Analytic continuation of $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$?

Question

Is there an analytic continuation of the generalised harmonic number $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$ to the positive reals x, for $k>1$?

I can’t find anything useful through Google, just some dead-ends relating to the polygamma function, but not actually yielding $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$ at the positive integers, and therefore incorrect. If possible, I’d appreciate a reference to the relevant literature as well.

UPDATE:

@metamorphy points to the following link:

Wikipedia - polygamma function - recurrence relation

Using my own variables rather than Wikipedia's, this gives

$$H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}=\zeta(k)-\frac{\psi^{(k-1)}(x+1)}{(-1)^k(k-1)!}$$

However this continuation only works for integer values of $k$, where I was hoping for a function that works for all real $k>1$.

Using Polygamma Functions in combination with the Riemann Zeta Function is one possibility. — mrtaurho, Apr 23 '20 at 18:47
@MariusS.L. Hey there; no need to be mean! If you don't know what exactly you're looking for it's quite possible to miss the (maybe) most obvious search possibilites. — mrtaurho, Apr 23 '20 at 19:00
Hi @Marius. Respectfully, the idea of analytic continuation does indeed make sense - it’s how you get the (continuous) gamma function from the (integer-only) factorial function. It isn’t the zeta function I’m after, it’s an analytic continuation of the generalised harmonic number. — Richard Burke-Ward, Apr 23 '20 at 19:01
Hi @metamorphy. Um, yeah, that’ll be the one... Feeling very stupid for limiting searches to ‘harmonic’! I’ll post the answer and close it. — Richard Burke-Ward, Apr 23 '20 at 19:06
Sorry, I didn't intend to be mean. The Gamma function is an integral, or a limit. I still don't see what $\sum_{k=1}^{3/2} 1$ should be. This notation needs a definition - or an integral. — Marius S.L., Apr 23 '20 at 19:06
Hi @Marius. Don’t worry, I didn’t take it as mean, and I apologise if I sounded snippy. Perhaps this will help: https://math.stackexchange.com/questions/3058500/analytic-continuation-of-harmonic-series. Or just search for ‘analytic continuation’ to get a more general understanding of the principle. All the best! — Richard Burke-Ward, Apr 24 '20 at 07:41
Hi @metamorphy, please see my update to the question. The solution you found only works for integer values of the exponent $k$. — Richard Burke-Ward, Apr 24 '20 at 09:21
Sorry, I still don't quite understand. I get the gamma function, but 'interpolating' $(-1)^k$ by $e^{i \pi k}$ gives a complex result for non-integer $k$, whereas $H_x^{(k)}$ is real for real $k$. I tried using Euler's formula instead, but the results for non-integer $k$ were way off-beam. — Richard Burke-Ward, Apr 24 '20 at 10:23
My comment is outdated by my answer below, so I've removed it. — metamorphy, Apr 24 '20 at 10:23

metamorphy · Accepted Answer · 2020-04-24T10:43:30.367

This answer is about an analytic continuation of $H_z^{(s)}$. From the Wikipedia article, one obtains $$H_z^{(s)}=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\frac{1-e^{-zt}}{e^t-1}\,dt$$ which gives an analytic continuation of $H_z^{(s)}$ to $\Re s>0$ and $\Re z>0$ (in fact even $\Re z>-1$).

The same way, an analytic continuation of $H_z^{(s)}$ to $s\in\mathbb{C}\setminus\mathbb{Z}_{>0}$ (and still $\Re z>0$) is given by $$H_z^{(s)}=\frac{\Gamma(1-s)}{2\pi\mathrm{i}}\int_\lambda t^{s-1}\frac{1-e^{zt}}{e^{-t}-1}\,dt$$ where the contour $\lambda$ encircles (closely) the negative real axis (the branch cut of $t^{s-1}$). This is easily shown by taking the limit of "closely" for $\Re s>0$ (and still $s\notin\mathbb{Z}_{>0}$) and using the formula above.

Finally, to handle $\Re z\leqslant 0$, one can simply use the functional equation connecting $H_z^{(s)}$ and $H_{z+1}^{(s)}$.

Analytic continuation of $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$?

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