The sum of fractional powers $\sum\limits_{k=1}^x k^t$.

Question

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.)

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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.)

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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.

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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).

Answer 1

This is just a "what if ?" consideration, not an answer, and I just guess that it might be of some help to your scope. So, flanking the analysis you are conducting, you may consider this alternative development for $S_x(t)$.

• 1st development $$ \begin{gathered} S_x (t) = \sum\limits_{k = 1}^x {k^{\,t} } = \sum\nolimits_{\;k = 1}^{\;x + 1} {k^{\,t} } = \frac{{B_{\,t + 1} (x + 1) - B_{\,t + 1} (1)}} {{t + 1}} = \quad \quad \left( \text{1} \right) \hfill \\ = \sum\nolimits_{\;k = 0}^{\;x} {\left( {k + 1} \right)^{\,t} } = \sum\nolimits_{\;k = 0}^{\;x} {\sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered} t \hfill \\ j \hfill \\ \end{gathered} \right)k^{\,j} } } = \sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered} t \hfill \\ j \hfill \\ \end{gathered} \right)\sum\nolimits_{\;k = 0}^{\;x} {k^{\,j} } } = \hfill \\ = \sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered} t \hfill \\ j \hfill \\ \end{gathered} \right)\left( {\frac{{B_{\,j + 1} (x) - B_{\,j + 1} (0)}} {{j + 1}}} \right)} = \quad \quad \left( 2 \right) \hfill \\ = \sum\nolimits_{\;k = 0}^{\;x} {\sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,l\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\left( \begin{gathered} t \hfill \\ j \hfill \\ \end{gathered} \right)\left\{ \begin{gathered} j \\ l \\ \end{gathered} \right\}k^{\,\underline {\,l\,} } } } = \hfill \\ = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,l\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\left( \begin{gathered} t \hfill \\ j \hfill \\ \end{gathered} \right)\left\{ \begin{gathered} j \\ l \\ \end{gathered} \right\}\frac{{x^{\,\underline {\,l + 1\,} } }} {{l + 1}}} = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,l\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\frac{{t^{\,\underline {\,j\,} } }} {{j!}}\left\{ \begin{gathered} j \\ l \\ \end{gathered} \right\}\frac{{x^{\,\underline {\,l + 1\,} } }} {{l + 1}}} \quad \quad \left( 3 \right) \hfill \\ \end{gathered} $$ where the symbol $\sum\nolimits_{\;k = 1}^{\;x + 1} {}$ indicates the indefinite sum , computed between the indicated bounds, and the curly backets the Stirling N. of 2nd kind.
  To the purpose of derivating vs. $t$ and $x$ , you may replace the falling factorials $t^{\,\underline {\,j\,} } $ and$x^{\,\underline {\,l + 1\,} } $ with the corresponding Stirling devopment in $t^n$ and $x^m$ or with their expression through the Gamma function.
• 2nd development
  You can also write $S_x(t)$ in terms of the Hurwitz zeta function $$ \begin{gathered} S_x (t) = \sum\limits_{k = 1}^x {k^{\,t} } = \sum\nolimits_{\;k = 1}^{\;x + 1} {k^{\,t} } = \hfill \\ = \sum\nolimits_{\;k = 1}^{\;\infty } {k^{\,t} } - \sum\nolimits_{\;k = x + 1}^{\;\infty } {k^{\,t} } = \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + 1} \right)^{\,t} } - \sum\nolimits_{\;j = 0}^{\;\infty } {\left( {j + x + 1} \right)^{\,t} } = \hfill \\ = \zeta ( - t,1) - \zeta ( - t,x + 1)\quad \quad \left( 4 \right) \hfill \\ \end{gathered} $$
• Note concerning the handling of sums and products with non-integer bounds
  First let's note that $$ \begin{gathered} S_x (t) = \sum\limits_{k = 1}^x {k^{\,t} } \quad \Rightarrow \hfill \\ \Rightarrow \quad x^{\,t} = S_{x + 1} (t) - S_{x + 1} (t) = \left( {S_{x + 1} (t) + c(x + 1)} \right) - \left( {S_x (t) + c(x)} \right) \hfill \\ \end{gathered} $$ and $$ \begin{gathered} M_x (t) = \prod\limits_{k = 1}^x {k^{\,k^{\,t} } } = \prod\nolimits_{\;k = 1\;}^{\;x + 1} {k^{\,k^{\,t} } } = \prod\nolimits_{\;k = 0\;}^{\;x} {\left( {k + 1} \right)^{\,\left( {k + 1} \right)^{\,t} } } \quad \Rightarrow \hfill \\ \Rightarrow \quad \left( {x + 1} \right)^{\,\left( {x + 1} \right)^{\,t} } = \frac{{M_{x + 1} (t)}} {{M_x (t)}} = \frac{{c(x + 1)M_{x + 1} (t)}} {{c(x)M_x (t)}} \hfill \\ \end{gathered} $$ with $$ c(x)\;:\quad \text{any}\,\text{periodic}\,\text{function}\text{,}\,\text{with}\,\text{period}\,\;1 $$ Then let's take for example the starting base of your development, we get the following two different "definitions" for $B_t(x+1)$ $$ \begin{array}{*{20}c} {S_x (t) = \frac{{B_{\,t + 1} (x + 1) - B_{\,t + 1} (1)}} {{t + 1}}} \hfill & \begin{gathered} \hfill \\ = \hfill \\ \hfill \\ \end{gathered} \hfill & \begin{gathered} = \sum\limits_{k = 1}^x {k^{\,t} } = \sum\nolimits_{\;k = 1}^{\;x + 1} {k^{\,t} } = \hfill \\ = \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + 1} \right)^{\,t} } - \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t} } \hfill \\ \end{gathered} \hfill \\ \hline \begin{gathered} \quad \quad \quad \quad \Downarrow \hfill \\ \frac{\partial } {{\partial \,x}}S_x (t) = \hfill \\ = \frac{1} {{t + 1}}\frac{\partial } {{\partial \,x}}B_{\,t + 1} (x + 1) = \hfill \\ = B_{\,t} (x + 1) = \hfill \\ = - t\sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t - 1} } \hfill \\ \end{gathered} \hfill & \begin{gathered} | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ \end{gathered} \hfill & \begin{gathered} \quad \quad \quad \quad \Downarrow \hfill \\ \frac{{B_{\,t + 1} (x + 1)}} {{t + 1}} = f(t + 1) - \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t} } \hfill \\ \quad \quad \quad \quad \Downarrow \hfill \\ \hfill \\ B_{\,t} (x + 1) = \hfill \\ = t\,f(t) - t\sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t - 1} } \hfill \\ \end{gathered} \hfill \\ \end{array} $$ where
• the derivate in $x$ is first taken in extending to real index the known property for integer index, and then by derivating the espression of $S(x)$ as difference of the two sums;
• $f(t)$ can be any function in $t$, and in particular it could be $B_t(1)$, which in turn can be taken as $t\;\zeta (1 - t)$, as it is in many papers concerning the extension of Bernoulli polynomials.

Thus it is evident that such mathematical entities shall be handled with great care, and specially when taking derivatives.

Answer 2

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.

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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 $m=0$.