Indefinite sum

In the calculus of finite differences, the indefinite sum operator (also known as the antidifference operator), denoted by ${\textstyle \sum _{x}}$ or ${\displaystyle \Delta ^{-1}}$,^[1][2] is the linear operator that is the inverse of the forward difference operator ${\displaystyle \Delta }$. It relates to the forward difference operator as the indefinite integral relates to the derivative. Thus,^[3]

${\displaystyle \Delta \sum _{x}f(x)=f(x)\,.}$

More explicitly, if ${\textstyle \sum _{x}f(x)=F(x)}$, then

${\displaystyle F(x+1)-F(x)=f(x)\,.}$

The solution is not unique; it is determined only up to an additive periodic function with period 1. Therefore, each indefinite sum represents a family of functions.

Indefinite sums can be used to calculate definite sums with the formula:^[4]

${\displaystyle \sum _{k=a}^{b}f(k)=\Delta ^{-1}f(b+1)-\Delta ^{-1}f(a).}$

The inverse forward difference operator, ${\displaystyle \Delta ^{-1}}$, extends the summation up to ${\displaystyle x-1}$:

${\displaystyle \sum _{k=0}^{x-1}f(k).}$

Some authors analytically extend summation for which the upper limit is the argument without a shift:^[5][6][7]

${\displaystyle \sum _{k=1}^{x}f(k).}$

In this case, a closed-form expression ${\displaystyle F(x)}$ for the sum is a solution of

${\displaystyle F(x+1)-F(x)=f(x+1),}$

which is called the telescoping equation.^[8] It is the inverse of the backward difference operator ${\displaystyle \nabla }$, ${\displaystyle \nabla ^{-1}}$:

${\displaystyle F(x)-F(x-1)=f(x),}$

It is related to the forward antidifference operator using the fundamental theorem of the calculus of finite differences.

The functional equation ${\displaystyle F(x+1)-F(x)=f(x)}$ does not have a unique solution. If ${\displaystyle F_{1}(x)}$ is a particular solution, then for any function ${\displaystyle C(x)}$ satisfying ${\displaystyle C(x+1)=C(x)}$ (i.e., any 1-periodic function), the function ${\displaystyle F_{2}(x)=F_{1}(x)+C(x)}$ is also a solution. Therefore, the indefinite sum operator defines a family of functions differing by an arbitrary 1-periodic component, ${\displaystyle C(x)}$.

To select the unique canonical solution up to an additive constant ${\displaystyle C}$ (instead of up to the additive 1-periodic function ${\displaystyle C(x)}$) one must impose additional constraints.

Suppose ${\displaystyle f(z)}$ is analytic in a vertical strip containing the real axis, and let ${\displaystyle F(z)}$ be an analytic solution of ${\displaystyle F(z+1)-F(z)=f(z)}$ in that strip. To ensure uniqueness, we require ${\displaystyle F(z)}$ to be of minimal growth, specifically to be of exponential type less than ${\displaystyle 2\pi }$ in the imaginary direction. That is, there exist constants ${\displaystyle M>0}$ and ${\displaystyle \epsilon >0}$ such that $${\displaystyle |F(z)|\leq Me^{(2\pi -\epsilon )|\Im (z)|}}$$ as ${\displaystyle |\Im (z)|\to \infty }$.^[9][10]

Now let ${\displaystyle F_{1}(z)}$ and ${\displaystyle F_{2}(z)}$ be two analytic solutions satisfying this growth condition. Their difference ${\displaystyle C(z)=F_{1}(z)-F_{2}(z)}$ is then analytic, 1‑periodic (i.e., ${\displaystyle C(z+1)=C(z)}$), and inherits the same exponential type less than ${\displaystyle 2\pi }$.

A fundamental result in complex analysis states that a non‑constant 1‑periodic entire function must have exponential type at least ${\displaystyle 2\pi }$. This follows from its Fourier series expansion: if ${\displaystyle C(z)}$ is non‑constant, its Fourier series contains a term ${\displaystyle a_{n}e^{2\pi inz}}$ with ${\displaystyle n\neq 0}$, which has type ${\displaystyle 2\pi |n|\geq 2\pi }$. Since ${\displaystyle C(z)}$ has type strictly less than ${\displaystyle 2\pi }$, it cannot contain any such term and therefore must be constant.

Consequently, under this minimal growth condition, any two solutions differ by at most a constant. Hence ${\displaystyle F(z)}$ is unique up to an additive constant ${\displaystyle C}$.

The indefinite product operator, denoted by ${\displaystyle \prod _{x}}$, is the multiplicative analogue of the indefinite sum. If ${\displaystyle \prod _{x}f(x)=F(x)}$, then:

${\displaystyle {\frac {F(x+1)}{F(x)}}=f(x).}$

Its common discrete analog is ${\displaystyle \prod _{k=1}^{x-1}f(k)}$. The two operators are related by:

${\displaystyle \prod _{x}f(x)=\exp \left(\sum _{x}\ln f(x)\right),}$

${\displaystyle \sum _{x}f(x)=\ln \left(\prod _{x}\exp(f(x))\right).}$

The Laplace summation formula is a formal asymptotic expansion (generally convergent only for polynomials) of the inverse forward difference ${\displaystyle \Delta ^{-1}f(x)}$:^[11][12]

${\displaystyle \sum _{x}f(x)=\int _{0}^{x}f(t)dt-\sum _{k=1}^{\infty }{\frac {c_{k}\Delta ^{k-1}f(x)}{k!}}+C}$

where ${\displaystyle c_{k}=\int _{0}^{1}(x)_{k}dx}$ are the Cauchy numbers of the first kind.

${\displaystyle (x)_{k}={\frac {\Gamma (x+1)}{\Gamma (x-k+1)}}}$ is the falling factorial.

The inverse forward difference operator, ${\displaystyle \Delta ^{-1}f(x)}$, can be expressed formally (generally convergent only for polynomials) by its Newton series expansion:

${\displaystyle \sum _{x}f(x)=\sum _{k=1}^{\infty }{\binom {x}{k}}\Delta ^{k-1}f\left(0\right)+C=\sum _{k=1}^{\infty }{\frac {\Delta ^{k-1}f(0)}{k!}}(x)_{k}+C,}$

Given that ${\displaystyle f(x)}$ can be represented by its Maclaurin Series expansion, the Taylor series about ${\displaystyle 0}$, it is sometimes possible to represent the indefinite sum using Bernoulli polynomials because ${\displaystyle \sum _{x}x^{a}={\frac {B_{a+1}(x)}{a+1}}+C}$:

${\displaystyle \sum _{x}f(x)=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}B_{n}(x)+C.}$

If ${\displaystyle f(x)}$ is analytic on the right half-plane and satisfies the decay condition ${\displaystyle \lim _{x\to {+\infty }}f(x)=0}$, the analytic continuation of ${\displaystyle \nabla ^{-1}f(x)=\sum _{k=1}^{x}f(k)}$ is given by:^[5]

${\displaystyle \nabla ^{-1}f(x)=\sum _{n=1}^{\infty }\left(f(n)-f(n+x)\right)+C.}$

This formula is derived from axioms presented in the paper based on fractional sums, which uniquely extends the definition of the summation to complex limits. The decay condition ${\displaystyle \lim _{x\to {+\infty }}f(x)=0}$ represents the simplest case of the general asymptotic requirements for the function ${\displaystyle f(x)}$.

The Euler–Maclaurin formula extends ${\displaystyle \nabla ^{-1}f(x)=\sum _{k=1}^{x}f(k)}$:^[6][9] $${\displaystyle {\begin{aligned}\nabla ^{-1}f(x)&=\int _{1}^{x}f(t)dt+{\frac {f(1)+f(x)}{2}}\\&\quad +\sum _{k=1}^{p}{\frac {B_{2k}}{(2k)!}}\left(f^{(2k-1)}(x)-f^{(2k-1)}(1)\right)+R_{p}+C\end{aligned}}}$$ where ${\displaystyle B_{2k}}$ are the even Bernoulli numbers, ${\displaystyle p}$ is an arbitrary positive integer, and ${\displaystyle R_{p}}$ is the remainder term given by:

${\displaystyle R_{p}=(-1)^{p+1}\int _{1}^{x}f^{(p)}(t){\frac {P_{p}(t)}{p!}}\,dt,}$

with ${\displaystyle P_{p}(t)=B_{p}(t-\lfloor t\rfloor )}$ being the periodized Bernoulli function related to the Bernoulli polynomials.

The indefinite sum ${\displaystyle \nabla ^{-1}f(x)=\sum _{k=1}^{x}f(k)}$ can be analytically continued by applying the standard Abel-Plana formula to the finite sum ${\displaystyle \sum _{k=1}^{n}f(k)}$ and then analytically continuing the integer limit ${\displaystyle n}$ to the variable ${\displaystyle x}$. This yields the formula:^[7] $${\displaystyle {\begin{aligned}\nabla ^{-1}f(x)&=\int _{1}^{x}f(t)dt+{\frac {f(1)+f(x)}{2}}\\&\quad +i\int _{0}^{\infty }{\frac {\left(f(x-it)-f(1-it)\right)-\left(f(x+it)-f(1+it)\right)}{e^{2\pi t}-1}}dt+C\end{aligned}}}$$

This analytic continuation is valid when the conditions for the original formula are met. The sufficient conditions are:^[9][10]

1. Analyticity: ${\displaystyle f(z)}$ must be analytic in the closed vertical strip between ${\displaystyle \Re (z)=1}$ and ${\displaystyle \Re (z)=\Re (x)}$. The formula provides analytic continuation up to, but not beyond, the nearest singularities of ${\displaystyle f}$ to the line ${\displaystyle \Re (z)=1}$.
2. Growth: ${\displaystyle f(z)}$ must be of exponential type less than ${\displaystyle 2\pi }$ in this strip, satisfying ${\displaystyle |f(z)|\leq Me^{(2\pi -\epsilon )|\Im (z)|}}$ for some ${\displaystyle M>0}$, ${\displaystyle \epsilon >0}$ as ${\displaystyle |\Im (z)|\to \infty }$.

The constant term ${\displaystyle C}$, in the context of indefinite sums naturally extending the discrete summation, is often defined based on the respective empty sum.

For the inverse forward difference, ${\displaystyle \Delta ^{-1}f(x)}$, the typical summation equivalent is ${\displaystyle \sum _{k=0}^{x-1}f(k)}$ so the empty sum is when ${\displaystyle \Delta ^{-1}f(0)=0}$ as it correlates to ${\displaystyle \sum _{k=0}^{-1}f(k).}$

For the inverse backward difference, ${\displaystyle \nabla ^{-1}f(x)}$, the typical summation equivalent is ${\displaystyle \sum _{k=1}^{x}f(k)}$ so the empty sum is when ${\displaystyle \nabla ^{-1}f(0)=0}$ as it correlates to ${\displaystyle \sum _{k=1}^{0}f(k).}$

In older texts relating to Bernoulli polynomials (predating more modern analytic techniques) the constant ${\displaystyle C}$ was often fixed using integral conditions.

Let

${\displaystyle F(x)=\sum _{x}f(x)+C}$

Then the constant ${\displaystyle C}$ is fixed from the condition

${\displaystyle \int _{0}^{1}F(x)\,dx=0}$

or

${\displaystyle \int _{1}^{2}F(x)\,dx=0}$

Alternatively, Ramanujan summation can be used:

${\displaystyle \sum _{x\geq 1}^{\Re }f(x)=-f(0)-F(0)}$

or at 1

${\displaystyle \sum _{x\geq 1}^{\Re }f(x)=-F(1)}$

respectively.^[13][14]

Indefinite summation by parts:^[15]

${\displaystyle \sum _{x}f(x)\Delta g(x)=f(x)g(x)-\sum _{x}(g(x)+\Delta g(x))\Delta f(x)}$

${\displaystyle \sum _{x}f(x)\Delta g(x)+\sum _{x}g(x)\Delta f(x)=f(x)g(x)-\sum _{x}\Delta f(x)\Delta g(x)}$

Definite summation by parts:

${\displaystyle \sum _{i=a}^{b}f(i)\Delta g(i)=f(b+1)g(b+1)-f(a)g(a)-\sum _{i=a}^{b}g(i+1)\Delta f(i)}$

If ${\displaystyle T}$ is a period of function ${\displaystyle f(x)}$ then

${\displaystyle \sum _{x}f(Tx)=xf(Tx)+C.}$

If ${\displaystyle T}$ is an antiperiod of function ${\displaystyle f(x)}$, that is ${\displaystyle f(x+T)=-f(x)}$ then

${\displaystyle \sum _{x}f(Tx)=-{\frac {1}{2}}f(Tx)+C.}$

The unique analytic continuation of ${\displaystyle F(z)=\nabla ^{-1}f(z)}$ defined as ${\displaystyle F(x)-F(x-1)=f(x)}$ with exponential type less than ${\displaystyle 2\pi }$ in the imaginary direction where ${\displaystyle f(z)}$ is entire and the constant term ${\displaystyle C}$ is chosen such that ${\displaystyle F(0)=0}$ (the empty sum condition), ${\displaystyle F(z)}$ satisfies a reflection formula.

If ${\displaystyle f(z)}$ is an odd function (${\displaystyle f(-z)=-f(z)}$), the unique analytic continuation ${\displaystyle F(z)}$ satisfies:

${\displaystyle F(z)=F(-1-z).}$

This represents a point symmetry about ${\displaystyle z=-1/2}$.

If ${\displaystyle f(z)}$ is an even function (${\displaystyle f(-z)=f(z)}$), the unique analytic continuation ${\displaystyle F(z)}$ satisfies:

${\displaystyle F(z)+F(-1-z)=F(-1)}$.

For positive integer exponents, Faulhaber's formula can be used. Note that ${\displaystyle x}$ in the result of Faulhaber's formula must be replaced with ${\displaystyle x-1}$ due to the offset, as Faulhaber's formula finds ${\displaystyle \nabla ^{-1}}$ rather than ${\displaystyle \Delta ^{-1}}$.

For negative integer exponents, the indefinite sum is closely related to the polygamma function:

${\displaystyle \sum _{x}{\frac {1}{x^{a}}}={\frac {(-1)^{a-1}\psi ^{(a-1)}(x)}{(a-1)!}}+C,\,a\in \mathbb {N} }$

For fractions not listed in this section, one may use the polygamma function with partial fraction decomposition. More generally,

${\displaystyle \sum _{x}x^{a}={\begin{cases}{\frac {B_{a+1}(x)}{a+1}}+C,&{\text{if }}a\neq -1\\\psi (x)+C,&{\text{if }}a=-1\end{cases}}={\begin{cases}-\zeta (-a,x)+C,&{\text{if }}a\neq -1\\\psi (x)+C,&{\text{if }}a=-1\end{cases}}}$

where ${\displaystyle B_{a}(x)}$ are the Bernoulli polynomials, ${\displaystyle \zeta (s,a)}$ is the Hurwitz zeta function, and ${\displaystyle \psi (z)}$ is the digamma function. This is related to the generalized harmonic numbers.

As the generalized harmonic numbers use reciprocal powers, ${\displaystyle a}$ must be substituted for ${\displaystyle -a}$, and the most common form uses the inverse of the backward difference offset:

${\displaystyle \nabla ^{-1}x^{a}={H_{x}^{(-a)}}=\zeta (-a)-\zeta (-a,x+1).}$

Here, ${\displaystyle \zeta (-a)}$ is the constant ${\displaystyle C}$.

The Bernoulli polynomials are also related via a partial derivative with respect to ${\displaystyle x}$:

${\displaystyle {\frac {\partial }{\partial x}}\left(\sum _{x}x^{a}\right)=B_{a}(x)=-a\zeta (1-a,x).}$

Similarly, using the inverse of the backwards difference operator may be considered more natural, as:

${\displaystyle {\frac {\partial }{\partial x}}\left(\nabla ^{-1}x^{a}\right){\bigg |}_{x=0}=-a\zeta (1-a,x+1){\bigg |}_{x=0}=-a\zeta (1-a)=B_{a}.}$

Further generalization comes from use of the Lerch transcendent:

${\displaystyle \sum _{x}{\frac {z^{x}}{(x+a)^{s}}}=-z^{x}\,\Phi (z,s,x+a)+C,}$

which generalizes the generalized harmonic numbers as ${\displaystyle z\Phi \left(z,s,a+1\right)-z^{x+1}\Phi \left(z,s,x+1+a\right)}$ when taking ${\displaystyle \nabla ^{-1}}$. Additionally, the partial derivative is given by

${\displaystyle {\frac {\partial }{\partial x}}\left(-z^{x}\Phi \left(z,s,x+a\right)\right)=z^{x}\left(s\Phi \left(z,s+1,x+a\right)-\ln(z)\Phi \left(z,s,x+a\right)\right).}$

${\displaystyle \sum _{x}B_{a}(x)=(x-1)B_{a}(x)-{\frac {a}{a+1}}B_{a+1}(x)+C}$

${\displaystyle \sum _{x}a^{x}={\frac {a^{x}}{a-1}}+C}$

${\displaystyle \sum _{x}\log _{b}x=\log _{b}\Gamma (x)+C}$

${\displaystyle \sum _{x}\log _{b}ax=\log _{b}(a^{x-1}\Gamma (x))+C}$

${\displaystyle \sum _{x}\sinh ax={\frac {1}{2}}\operatorname {csch} \left({\frac {a}{2}}\right)\cosh \left({\frac {a}{2}}-ax\right)+C}$

${\displaystyle \sum _{x}\cosh ax={\frac {1}{2}}\operatorname {csch} \left({\frac {a}{2}}\right)\sinh \left(ax-{\frac {a}{2}}\right)+C}$

${\displaystyle \sum _{x}\tanh ax={\frac {1}{a}}\psi _{e^{a}}\left(x-{\frac {i\pi }{2a}}\right)+{\frac {1}{a}}\psi _{e^{a}}\left(x+{\frac {i\pi }{2a}}\right)-x+C}$

where ${\displaystyle \psi _{q}(x)}$ is the q-digamma function.

${\displaystyle \sum _{x}\sin ax=-{\frac {1}{2}}\csc \left({\frac {a}{2}}\right)\cos \left({\frac {a}{2}}-ax\right)+C\,,\,\,a\neq 2n\pi }$

${\displaystyle \sum _{x}\cos ax={\frac {1}{2}}\csc \left({\frac {a}{2}}\right)\sin \left(ax-{\frac {a}{2}}\right)+C\,,\,\,a\neq 2n\pi }$

${\displaystyle \sum _{x}\sin ^{2}ax={\frac {x}{2}}+{\frac {1}{4}}\csc(a)\sin(a-2ax)+C\,\,,\,\,a\neq n\pi }$

${\displaystyle \sum _{x}\cos ^{2}ax={\frac {x}{2}}-{\frac {1}{4}}\csc(a)\sin(a-2ax)+C\,\,,\,\,a\neq n\pi }$

${\displaystyle \sum _{x}\tan ax=ix-{\frac {1}{a}}\psi _{e^{2ia}}\left(x-{\frac {\pi }{2a}}\right)+C\,,\,\,a\neq {\frac {n\pi }{2}}}$

where ${\displaystyle \psi _{q}(x)}$ is the q-digamma function.

$${\displaystyle {\begin{aligned}\sum _{x}\tan x&=ix-\psi _{e^{2i}}\left(x+{\frac {\pi }{2}}\right)+C\\&=-\sum _{k=1}^{\infty }\left(\psi \left(k\pi -{\frac {\pi }{2}}+1-x\right)+\psi \left(k\pi -{\frac {\pi }{2}}+x\right)\right.\\&\quad \left.-\psi \left(k\pi -{\frac {\pi }{2}}+1\right)-\psi \left(k\pi -{\frac {\pi }{2}}\right)\right)+C\end{aligned}}}$$

${\displaystyle \sum _{x}\cot ax=-ix-{\frac {i\psi _{e^{2ia}}(x)}{a}}+C\,,\,\,a\neq {\frac {n\pi }{2}}}$

$${\displaystyle {\begin{aligned}\sum _{x}\operatorname {sinc} x&=\operatorname {sinc} (x-1)\left({\frac {1}{2}}+(x-1)\right.\\&\quad \left.\times \left(\ln(2)+{\frac {\psi ({\frac {x-1}{2}})+\psi ({\frac {1-x}{2}})}{2}}\right.\right.\\&\quad \quad \left.\left.-{\frac {\psi (x-1)+\psi (1-x)}{2}}\right)\right)+C\end{aligned}}}$$

where ${\displaystyle \operatorname {sinc} (x)}$ is the normalized sinc function.

${\displaystyle \sum _{x}\psi (x)=(x-1)\psi (x)-x+C}$

${\displaystyle \sum _{x}\Gamma (x)=(-1)^{x+1}\Gamma (x){\frac {\Gamma (1-x,-1)}{e}}+C}$

where ${\displaystyle \Gamma (s,x)}$ is the incomplete gamma function.

${\displaystyle \sum _{x}(x)_{a}={\frac {(x)_{a+1}}{a+1}}+C}$

where ${\displaystyle (x)_{a}}$ is the falling factorial.

${\displaystyle \sum _{x}\operatorname {sexp} _{a}(x)=\ln _{a}{\frac {(\operatorname {sexp} _{a}(x))'}{(\ln a)^{x}}}+C}$

(see super-exponential function)

• Indefinite product
• Time scale calculus
• List of derivatives and integrals in alternative calculi

• "Difference Equations: An Introduction with Applications", Walter G. Kelley, Allan C. Peterson, Academic Press, 2001, ISBN 0-12-403330-X
• Markus Müller. How to Add a Non-Integer Number of Terms, and How to Produce Unusual Infinite Summations
• Markus Mueller, Dierk Schleicher. Fractional Sums and Euler-like Identities
• S. P. Polyakov. Indefinite summation of rational functions with additional minimization of the summable part. Programmirovanie, 2008, Vol. 34, No. 2.
• "Finite-Difference Equations And Simulations", Francis B. Hildebrand, Prenctice-Hall, 1968