## Preamble

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n^k = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \zeta(-k)$$ Notice, however, that the inner sum is divergent, and it has been replaced by its analytical continuation.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 $\frac{d^x}{dn^x} e^{-n} = \frac{(-1)^x}{x!}$

even the integralappears 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 . $\endgroup$