to non-trivial equations (as shown in the application examples above).

REPLY [2 votes]: Applying the Euler-Maclaurin Sum Formula
The Euler-Maclaurin Sum Formula can be applied to $k^t$ to get the approximation
$$
\sum_{k=1}^nk^t=\zeta(-t)+\frac1{t+1}n^{t+1}+\frac12n^t+\frac{t}{12}n^{t-1}-\frac{t^3-3t^2+2t}{720}n^{t-3}+O\!\left(n^{t-5}\right)
$$
When $t\lt-1$, this describes how the series for $\zeta(-t)$ converges.

Possible Extension to Non-Integral Summation Limits
Consider
$$
\begin{align}
\lim_{\delta\to0}\frac1\delta\left(\sum_{k=1}^{n+\delta}k^t\color{#C00000}{-\sum_{k=1}^{m+\delta}k^t}\right)
&=\lim_{\delta\to0}\frac1\delta\sum_{k=m+1+\delta}^{n+\delta}k^t\\
&=\lim_{\delta\to0}\frac1\delta\sum_{k=m+1}^n(k+\delta)^t\\
&=t\sum_{k=m+1}^nk^{t-1}
\end{align}
$$
Thus, if we give a meaning to taking a derivative with respect to the upper limit of summation, it would give
$$
\frac{\mathrm{d}}{\mathrm{d}n}\sum_{k=1}^nk^t=t\sum_{k=1}^nk^{t-1}\color{#C00000}{+C}
$$
where $C$ is related to the behavior near $m=0$.