Product integral should not be confused with Volterra integral equation.
A product integral is any product-based counterpart of the usual sum-based integral of calculus. The product integral was developed by the mathematician Vito Volterra in 1887 to solve systems of linear differential equations.[1] [2]
f:[a,b]\toR
b | |
\int | |
a |
f(x)dx=\lim\Delta\sumf(xi)\Deltax,
[a,b]
b | |
\prod | |
a |
(1+f(x)dx)=\lim\Delta\prod(1+f(xi)\Deltax).
For the case of
f:[a,b]\to\R
f:[a,b]\toA
A
A
For the commutative case, three distinct definitions are commonplace in the literature, referred to as Type-I, Type-II or geometric, and type-III or bigeometric.[3] [4] Such integrals have found use in epidemiology (the Kaplan–Meier estimator) and stochastic population dynamics. The geometric integral, together with the geometric derivative, is useful in image analysis[5] and in the study of growth/decay phenomena (e.g., in economic growth, bacterial growth, and radioactive decay).[6] [7] The bigeometric integral, together with the bigeometric derivative, is useful in some applications of fractals,[8] [9] [10] [11] and in the theory of elasticity in economics.[4] [12]
The non-commutative case commonly arises in quantum mechanics and quantum field theory. The integrand is generally an operator belonging to some non-commutative algebra. In this case, one must be careful to establish a path-ordering while integrating. A typical result is the ordered exponential. The Magnus expansion provides one technique for computing the Volterra integral. Examples include the Dyson expansion, the integrals that occur in the operator product expansion and the Wilson line, a product integral over a gauge field. The Wilson loop is the trace of a Wilson line. The product integral also occurs in control theory, as the Peano–Baker series describing state transitions in linear systems written in a master equation type form.
The Volterra product integral is most useful when applied to matrix-valued functions or functions with values in a Banach algebra. When applied to scalars belonging to a non-commutative field, to matrixes, and to operators, i.e. to mathematical objects that don't commute, the Volterra integral splits in two definitions.[13]
The left product integral is
1 | |
P(A,D)=\prod | |
i=m |
(1+A(\xii)\Deltati)=(1+A(\xim)\Deltatm) … (1+A(\xi1)\Deltat1)
b | |
\prod | |
a |
(1+A(t)dt)=\lim | |
max\Deltati\to0 |
P(A,D)
The right product integral
m | |
P(A,D) | |
i=1 |
(1+A(\xii)\Deltati)=(1+A(\xi1)\Deltat1) … (1+A(\xim)\Deltatm)
(1+A(t)dt)
b | |
\prod | |
a |
=\lim | |
max\Deltati\to0 |
P(A,D)*
Where
1
The Magnus expansion provides a technique for computing the product integral. It defines a continuous-time version of the Baker–Campbell–Hausdorff formula.
The product integral satisfies a collection of properties defining a one-parameter continuous group; these are stated in two articles showing applications: the Dyson series and the Peano–Baker series.
The commutative case is vastly simpler, and, as a result, a large variety of distinct notations and definitions have appeared. Three distinct styles are popular in the literature. This subsection adopts the product
style\prod
style\int
When the function to be integrated is valued in the real numbers, then the theory reduces exactly to the theory of Lebesgue integration.
The type I product integral corresponds to Volterra's original definition.[2] [14] [15] The following relationship exists for scalar functions
f:[a,b]\toR
b | |
\prod | |
a |
(1+f(x)dx)=
b | |
\exp\left(\int | |
a |
f(x)dx\right),
b | |
\prod | |
a |
f(x)dx=\lim\Delta
\Deltax | |
\prod{f(x | |
i) |
b | |
style\prod | |
i=a |
i,a,b\inZ
b | |
style\int | |
a |
dx
x\in[a,b]
multiplicative | |||||||||||||||
discrete |
f(i) |
f(i) | |||||||||||||
continuous |
f(x)dx |
f(x)dx |
It is very useful in stochastics, where the log-likelihood (i.e. the logarithm of a product integral of independent random variables) equals the integral of the logarithm of these (infinitesimally many) random variables:
ln
b | |
\prod | |
a |
p(x)dx=
b | |
\int | |
a |
lnp(x)dx.
b | |
\prod | |
a |
f(x)d(ln=
ln(b) | |
\exp\left(\int | |
ln(a) |
lnf(ex)dx\right),
For the commutative case, the following results hold for the type II product integral (the geometric integral).
b | |
\prod | |
a |
cdx=cb-a,
b | |
\prod | |
a |
xdx=
bb | |
aa |
{\rme}a-b,
b | |
\prod | |
0 |
xdx=bb{\rme}-b,
b | |
\prod | |
a |
\left(f(x)k\right)dx=
b | |
\left(\prod | |
a |
f(x)dx\right)k,
b | |
\prod | |
a |
\left(cf(x)\right)dx=
| ||||||||||
c |
,
The geometric integral (type II above) plays a central role in the geometric calculus,[16] [3] [17] which is a multiplicative calculus. The inverse of the geometric integral, which is the geometric derivative, denoted
f*(x)
| ||||
f |
\right)
b | |
\prod | |
a |
f*(x)dx=
b | ||
\prod | \exp\left( | |
a |
f'(x) | |
f(x) |
dx\right)=
f(b) | |
f(a) |
,
(fg)*=f*g*.
(f/g)*=f*/g*.
\sqrt[n]{X1X2 … Xn}\underset{n\toinfty}{\longrightarrow}\prodxXdF(x),
where X is a random variable with probability distribution F(x).
Compare with the standard law of large numbers:
X1+X2+ … +Xn | |
n |
\underset{n\toinfty}{\longrightarrow}\intXdF(x).
When the integrand takes values in the real numbers, then the product intervals become easy to work with by using simple functions. Just as in the case of Lebesgue version of (classical) integrals, one can compute product integrals by approximating them with the product integrals of simple functions. The case of Type II geometric integrals reduces to exactly the case of classical Lebesgue integration.
Because simple functions generalize step functions, in what follows we will only consider the special case of simple functions that are step functions. This will also make it easier to compare the Lebesgue definition with the Riemann definition.
Given a step function
f:[a,b]\toR
a=y0<y1<...<ym
a=x0<x1<...<xn=b, x0\let0\lex1,x1\let1\lex2,...,xn-1\letn-1\lexn,
one approximation of the "Riemann definition" of the type I product integral is given by[18]
n-1 | |
\prod | |
k=0 |
\left[(1+f(tk)) ⋅ (xk+1-xk)\right].
The (type I) product integral was defined to be, roughly speaking, the limit of these products by Ludwig Schlesinger in a 1931 article.
Another approximation of the "Riemann definition" of the type I product integral is defined as
n-1 | |
\prod | |
k=0 |
\exp(f(tk) ⋅ (xk+1-xk)).
When
f
It turns out that[20] for any product-integrable function
f
b | |
\prod | |
a |
(1+f(x)dx)\overset{def}{=}
m-1 | |
\prod | |
k=0 |
\exp(f(sk) ⋅ (yk+1-yk)),
where
y0<a=s0<y1<...<yn-1<sn-1<yn=b
a=y0<y1<...<ym
f
This generalizes to arbitrary measure spaces readily. If
X
\mu
f(x)=
n | |
\sum | |
k=1 |
ak
I | |
Ak |
(x)
A0,A1,...,Am-1\subseteqX
\prodX(1+f(x)d\mu(x))\overset{def}{=}
m-1 | |
\prod | |
k=0 |
\exp(ak\mu(Ak)),
since
ak
f
Ak
X=R
\mu
Ak
f
Taking logarithms of both sides of the above definition, one gets that for any product-integrable simple function
f
ln\left(\prodX(1+f(x)d\mu(x))\right)=ln\left(
m-1 | |
\prod | |
k=0 |
\exp(ak\mu(Ak))\right)=
m-1 | |
\sum | |
k=0 |
ak\mu(Ak)=\intXf(x)d\mu(x)\iff
\prodX(1+f(x)d\mu(x))=\exp\left(\intXf(x)d\mu(x)\right),
where we used the definition of integral for simple functions. Moreover, because continuous functions like
\exp
f
\prodX(1+f(x)d\mu(x))=\exp\left(\intXf(x)d\mu(x)\right)
holds generally for any product-integrable
f
The Type I integral is multiplicative as a set function,[22] which can be shown using the above property. More specifically, given a product-integrable function
f
{\calV}f
B\subseteqX
{\calV}f(B)\overset{def}{=}\prodB(1+f(x)d\mu(x))\overset{def}{=}\prodX(1+(f ⋅ IB)(x)d\mu(x)),
where
IB(x)
B
B1,B2
\begin{align} {\calV}f(B1\sqcupB2)&=
\prod | |
B1\sqcupB2 |
(1+f(x)d\mu(x))\\ &=\exp\left(
\int | |
B1\sqcupB2 |
f(x)d\mu(x)\right)\\ &=\exp\left(
\int | |
B1 |
f(x)d\mu(x)+
\int | |
B2 |
f(x)d\mu(x)\right)\\ &=\exp\left(
\int | |
B1 |
f(x)d\mu(x)\right)\exp\left(
\int | |
B2 |
f(x)d\mu(x)\right)\\ &=
\prod | |
B1 |
(1+f(x)d\mu(x))
\prod | |
B2 |
(1+f(x)d\mu(x))\\ &={\calV}f(B1){\calV}f(B2). \end{align}
This property can be contrasted with measures, which are sigma-additive set functions.
However, the Type I integral is not multiplicative as a functional. Given two product-integrable functions
f,g
A
\prodA(1+(fg)(x)d\mu(x)) ≠ \prodA(1+f(x)d\mu(x))\prodA(1+g(x)d\mu(x)).
If
X
\mu
f(x)=
n | |
\sum | |
k=1 |
ak
I | |
Ak |
(x)
A0,A1,...,Am-1\subseteqX
\prodXf(x)d\mu(x)\overset{def}{=}
m-1 | |
\prod | |
k=0 |
\mu(Ak) | |
a | |
k |
.
This can be seen to generalize the definition given above.
Taking logarithms of both sides, we see that for any product-integrable simple function
f
ln\left(\prodXf(x)d\mu(x)\right)=
m-1 | |
\sum | |
k=0 |
ln(ak)\mu(Ak)=\intXlnf(x)d\mu(x)\iff\prodXf(x)d\mu(x)=\exp\left(\intXlnf(x)d\mu(x)\right),
where the definition of the Lebesgue integral for simple functions was used. This observation, analogous to the one already made for Type II integrals above, allows one to entirely reduce the "Lebesgue theory of type II geometric integrals" to the Lebesgue theory of (classical) integrals. In other words, because continuous functions like
\exp
ln
f
\prodXf(x)d\mu(x)=\exp\left(\intXlnf(x)d\mu(x)\right)
holds generally for any product-integrable
f