Consider the operator $\frac D{e^D-1}$ which we will call "shadow":

$$\frac {D}{e^D-1}f(x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } e^{-iwx}\frac{-iw}{e^{-i w}-1}\int_{-\infty }^{+\infty } e^{i t w} f(t) \, dt \, dw$$

The integrals here should be understood as Fourier transforms.

Now, intuitively, why the following?

$$\left.\frac {D_x}{e^{D_x}-1} \left[\frac1\pi\ln \left(\frac{x+1/2 +\frac{z}{\pi }}{x+1/2 -\frac{z}{\pi }}\right)\right]\right|_{x=0}=\tan z$$

There are other examples where shadow converts trigonometric functions into inverse trigonometric, logarithms to exponents, etc:

$$\left.\frac {D_x}{e^{D_x}-1} \left[\frac1{\pi }\ln \left(\frac{x+1-\frac{z}{\pi }}{x+\frac{z}{\pi }}\right)\right]\right|_{x=0}=\cot z$$

  • $\begingroup$ Cant we expand $\frac{D}{e^D-1}$ as $\sum_{k=0}^{\infty} {B^{-}_{k}} \frac{D^k}{k!}$? $\endgroup$
    – Alapan Das
    Jan 1, 2021 at 12:54
  • $\begingroup$ @AlapanDas. Yes. Currently there are some mistakes in the formulas, I am fixing this. $\endgroup$
    – Anixx
    Jan 1, 2021 at 12:57
  • $\begingroup$ @AlapanDas fixed typos $\endgroup$
    – Anixx
    Jan 1, 2021 at 13:16
  • 1
    $\begingroup$ I think the second example may not be the most enlightening, given the relation $\cot z = \tan \left(\frac{\pi}{2} - z\right)$. Do you happen to have another? $\endgroup$
    – user44191
    Jan 1, 2021 at 14:02
  • $\begingroup$ @AlapanDas yes, with hyperbolic arctangent, but it also can be derived from this example, I think. Also, with logarithm and exponent, but looks not nice. $\endgroup$
    – Anixx
    Jan 1, 2021 at 14:12

2 Answers 2


This is basically a lightly transformed version of Euler's cotangent partial fraction expansion $$ \pi \cot(\pi z) = \frac{1}{z} + \sum_{n=1}^\infty \frac{1}{z-n} + \frac{1}{z+n}$$ (the log derivative of his famous sine product formula $\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \big(1-\frac{z^2}{n^2}\big)$). By telescoping series one can rewrite this as $$ \pi \cot(\pi z) = \sum_{n=0}^\infty \frac{1}{z-n-1} + \frac{1}{z+n}.$$ By Taylor's theorem, $e^{nD_x}$ is the operation of translation by $n$, so formally by geometric series we have $$ \left.\frac{1}{1-e^{D_x}} f\, \right|_{x=0} = \sum_{n=0}^\infty \left.e^{nD_x} f\right|_{x=0} = \sum_{n=0}^\infty f(n)$$ (which incidentally helps explain the Euler-Maclaurin formula) and so $$ \pi \cot(\pi z) = \left.\frac{1}{1-e^{D_x}} \left(\frac{1}{z-x-1} + \frac{1}{z+x}\right) \right|_{x=0}$$ or equivalently $$ \pi \cot(\pi z) = - \left.\frac{D_x}{1-e^{D_x}} \ln \frac{x+z}{x+1-z} \right|_{x=0}.$$ This gives your identities after some simple rearrangements (and replacing $z$ with either $z/\pi$ or $z/\pi + 1/2$).

The primary reason for Euler's partial fraction identity is that the poles and residues of the cotangent function are easily identified and computed. The reason they can be collapsed into an expression involving the summation operator $\frac{1}{1-e^{D_x}}$ is that these poles and residues enjoy a translation invariance, which ultimately comes from the periodicity of the cotangent function. I would imagine there are similar identities for the Weierstrass $\wp$ function, which is doubly periodic with very specific pole behavior.

  • $\begingroup$ Thanks. We here have a case where essentially inverse trigonometric functions are linked to trigonometric functions. This is a very interesting topic for me, especially given this "shadow operator" (or Todd operator) plays a crucial role here: mathoverflow.net/questions/115743/an-algebra-of-integrals/… So I was asking for a deeper intuition, maybe of algebraic nature. $\endgroup$
    – Anixx
    Jan 5, 2021 at 19:35

The op $$T_x = \frac{D_x}{e^{D_x}-1} = e^{b.D_x},$$

where $(b.)^n = b_n$ are the Bernoulli numbers, is (mod signs) often referred to as the Todd operator (maybe originally given that name by Hirzebruch, who used it to construct his Todd characteristic class).

It has a discretizing (or derivational) property that can be expressed in the following useful ways

$$f(x) = T_x T_x^{-1} f(x) = \frac{D}{e^D-1} \frac{e^D-1}{D} f(x) = T_x \int_{x}^{x+1} f(t) dt$$

$$ = e^{b.D} \;\int_{x}^{x+1} f(t) dt = \int_{b.+x}^{b.+x+1} f(t) dt =\int_{B.(x)}^{B.(x)+1} f(t) dt$$

$$ = F(B.(x)+1) - F(B.(x)) = F(B.(x+1)) - F(B.(x)) = D_x \; F(x),$$


$$B_n(x) = (b.+x)^n = \sum_{k=0}^n \binom{n}{k} \; b_k \; x^{n-k}$$

are the celebrated Appell Bernoulli polynomials, with the e.g.f. $e^{B.(x)t}= e^{(b.+x)t} = \frac{t}{e^t-1}e^{xt}$, and $F(x)$ is the indefinite integral/primitive of $f(x)$. The last equality illustrates the derivational property of the Bernoulli polynomials and completely defines them.

This leads to

$$\sum_{k=0}^n f(x+k) = T \; \int_{x}^{x+n+1} f(t) dt $$

$$ = e^{b.D} \; \int_{x}^{x+n+1} f(t) dt = \int_{B.(x)}^{B.(x+n+1)} f(t) dt$$

$$ = F(B.(x+n+1)) - F(B.(x)),$$

and, in particular, the string of relations

$$\sum_{k=0}^n (x+k)^s =T_x \; \int_{x}^{x+n+1} t^{s} dt $$

$$= e^{b.D} \int_{x}^{x+n+1} t^{s} dt = \int_{B.(x)}^{B.(x+n+1)} t^s dt$$

$$ = T_x \; \frac{(x+n+1+)^{s+1} -x^{s+1}}{s+1} = e^{b.D} \frac{(x+n+1+)^{s+1} -x^{s+1}}{s+1}$$

$$ = \frac{(B.(x+1+n))^{s+1} -(B.(x))^{s+1}}{s+1} = \frac{B_{s+1}(x+1+n) - B_{s+1}(x)}{s+1}$$

$$ = \sum_{k=0}^n \frac{B_{s+1}(x+1+k) - B_{s+1}(x+k)}{s+1}$$

$$ = \sum_{k=0}^n \frac{(B.(x+1+k))^{s+1} - (B.(x+k))^{s+1}}{s+1}$$

$$ = \sum_{k=0}^n D_x \; \frac{(x+k)^{s+1}}{s+1}.$$

If you appropriately take the limit $s \to -1$, you arrive at a relation to the natural logarithm from whence, along with the series expansions of the trig functions in Terry Tao's answer, you can tease out your particular formulas.

For a more sophisticated illustrative application of the discretizing formula, see Eqn. 1, "the Khovanskii-Pukhlikov formula, the combinatorial counterpart to the Hirzebruch-Riemann-Roch formula (HRR) for a smooth toric variety X with a very ample divisor D ... " on page 2 of the "$T_y$-operator on integrals over lattice polytopes" by Goda, Kamimura, and Ohmoto.

Note also the umbral inverse sequence to the Bernoulli polynomials, the Appell power polynomials

$$\hat{B}_n(x) = \frac{(x+1)^{n+1}-x^{n+1}}{n+1},$$

with the .e.g.f. $\frac{e^t-1}{t}\; e^{xt}$, is defined also by the umbral compositional inversion

$$B_n(\hat{B}.(x)) = x^n = \hat{B}_n(B.(x)),$$

so the

  1. derivational property of the Appell Bernoulli polynomials

$$ \frac{(B_.(x)+1)^{n+1}}{n+1} - \frac{(B.(x))^{n+1}}{n+1} = \frac{(b.+x+1)^{n+1} - (b.+x)^{n+1}}{n+1}$$

$$ = \frac{B_{n+1}(x+1) - B_{n+1}(x)}{n+1} = \hat{B}_n(B.(x)) = x^n = D \; \frac{x^{n+1}}{n+1},$$

  1. reciprocal relationship of the defining e.g.f.s of the moments of the inverse pair of Appell polynomial sequences

$$B(t) =e^{b.t}= \frac{t}{e^t-1},$$

$$\hat{B}(t) = e^{\hat{b}.t}=\frac{e^t-1}{t}, $$

  1. reciprocity of the dual ops

$$T= B(D) = \frac{D}{e^D-1} = e^{b.D},$$

$$T^{-1}= \hat{B}(D) = \frac{e^D-1}{D} = e^{\hat{b}.D},$$

  1. dual polynomial generating properties of the ops

$$T \; x^n = \frac{D}{e^D-1} \; x^n = e^{b.D} \; x^n = (b. + x)^n = B_n(x), $$

$$ T^{-1} \; x^n = \frac{e^D-1}{D} \; x^n = e^{\hat{b.}D} x^n = (\hat{b.}+x)^n = \hat{B}_n(x),$$

  1. umbral compositional inverse relationship of the dual sets of polynomials

$$ B_n(\hat{B}.(x)) = T^{-1} \; T \; x^n = x^n = T \; T^{-1} \; x^n = \hat{B}_n(B.(x)),$$

  1. and the discretizing property of the Todd operator

$$ x^n = T \; T^{-1} x^n = T \; \int_{x}^{x+1} t^n \; dt$$

$$ = T \frac{(x+1)^{n+1} - x^{n+1}}{n+1}$$

$$ =\frac{(B.(x)+1)^{n+1} -(B.(x))^{n+1}}{n+1} = \hat{B}_n(B.(x))$$

are all intimately (and productively) interlinked, different facets of an Appell duality, and can be generalized via the Mellin transform. (The inverse $T^{-1}$ to the Todd operator $T$ is called the Bernoulli operator in "Recent Contributions to the Calculus of Finite Differences: A Survey" by Loeb and Rota.)

This isn't the whole story--the relationships run even deeper through a Weyl algebra, Graves/Lie/Pincherle commutator, and ladder ops--but this perspective already leads to fruitful further exploration. For example, we obtain to boot in the limit as $n \to +\infty$ for the discretizing sum a modified Hurwitz zeta function as the generalization (interpolation) of the Bernoulli polynomials,

$$ B_{-s}(x) = s \; \zeta(s+1,x),$$

which inherits the properties of an Appell sequence of polynomials.

The 'shadow' equation is somewhat restrictive since it assumes the FT of $f(x)$ exists, which is not a necessary condition for the discretizing property to apply; e.g., note the similar Laplace transform Abel-Plana formula.

With a different normalization for the FT,

$$FT(f(x)) = \tilde{f}(\omega) = \int_{-\infty}^{\infty} e^{-i 2\pi \omega x} f(x) \; dx,$$


$$f(b.+x) = e^{b.D_x} f(x) = \frac{D_x}{e^{D_x}-1} \; f(x) = \frac{D_x}{e^{D_x}-1} FT^{-1}[\tilde{f}(\omega)]$$

$$ = \frac{D_x}{e^{D_x}-1} \; \int_{-\infty}^{\infty} e^{i 2\pi \omega x} FT[f(x)] \; d\omega = \int_{-\infty}^{\infty} e^{i 2\pi \omega x} \frac{i 2\pi \omega}{e^{i 2\pi \omega}-1} FT[f(x)] \; d\omega. $$

Characterizing the action of the Todd operator using rather the Mellin transform interpolation a la Ramanujan/Hardy, gives an alternate, constructive route to the Hurwitz zeta function:

$$ B_{-s}(z) = (B.(z))^{-s} = (b.+z)^{-s} = e^{b.D_z} \; z^{-s}$$

$$ = e^{b.D_z} \int_{0}^{\infty} e^{-zt} \; \frac{t^{s-1}}{(s-1)!} \; dt$$

$$ = \int_{0}^{\infty} e^{-(b.+z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt$$

$$ =\int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$

$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s+1,z).$$

A series expansion for the Appell Bernoulli function for all real or complex $s$ and real or complex $z$ with $|z-1| < 1$ is given by the umbral binomial expansion

$$s \; \zeta(s+1,z) = B_{-s}(z)$$

$$ = (b.+z)^{-s} = (b. + 1 - 1 + z)^{-s} = (B.(1)+z-1)^{-s}$$

$$ = \sum_{n \geq 0} \binom{-s}{n} B_{-s-n}(1) \; (z-1)^n = \sum_{n \geq 0} \binom{-s}{n} (s+n) \; \zeta(s+n+1) \; (z-1)^n$$


$$(b.+1)^{-s} = (B.(1))^{-s} = B_{-s}(1) = s \; \zeta(s+1,1) = s \; \zeta(s+1)$$

with $\zeta(s)$, the Riemann zeta function.

  • $\begingroup$ Why did you point out the Appell anti-Bernoulli polynomials? This relationship is new to me, but I do not see how it is connected to the rest. Or did you simply show an example of the properties of Bernoulli polynomials? If Todd operator is connected to Umbrial calculus, it would be interesting. $\endgroup$
    – Anixx
    Jan 2, 2021 at 5:42
  • $\begingroup$ Thanks to pointing out the name of this operator. I did not know it. It plays an important role in the theory I am currently working on (there are two variants of the theory, the first variant is described here mathoverflow.net/questions/115743/an-algebra-of-integrals/… . In the theory variant1, $\text{reg }f(\omega_-+x)=T[f](x)$, very straightforward. In the theory variant2, $f(\omega_-+x)=T[f](x)|_{x\to\infty}$ and $f(x)|_{x\to\infty}=\int_{\omega_-}^{\omega_-+1}f(x)dx$. $\endgroup$
    – Anixx
    Jan 2, 2021 at 5:52
  • $\begingroup$ We've crossed paths on this before--see the refs in mathoverflow.net/questions/266046/… and in the latest version of "Computing the continuous discretely" by Beck and Robins the relation between the discretization and the Bernoulli polynomials is presented, but not quite in the form above. $\endgroup$ Jan 2, 2021 at 6:20
  • $\begingroup$ Explicitly write out the umbral compositional inversion of $\hat{B}_2(x)$ and mull over the sketch above, you should be able to see that the discretization, the umbral compositional inversion, the derivation property, and the sum/integral chain are all facets of the same mathematical machinery--calculus via the Bernoulli polynomials and their umbral inverse. Of course a function can be written as a Taylor series in a disc of analyticity. Then acting on the divided powers gives a formal derivative. I've got a monograph on this I'll put on the web soon, detailing and extending earlier notes. $\endgroup$ Jan 2, 2021 at 6:37
  • $\begingroup$ This is very interesting. I wonder if there could be formulation of umbrial calculus in terms of the linked theory of divergent series and integrals. $\endgroup$
    – Anixx
    Jan 2, 2021 at 6:39

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