The following is a rather unrigorous way to obtain the Euler-Maclaurin formula. Consider some $\sum_{n=1}^\infty f(n)$. We may rewrite this as $$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}n^k=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \sum_{n=0}^\infty n^k = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \zeta(-k)$$ Notice, however, that the inner sum is divergent, and it has been replaced by its analytical continuation. I consider this case in general here. The main thing I will take from that link is that, because we have replaced a divergent series by its analytical continuation, we must pick up the pole due to $\zeta$ at $k=-1$. I will return to this point later. For now, let us just look at the sum.

We have that $$\sum f(n) = \sum \frac{f^{k}(0)}{k!} \zeta(-k) = -\sum \frac{f^{(k)}(0)}{k!} \frac{B_{k+1}}{k+1} = -\sum_{k=0}^\infty \frac{f^{(2k-1)}(0)}{(2k)!} B_{2k}$$ We use that the Bernoulli numbers can be written in terms of the zeta function and that all odd Bernoulli numbers are zero except the first one. At this point, we have the E-M formula, but missing the integral. That term comes from the pole of the $\zeta$ function. Consider writing the sum as the contour integral $$\int_{c-i \infty}^{c + i \infty} \frac{1}{e^{2 \pi i k}-1}\frac{f^{(k)}(0)}{k!} \zeta(-k) dk$$ If we take $c<-1$ then picking up the extra residue causes this to evaluate to $-\sum_{k=0}^\infty \frac{f^{(2k-1)}(0)}{(2k)!} B_{2k}+ f^{(-1)}(0)$. If we interpret $f^{(-1)}(0)$ as being the integral $\int_0^\infty f(k)dk$, then we obtain all of the E-M formula.

The Question

I am interested in the following generalization. Consider $\sum_{n=1}^\infty f(n)n^x$. If we run the same argument I gave above with this new series, we will end up with $\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \zeta(-x-k)$. However, notice the location of the residue has changed! Now, the pole is at $k=-(1+x)$. Thus, based on the non-rigorous argument above, we should expect that now the pole contributes the term $(-1)^x \frac{f^{-(1+x)}(0)}{(1+x)!}$. Therefore, we should have that $$\sum_{n=1}^\infty f(n)n^x- \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \zeta(-x-k) = (-1)^x\frac{f^{-(1+x)}(0)}{x!}$$ However, notice that the object on the right involves evaluating the nth integral at non-integer values for $x$ not an integer. I am interested in how this type of fractional integral relates to other fractional integrals. For instance, it is equal to any specific definition of the fractional integral? Are there choices for $f$ that appear to make it unrelated to fractional integrals?

Some thoughts

If we take $f(n) = e^{-n}$ then both series converge. Doing numerical computations with Mathematica, it appears that $$\frac{\sum \frac{f(n)}{n^x} - \sum \frac{(-1)^k \zeta(x-k)}{k!}}{(-x)!} = 1$$ for all x positive and negative, which seems to be a good sign, since this implies $\frac{d^x}{dn^x} e^{-n} = \frac{(-1)^x}{x!}$

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    $\begingroup$ A way to arrive at the additional integral is by an extension of the pascal-matrix to an "infinitesimal subdiagonal", which via a path towards a "zeta"-matrix (providing the integrals of the bernoulli-polynomials) . Using this matrix, the Euler-Maclaurin-formula appears much naturally by change of summation and even the integral appears completely naturally via the harmonic series/$\zeta(1)$. It is perhaps an interesting alternative way to arrive at this; see my index at go.helms-net.de/math/index.htm and there my essay go.helms-net.de/math/binomial_new/EulerMacLaurin.pdf . $\endgroup$ Feb 12, 2023 at 8:34
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    $\begingroup$ (... continued...) The explanation of the "zeta-matrix" is not well included in that essay but can be found in go.helms-net.de/math/binomial_new/04_3_SummingOfLikePowers.pdf (a much explorative piece btw.) - - - Hope this is of interest here. $\endgroup$ Feb 12, 2023 at 8:38
  • $\begingroup$ You can replace $e^{-n}$ with $e^{e^{i\pi}n}$ so then you have $\frac{d^x}{dn^x}[e^{e^{i\pi}n}] = e^{i\pi x} e^{e^{i\pi}n}$ and that sheds light on fractionally whats happening here. I might be misunderstanding but I think the $x!$ correction you added was a typo. $\endgroup$ Jul 27, 2023 at 17:17

1 Answer 1


TLDR: Facts: Yes, those fractional derivatives do appear.

Opinion: there might be a natural choice of which one to use and this problem can give us an opinion on WHAT IS the natural fractional derivative.


There is an elementary construction for the operator you seek: Consider that

$$H[f] = f(s)+f(s+1)+f(s+2)+f(s+3) + \ ... = \frac{1}{1-e^{\frac{d}{ds}}} [f] $$

Now observe that:

$$ H^2[f] = \left( f(s)+f(s+1) +f(s+2) ... \right) + \left(f(s+1)+f(s+2) + .. \right) \ ... \rightarrow $$

$$ H^2[f] = f(s)+2f(s+1)+3f(s+2) +.... $$ So that $$f(1) + f(2) +f(3) ... = H^2[f] - H[f] $$

And in general

$$ H^k(s) = \sum_{n=0}^{\infty} \left[ \begin{pmatrix} n+1 \\ k-1 \end{pmatrix} f(s+n) \right] $$

So there is an alternating sum of these which yields your desired zeta weighted sums.

$$ \sum_{n=1}^{\infty} f(n) = (H^1[f] - H^0(f))_{@s=0} $$ $$ \sum_{n=1}^{\infty} nf(n) = (H^2[f] - H^1[f])_{@s=0} $$ $$ \sum_{n=1}^{\infty} n^2f(n) = (2H^3[f] - H^2[f]+ H^1(f))_{@s=0} $$

There is probably a simple combinatorial closed form for this (I'll get around to finding it later) so that

$$ \sum_{n=1}^{\infty} n^r f(n) = \sum_{k=0}^{\infty} t(r,k)H^{r-k}[f]_{@s=0} $$

So then you translate to the $\frac{d}{ds}$ realm by expanding:

$$ \sum_{n=1}^{\infty} n^r f(n) = \sum_{k=0}^{\infty} t(r,k) \left(\frac{1}{1-e^{\frac{d}{ds}}} \right)^{r-k}[f]_{@s=0} $$

As usual when $r=1$ this recovers the Euler Maclaurin formula. When $r$ is not an integer you will end up with fractional powers in the expansion of:

$$ \left(\frac{1}{1-e^{t}} \right)^{r-k} $$

Corresponding to the fractional derivatives you find. When $r$ is a non negative integer you can end up with multi-integrals (due to the presence of the $\frac{1}{x^n}$ terms in the series expansion of $\left( \frac{1}{1-e^x}\right)^n $. I briefly sketch that kind of phenomenon in bottom part of my answer here


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