Generalizations of the Euler–Maclaurin Summation Formula

Question

I'm using the Euler–Maclaurin formula in a research project I'm working on. While brilliant, the elementary proof found in Apostol - An Elementary View of Euler's Summation Formula does not give me enough detail.

Specifically, I would like to get an integral-residue kind of formula for functions which are continuously differentiable only on open intervals. To be precise:

Consider $f:(0,1) \to \mathbb{R}$ a continously differentiable function, and define $$R^N_f := \sum\limits_{m=1}^{N} f\left( \frac{m}{N} \right) - \int\limits_0^1 f(t)~ dt ~ . $$

If $f$ is also continuously differentiable on $[0,1]$, then the Euler–Maclaurin formula gives a closed-form expression for $R_f^N$ (which depends on an unknown point). If the integral on the RHS exists, but the function is not continuously differentiable on the closed interval — what can be said about the error term?

More generally speaking, if no such result exists, I'm interested in

1. Generalizations to broader function spaces than that of analytic functions.
2. Generalizations to Lebesgue integrals with respect to other measures.
3. Reminder theorems for continuously differentiable functions.

Answer 1

As for point (1), maybe the following references will be useful:

• The Euler–Maclaurin formula revisited, by D. Elliott

• The Euler–Maclaurin expansion and finite-part integrals, by G. Monegato, J.N. Lyness and references in it, especially [8] (I. Navot) and [9] (B.W. Ninham).

Answer 2

This isn't an answer but its too long I think to be a normal comment.

The Euler Maclaurin Formula can be seen as a correction to an equation for summation. We will develop it below (somewhat non rigorously) and then use it to motivate a class of generalizations of the Euler-maclaurin formula

Suppose we have some analytic $f(n)$

$$ f(n) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} n^k $$

Then it follows that

$$ \sum_{n=1}^{\infty} f(n) = \sum_{n=1}^{\infty} \left[ \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} n^k \right] $$

Interchanging the summations we have that:

$$ \sum_{n=0}^{\infty} f(n)= \sum_{k=0}^{\infty} \left[ \frac{f^{(k)}(0)}{k!} \sum_{n=1}^{\infty} n^k \right] $$

Now with a jump we can consider that: $\sum_{n=0}^{\infty} n^k = \zeta(-k)$ (which involves the bernoulli numbers for positive integers k) and so we have:

$$ \sum_{n=0}^{\infty} f(n) = \sum_{k=0}^{\infty} \left[ \frac{f^{(k)}(0)}{k!} \zeta(-k) \right] $$

A way to view the Infinite Euler Maclaurin formula is to ask "what correction terms do we need to add to this to make it true generally?" and that's a rather confusing way to go about it but its certainly a way to look at it.

Now the reason you might do it this way is because ultimately this entire set up hinged on $f$ being analytic and therefore having a power series representation. You might ask what if $f$ had a different series representation, theres a lot of ways to go here such as laurent series, puiseux series etc... I will consider one particular example a $\log$ power series and show how this leads to a new Euler-Maclaurin formula:

So instead of looking at functions of the form $\sum_{n=0}^{\infty} a_n x^n$ we instead want to consider functions of the form $\sum_{n=0}^{\infty} a_n \log(x)^n$. we give some examples below:

$$ x = \sum_{n=0}^{\infty} \frac{1}{n!} \log(x)^n $$ $$ \frac{1}{1-x} = -\frac{1}{x} + \frac{1}{2} + \sum_{n=1}^{\infty} \frac{B_n}{n!} \log(x)^n $$

$$ e^{x-1} = \sum_{n=0}^{\infty} B_n \log(x)^n $$

Where the first $B_n$ are the bernoulli numbers and the second $B_n$ are the bell numbers (I'm sorry I know its a little confusing, but just wanted to give some examples).

From here we basically can then repeat our argument. If $f(n)$ has a log power series then

$$ \sum_{n=1}^{\infty} f(n) = \sum_{n=1}^{\infty} \left[ \sum_{k=0}^{\infty} a_k \log(n)^k \right] $$

And here is where it gets interesting. Expressions such as

$$ \log(1)^s + \log(2)^s + \log(3)^s + \log(4)^s + ... = \sum_{n=1}^{\infty} \log(n)^s $$

Can be defined using a log-zeta function. Which we will call $\eta$. It is defined for real $s \ge 0$ as:

$$ \eta(s) = \frac{d^s}{dz^s} [ \zeta(-z)]_{@ z = 0} $$

Then it can be defined for the entire complex plane via analytic continuation. We also define an operator $O = x \frac{d}{dx}$ so for example $O[\log(x)] = x \frac{d}{dx} [ \log(x) ] = 1 $.

It then follows that:

$$ \sum_{n=1}^{\infty} f(n) = \sum_{k=0}^{\infty} \left[ \frac{ O^{(k)}[f(x)]_{@ x = 1} } {k!} \eta(k) \right] $$

Obviously our derivation is less than rigorous in the same ways that our original zeta resummation derivation was not rigorous. The Euler Maclaurin formula was the "correction" of the unrigorous derivation. Similarly we should expect there is a "correction" to this $\log$-zeta series resummation procedure we just cooked up, and this is a direction to generalize or stretch the Euler-Maclaurin formula.

Essentially, what I conjecture is that for every "series" representation of some class of functions you can come up with, there is a corresponding natural Euler-Maclaurin formula that corresponds to that series representation.

This at least attempts to address (1)

My colleague Hamza Chaudhary informed me there is a generalization called Darboux's formula which replaces the bernoulli polynomials with a sequence of integrally connected polynomials. I think that the logarithmic question above can be viewed as a special case of a generalization of Darboux's formula where instead of looking at sequences of polynomials you are more generally concerned with sequences of arbitrary functions from integration (we will need to start with $\int \frac{f}{x}$ to assemble the formula).

The General Theory

(I realized I mixed $\ln$ with $\log$ just assume they are both the same for the rest of this until I can edit this)

To deeply see what's going on we need to set up some slightly abstract machinery. We define the operator $O_Q(x) = \frac{1}{Q'(x)} \frac{d}{dx}$ So that $O_x(x) = \frac{d}{dx}$ and $O_{\log(x)} = x \frac{d}{dx}$. Observe that most "nice" functions (we can call them $Q$-lytic as a generalization of Analytic, so that standard power series are $x$-lytic and log power series are $\log$-lytic) are then given by the $Q$ power series below:

$$ f(x) = \sum_{n=0}^{\infty} \frac{O^n_Q[f]_{@x=a}\left(Q(x)-Q(a)\right)^n}{n!} $$

This can be expanded below:

$$ f(x) = f(a) + O_Q[f]_{@x=a}(Q(x)-Q(a)) + \frac{1}{2!}O_Q^2[f]_{@x=a}\left(Q(x)-Q(a)\right) ^2 + ... $$

To see concretely for $O_x$ and $O_{\log(x)}$ we then have respectively

$$ f(x) = f(a) + O_x[f]_{@x=a}(x-a) + \frac{1}{2!}O_x^2[f]_{@x=a}\left(x-a\right) ^2 + ... $$

$$ f(x) = f(a) + O_{\log(x)}[f]_{@x=a}(\ln(x)-\ln(a)) + \frac{1}{2!}O_{\log(x)}^2[f]_{@x=a}\left(\ln(x)-\ln(a)\right) ^2 + ... $$

The first being the standard taylor series centered at $a$ and the second being its logarithmic equivalent. So now there's a step that we can call commutative renormalization whereby $x$ will be replaced by an expression $T_Q(a,x)$ where $T$ is commutative and $Q(T_Q(a,x)) = Q(x)+Q(a)$ If we can find such a $T$ then:

$$ f(T_Q(a,x)) = f(a) + O_Q[f]_{@x=a}(Q(x)) + \frac{1}{2!}O_Q^2[f]_{@x=a}\left(Q(x)\right) ^2 + ... $$

It's easy to see that $T_Q(a,x)= Q^{-1}\left( Q(a) + Q(b) \right)$ so in the case of $Q(x) =x$ we have $T_Q(a,x) = a+x$. And the case of $Q(x) = \ln(x)$ we have $T_Q(a,x) = ax$. Now because $T_Q$ is commutative we have that all 4 of the following are true:

$$ f(T_Q(a,x)) = f(a) + O_Q[f]_{@x=a}(Q(x)) + \frac{1}{2!}O_Q^2[f]_{@x=a}\left(Q(x)\right) ^2 + ... $$

$$ f(T_Q(x,a)) = f(a) + O_Q[f]_{@x=a}(Q(x)) + \frac{1}{2!}O_Q^2[f]_{@x=a}\left(Q(x)\right) ^2 + ... $$

$$ f(T_Q(x,a)) = f(x) + O_Q[f](Q(a)) + \frac{1}{2!}O_Q^2[f]\left(Q(a)\right) ^2 + ... = e^{Q(a)O_Q} $$

$$ f(T_Q(a,x)) = f(x) + O_Q[f](Q(a)) + \frac{1}{2!}O_Q^2[f]\left(Q(a)\right) ^2 + ... = e^{Q(a)O_Q} $$

The last exponentials being an operator abuse of notation. Pay careful attention to the 3rd equation, this is where we will direct our attention. To see this concretely we have:

$$ f(a+x) = f(x) + O_x[f](a) + \frac{1}{2!}O_x^2[f]\left(a \right)^2 + ... = e^{a \frac{d}{dx}} $$

$$ f(ax) = f(x) + O_{\ln(x)}[f](\ln(a)) + \frac{1}{2!}O_{\ln(x)}^2[f]\left(\ln(a) \right) ^2 + ... = e^{\ln(a) x \frac{d}{dx}}= a^{x \frac{d}{dx}} $$

So now (and we start concrete then go abstract to keep it easy to pattern match) wish to evaluate

$$ f(0+x) + f(1+x) + f(2+x) + ... = \sum_{n=0}^{\infty} e^{n\frac{d}{dx}} = \frac{1}{1-e^{O_x}} = \frac{1}{1-e^{\frac{d}{dx}}}$$

At $x=1$ this gives us the summation we want $\sum_{n=1}^{\infty} f(n)$ . We also on the $\log$-side have

$$ f(1*x) + f(2*x) + f(3*x) + ... = \sum_{n=1}^{\infty} n^{x\frac{d}{dx}} = \zeta(-O_{\ln(x)}) = \zeta\left( -x \frac{d}{dx}\right) $$

At $x=1$ this also gives us the summation that we want $\sum_{n=1}^{\infty} f(n)$.

More generally:

$$ \sum_{n=a}^{\infty} f(T(n,x)) = \sum_{n=a}^{\infty} e^{Q(n)O_Q} $$

Now so far everything has been trivial but abstract so we will make this concrete and produce our actual Euler Maclaurin expansion. Recall the function $\frac{1}{1-e^x}$ has a laurent expansion

$\frac{1}{1-e^x} = -\frac{1}{x} + \frac{1}{2} - \frac{1}{12}x + ... $

So

$$ \sum_{n=1}^{\infty} f(n) = \frac{1}{1-e^{\frac{d}{dx}}}_{@x=1} = -\int_{1}^{\infty} f(x) dx + \frac{f(\infty) - f(1)}{2} - \frac{f'(\infty)-f'(1)}{12} + ... $$

Which is the Euler Maclaurin Formula as we know it. We also have:

$$ \sum_{n=1}^{\infty} f(n) = \zeta\left(-x \frac{d}{dx} \right)_{@x=1} = \zeta(0) \left( f(\infty) - f(1) \right) - \frac{\zeta'(0)}{1!} \left( O_{\ln(x)}[f]_{@x=\infty} - O_{\ln(x)}[f]_{@x=1} \right) + ... $$

Notice here we get a sketch of a proof for @Caleb Brigg's observation, because the laurent series for $\zeta(-s)$ at $s=0$ does not have a singularity, no integral term appears in the $O_{\ln}$ summation. On the other hand because $\frac{1}{1-e^s}$ has a pole of order $1$ at $s=0$ then there must be an integral term to account for it.

Generally speaking the laurent expansion of $\sum_{n=a}^{\infty} e^{Q(n)x}$ will suggest the coefficients and form of its corresponding Euler-Maclaurin formula equivalent.

Answer 3

There are a lot of variants of this formula in the book Interpolation by J. F. Steffensen. They are known as Laplace’s, Gauss’s, Lubbock’s and Woolhouse's Summation-Formulas.

Answer 4

This is mainly in response to @SidharthGhoshal's answer-- I'd like to give my informal approach, which might be helpful in analyzing his approach.

This is a residue-based approach. The idea is that when we replace a divergent series with its analytical continuation, we need to pick up the residue. Running through the E-M formula example, we take $$\sum_n f(n) = \sum_n \sum_k a_k n^k = \sum_k a_k \sum_n n^k$$ We replace the divergent sum $\sum n^k$ with its analytical continuation $ \zeta(-k)$. We need to pick up the residue of $\zeta(-k)$, which occurs at $k=-1$.Taking the argument from here and letting $a_k = \frac{f^{(k)}(0)}{k!}$

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.

If we run this same argument with $\sum \ln(n)^k$, the inner sum can be continued to $\frac{d^z}{dx^z} \zeta(x) \big|_{x=0}$ which has no poles, hence there doesn't need to be a correction term.

For a third example, consider a series of the form $\sum a_k e^{xn}$. For instance, let $f(n)= e^{-e^n}= \sum_{k=0}^\infty \frac{(-1)^k}{k!} e^{nk}$. In general, we have $$\sum_n f(n) = \sum_n \sum_k a_k e^{nk} = \sum_k a_k \frac{1}{1-e^k}$$ The inner sum has a pole at $k=0$ (and at $k=2 \pi i n$, but the other poles contribute only a negligible amount). If we let $a_k = (-1)^k g(k)$, then the residue becomes $2\pi i\cdot \text{Res}\left( \frac{\csc(\pi z)}{2i} \frac{1}{1-e^z} g(k)\right) = \frac{g(0)}{2} - g'(0)$.

In the specific case of $\sum_{n=0}^\infty e^{-e^n}$ we have $\sum_{n=0}^\infty e^{-e^n} \approx \sum_{k=1}^\infty \frac{(-1)^k}{k!} \frac{1}{1-e^k} + (\frac{1}{2} - \gamma)$ where the second term is the main correction term. The remaining terms for exact equality come from the other poles at $2 \pi i n$.

This last example where $f(n) = \sum a_k e^{n x}$ is much more difficult than the other cases, since there are infinitely many correction terms, though in many cases, all but the $k=0$ terms are quite small.

Answer 5

let $F$ be zero on rationals and one on irrational numbers. The Lebesgue integral on $[0,1]$ is $1$. Yet the trapezoidal rule gives $0$, if you use Euler Maclaurin formula.