$\newcommand\R{\mathbb R}$Let $f\colon\R^p\to\R$ be a continuous function. For $u=(u_1,\dots,u_p)$ and $v=(v_1,\dots,v_p)$ in $\R^p$, let $[u,v]:=\prod_{r=1}^p[u_r,v_r]$; $u\wedge v:=\big(\min(u_1,v_1),\dots,\min(u_p,v_p)\big)$; $u\vee v:=\big(\max(u_1,v_1),\dots,\max(u_p,v_p)\big)$; $$\int_u^v dx\, f(x):= (-1)^{\sum_{r=1}^p\,1(u_r>v_r) }\int_{[u\wedge v,u\vee v]}dx\,f(x).$$ Let $F\colon\R^p\to\R$ be any antiderivative of $f$, in the sense that $$D_1\cdots D_p F=f,$$ where $D_j$ is the operator of the partial differentiation with respect to the $j$th argument; it is assumed that the result of this repeated partial differentiation does not depend on the order of the arguments with respect to which the partial derivatives are taken. Let $[p]:=\{1,\dots,p\}$. For each set $J\subseteq[p]$, let $|J|$ denote the cardinality of $J$.

Then it is not hard to establish the following multidimensional generalization of the fundamental theorem of calculus (Lemma 5.1): \begin{equation} \int_u^v dx\, f(x)=\sum_{J\subseteq[p]}(-1)^{p-|J|}F(v_J), \end{equation} where $v_J:=\big(v_1\,1(1\in J)+u_1\,1(1\notin J),\dots,v_p\,1(p\in J)+u_p\,1(p\notin J)\big)$.

Has anyone seen this or similar statement elsewhere? (I am only asking about references, not proofs.)

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    $\begingroup$ Is this not Stokes’ Theorem? $\endgroup$ Commented Jan 12, 2021 at 21:54
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    $\begingroup$ @RiversMcForge : I don't see a relation with Stokes' theorem. $\endgroup$ Commented Jan 12, 2021 at 21:57
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    $\begingroup$ It probably generalizes to polytopes (I know this is not what you are asking, but it is just a comment). $\endgroup$
    – Malkoun
    Commented Jan 12, 2021 at 22:48
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    $\begingroup$ @Iosif Stokes theorem relies the integral of $D_ig$ over, say, the box and the integral of $g$ over it's boundary. This should be Stokes theorem applied $p$ times. $\endgroup$ Commented Jan 13, 2021 at 16:57
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    $\begingroup$ @IosifPinelis I do not know, but I would not expect this: in Stokes theorem the dimensions in LHS and RHS differ by 1, here they differ by $p$. $\endgroup$ Commented Jan 13, 2021 at 21:11

2 Answers 2


For an elementary fact like this, which may have been reinvented a thousand times, it is hard to find the first paper where this appeared. However, let me give some missing context. There is a whole industry in constructive quantum field theory and statistical mechanics about related "smart" interpolation formulas or Taylor formulas with integral remainders. These are used to perform so-called cluster expansions. For the OP's identity, there is no loss of generality in taking $u=(0,0,\ldots,0)$ and $v=(1,1,\ldots,1)$. In this case, via Möbius inversion in the Boolean lattice, the formula comes from the following identity.

Let $L$ be a finite set. Let $f:\mathbb{R}^L\rightarrow \mathbb{R}$, $\mathbf{x}=(x_{\ell})_{\ell\in L}\mapsto f(\mathbf{x})$ be a sufficiently smooth function, and let $\mathbf{1}=(1,\ldots,1)\in\mathbb{R}^L$, then $$ f(\mathbf{1})=\sum_{A\subseteq L}\int_{[0,1]^A}d\mathbf{h} \left[\left(\prod_{\ell\in A}\frac{\partial}{\partial x_{\ell}}\right)f\right](\psi_A(\mathbf{h})) $$ where $\psi_A(\mathbf{h})$ is the element $\mathbf{x}=(x_{\ell})_{\ell\in L}$ of $\mathbb{R}^L$ defined from the element $\mathbf{h}=(h_{\ell})_{\ell\in A}$ in $[0,1]^A$ by the rule: $x_{\ell}=0$ if $\ell\notin A$ and $x_{\ell}=h_{\ell}$ if $\ell\in A$. Of course one needs to 1) apply this to all $L$'s which are subsets of $[p]$, 2) use Möbius inversion in the Boolean lattice, and 3) specialize to $L=[p]$, and this gives the OP's identity.

The above formula is the most naive one of its kind used to do a "pair of cubes" cluster expansion. See formula III.1 in the article

A. Abdesselam and V. Rivasseau, "Trees, forests and jungles: a botanical garden for cluster expansions".

It is also explained in words on page 115 of the book

V. Rivasseau, "From Perturbative to Constructive Renormalization".

Now the formula is a particular case of a much more powerful one, namely, Lemma 1 in

A. Abdesselam and V. Rivasseau, "An explicit large versus small field multiscale cluster expansion",

where one sums over "allowed" sequences $(\ell_1,\ldots,\ell_k)$ of arbitrary length of elements of $L$, instead of subsets of $L$. The notion of allowed is based on an arbitrary stopping rule. The above identity corresponds to "allowed"$=$"without repeats", or the stopping rule that one should not tack on an $\ell$ at the end of a sequence where it already appeared. By playing with this kind of choice of stopping rule one can use Lemma 1 of my article with Rivasseau, to prove the Hermite-Genocchi formula, the anisotropic Taylor formula by Hairer in Appendix A of "A theory of regularity structures" and many other things. When $f$ is the exponential of a linear form for instance, one can obtain various algebraic identities as in the MO posts

rational function identity

Identity involving sum over permutations

I forgot to mention, one can use Lemma 1 to derive the Taylor formula from calculus 1. This corresponds to $L$ having one element and defining allowed sequences as the ones of length at most $n$. See


  • $\begingroup$ Thank you for this wealth of information. $\endgroup$ Commented Jan 13, 2021 at 21:33

The $p=2$ dimensional case is an exercise in Rogawski's calculus textbook. It is exercise 47 on page 885, section 15.1 (Integration in Several Variables) in the 2008 Early Transcendentals edition.

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    $\begingroup$ The $p=2$ case is also stated in the proof of Lemma 6.2 of this 1999 paper of Carbery and Wright: mathscinet.ams.org/mathscinet-getitem?mr=1683156 , with the three word proof "By Stokes' theorem". $\endgroup$
    – Terry Tao
    Commented Jan 13, 2021 at 17:10
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    $\begingroup$ Correction: that paper is by Carbery, Christ, and Wright. The general case is implicit in a followup paper mathscinet.ams.org/mathscinet-getitem?mr=1928871 of Katz, Krop, and Maggioni in 2002, proved by "a simple application of the fundamental theorem of calculus". $\endgroup$
    – Terry Tao
    Commented Jan 13, 2021 at 17:30
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    $\begingroup$ @ZachTeitler : Thank you for the reference. $\endgroup$ Commented Jan 13, 2021 at 20:50
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    $\begingroup$ @TerryTao : Thank you for the references. $\endgroup$ Commented Jan 13, 2021 at 20:51
  • $\begingroup$ The $p = 2$ case is also well-known for people who work in wave equations (for I hope obvious reasons). For example, it is used lots when studying the spherically symmetric Einstein-scalar-field model in mathematical relativity; I'd say every other paper there uses this formula either explicitly or implicitly. $\endgroup$ Commented Jan 14, 2021 at 2:17

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