Type I: Volterra integral

${\displaystyle \prod _{a}^{b}{\big (}1+f(x)\,dx{\big )}=\lim _{\Delta x\to 0}\prod {\big (}1+f(x_{i})\,\Delta x{\big )}.}$

The type I product integral corresponds to Volterra's original definition.^[2][14][15] The following relationship exists for scalar functions ${\displaystyle f:[a,b]\to \mathbb {R} }$:

${\displaystyle \prod _{a}^{b}{\big (}1+f(x)\,dx{\big )}=\exp \left(\int _{a}^{b}f(x)\,dx\right),}$

which is not a multiplicative operator. (So the concepts of product integral and multiplicative integral are not the same).

The Volterra product integral is most useful when applied to matrix-valued functions or functions with values in a Banach algebra, where the last equality is no longer true (see the references below).

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 ^[16]

Left Product integral

${\displaystyle P(A,D)=\prod _{i=m}^{1}(\mathbb {1} +A(\xi _{i})\Delta t_{i})=(\mathbb {1} +A(\xi _{m})\Delta t_{m})\cdots (\mathbb {1} +A(\xi _{1})\Delta t_{1})}$

With the notation of left products (i.e. normal products applied from left)

${\displaystyle \prod _{a}^{b}(\mathbb {1} +A(t)dt)=\lim _{\max \Delta t_{i}\to 0}P(A,D)}$

Right Product Integral

${\displaystyle P(A,D)^{*}=\prod _{i=1}^{m}(\mathbb {1} +A(\xi _{i})\Delta t_{i})=(\mathbb {1} +A(\xi _{1})\Delta t_{1})\cdots (\mathbb {1} +A(\xi _{m})\Delta t_{m})}$

With the notation of right products (i.e. applied from right)

${\displaystyle (\mathbb {1} +A(t)dt)\prod _{a}^{b}=\lim _{\max \Delta t_{i}\to 0}P(A,D)^{*}}$

Where ${\displaystyle \mathbb {1} }$ is the identity matrix and D is a partition of the interval [a,b] in the Riemann sense, i.e. the limit is over the maximum interval in the partition. Note how in this case time ordering comes evident in the definitions.

For scalar functions, the derivative in the Volterra system is the logarithmic derivative, and so the Volterra system is not a multiplicative calculus and is not a non-Newtonian calculus.^[2]

Type II: geometric integral

${\displaystyle \prod _{a}^{b}f(x)^{dx}=\lim _{\Delta x\to 0}\prod {f(x_{i})^{\Delta x}}=\exp \left(\int _{a}^{b}\ln f(x)\,dx\right),}$

which is called the geometric integral and is a multiplicative operator.

This definition of the product integral is the continuous analog of the discrete product operator

${\displaystyle \prod _{i=a}^{b}}$

(with ${\displaystyle i,a,b\in \mathbb {Z} }$) and the multiplicative analog to the (normal/standard/additive) integral

${\displaystyle \int _{a}^{b}dx}$

(with ${\displaystyle x\in [a,b]}$):

	additive	multiplicative
discrete	${\displaystyle \sum _{i=a}^{b}f(i)}$	${\displaystyle \prod _{i=a}^{b}f(i)}$
continuous	${\displaystyle \int _{a}^{b}f(x)\,dx}$	${\displaystyle \prod _{a}^{b}f(x)^{dx}}$

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:

${\displaystyle \ln \prod _{a}^{b}p(x)^{dx}=\int _{a}^{b}\ln p(x)\,dx.}$

Type III: bigeometric integral

${\displaystyle \prod _{a}^{b}f(x)^{d(\ln x)}=\exp \left(\int _{\ln(a)}^{\ln(b)}\ln f(e^{x})\,dx\right),}$

The type III product integral is called the bigeometric integral and is a multiplicative operator.

Type I: Volterra integral

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 ${\displaystyle f:[a,b]\to \mathbb {R} }$ with corresponding partition ${\displaystyle a=y_{0}<y_{1}<\dots <y_{m}}$ and a tagged partition

${\displaystyle a=x_{0}<x_{1}<\dots <x_{n}=b,\quad x_{0}\leq t_{0}\leq x_{1},x_{1}\leq t_{1}\leq x_{2},\dots ,x_{n-1}\leq t_{n-1}\leq x_{n},}$

one approximation of the "Riemann definition" of the type I product integral is given by^[18]

${\displaystyle \prod _{k=0}^{n-1}\left[{\big (}1+f(t_{k}){\big )}\cdot (x_{k+1}-x_{k})\right].}$

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

${\displaystyle \prod _{k=0}^{n-1}\exp {\big (}f(t_{k})\cdot (x_{k+1}-x_{k}){\big )}.}$

When ${\displaystyle f}$ is a constant function, the limit of the first type of approximation is equal to the second type of approximation.^[19] Notice that in general, for a step function, the value of the second type of approximation doesn't depend on the partition, as long as the partition is a refinement of the partition defining the step function, whereas the value of the first type of approximation does depend on the fineness of the partition, even when it is a refinement of the partition defining the step function.

It turns out that^[20] for any product-integrable function ${\displaystyle f}$, the limit of the first type of approximation equals the limit of the second type of approximation. Since, for step functions, the value of the second type of approximation doesn't depend on the fineness of the partition for partitions "fine enough", it makes sense to define^[21] the "Lebesgue (type I) product integral" of a step function as

${\displaystyle \prod _{a}^{b}{\big (}1+f(x)\,dx{\big )}{\overset {def}{=}}\prod _{k=0}^{m-1}\exp {\big (}f(s_{k})\cdot (y_{k+1}-y_{k}){\big )},}$

where ${\displaystyle y_{0}<a=s_{0}<y_{1}<\dots <y_{n-1}<s_{n-1}<y_{n}=b}$ is a tagged partition, and again ${\displaystyle a=y_{0}<y_{1}<\dots <y_{m}}$ is the partition corresponding to the step function ${\displaystyle f}$. (In contrast, the corresponding quantity would not be unambiguously defined using the first type of approximation.)

This generalizes to arbitrary measure spaces readily. If ${\displaystyle X}$ is a measure space with measure ${\displaystyle \mu }$, then for any product-integrable simple function ${\displaystyle f(x)=\sum _{k=1}^{n}a_{k}I_{A_{k}}(x)}$ (i.e. a conical combination of the indicator functions for some disjoint measurable sets ${\displaystyle A_{0},A_{1},\dots ,A_{m-1}\subseteq X}$), its type I product integral is defined to be

${\displaystyle \prod _{X}{\big (}1+f(x)\,d\mu (x){\big )}{\overset {def}{=}}\prod _{k=0}^{m-1}\exp {\big (}a_{k}\mu (A_{k}){\big )},}$

since ${\displaystyle a_{k}}$ is the value of ${\displaystyle f}$ at any point of ${\displaystyle A_{k}}$. In the special case where ${\displaystyle X=\mathbb {R} }$, ${\displaystyle \mu }$ is Lebesgue measure, and all of the measurable sets ${\displaystyle A_{k}}$ are intervals, one can verify that this is equal to the definition given above for that special case. Analogous to the theory of Lebesgue (classical) integrals, the Volterra product integral of any product-integrable function ${\displaystyle f}$ can be written as the limit of an increasing sequence of Volterra product integrals of product-integrable simple functions.

Taking logarithms of both sides of the above definition, one gets that for any product-integrable simple function ${\displaystyle f}$:

${\displaystyle \ln \left(\prod _{X}{\big (}1+f(x)\,d\mu (x){\big )}\right)=\ln \left(\prod _{k=0}^{m-1}\exp {\big (}a_{k}\mu (A_{k}){\big )}\right)=\sum _{k=0}^{m-1}a_{k}\mu (A_{k})=\int _{X}f(x)\,d\mu (x)\iff }$

${\displaystyle \prod _{X}{\big (}1+f(x)\,d\mu (x){\big )}=\exp \left(\int _{X}f(x)\,d\mu (x)\right),}$

where we used the definition of integral for simple functions. Moreover, because continuous functions like ${\displaystyle \exp }$ can be interchanged with limits, and the product integral of any product-integrable function ${\displaystyle f}$ is equal to the limit of product integrals of simple functions, it follows that the relationship

${\displaystyle \prod _{X}{\big (}1+f(x)\,d\mu (x){\big )}=\exp \left(\int _{X}f(x)\,d\mu (x)\right)}$

holds generally for any product-integrable ${\displaystyle f}$. This clearly generalizes the property mentioned above.

The Volterra product integral is multiplicative as a set function,^[22] which can be shown using the above property. More specifically, given a product-integrable function ${\displaystyle f}$ one can define a set function ${\displaystyle {\cal {V}}_{f}}$ by defining, for every measurable set ${\displaystyle B\subseteq X}$,

${\displaystyle {\cal {V}}_{f}(B){\overset {def}{=}}\prod _{B}{\big (}1+f(x)\,d\mu (x){\big )}{\overset {def}{=}}\prod _{X}{\big (}1+(f\cdot I_{B})(x)\,d\mu (x){\big )},}$

where ${\displaystyle I_{B}(x)}$ denotes the indicator function of ${\displaystyle B}$. Then for any two disjoint measurable sets ${\displaystyle B_{1},B_{2}}$ one has

${\displaystyle {\begin{aligned}{\cal {V}}_{f}(B_{1}\sqcup B_{2})&=\prod _{B_{1}\sqcup B_{2}}{\big (}1+f(x)\,d\mu (x){\big )}\\&=\exp \left(\int _{B_{1}\sqcup B_{2}}f(x)\,d\mu (x)\right)\\&=\exp \left(\int _{B_{1}}f(x)\,d\mu (x)+\int _{B_{2}}f(x)\,d\mu (x)\right)\\&=\exp \left(\int _{B_{1}}f(x)\,d\mu (x)\right)\exp \left(\int _{B_{2}}f(x)\,d\mu (x)\right)\\&=\prod _{B_{1}}(1+f(x)d\mu (x))\prod _{B_{2}}(1+f(x)\,d\mu (x))\\&={\cal {V}}_{f}(B_{1}){\cal {V}}_{f}(B_{2}).\end{aligned}}}$

This property can be contrasted with measures, which are additive set functions.

However the Volterra product integral is not multiplicative as a functional. Given two product-integrable functions ${\displaystyle f,g}$, and a measurable set ${\displaystyle A}$, it is generally the case that

${\displaystyle \prod _{A}{\big (}1+(fg)(x)\,d\mu (x){\big )}\neq \prod _{A}{\big (}1+f(x)\,d\mu (x){\big )}\prod _{A}{\big (}1+g(x)\,d\mu (x){\big )}.}$

Type II: geometric integral

If ${\displaystyle X}$ is a measure space with measure ${\displaystyle \mu }$, then for any product-integrable simple function ${\displaystyle f(x)=\sum _{k=1}^{n}a_{k}I_{A_{k}}(x)}$ (i.e. a conical combination of the indicator functions for some disjoint measurable sets ${\displaystyle A_{0},A_{1},\dots ,A_{m-1}\subseteq X}$), its type II product integral is defined to be

${\displaystyle \prod _{X}f(x)^{d\mu (x)}{\overset {def}{=}}\prod _{k=0}^{m-1}a_{k}^{\mu (A_{k})}.}$

This can be seen to generalize the definition given above.

Taking logarithms of both sides, we see that for any product-integrable simple function ${\displaystyle f}$:

${\displaystyle \ln \left(\prod _{X}f(x)^{d\mu (x)}\right)=\sum _{k=0}^{m-1}\ln(a_{k})\mu (A_{k})=\int _{X}\ln f(x)\,d\mu (x)\iff \prod _{X}f(x)^{d\mu (x)}=\exp \left(\int _{X}\ln f(x)\,d\mu (x)\right),}$

where we have used the definition of the Lebesgue integral for simple functions. This observation, analogous to the one already made above, allows one to entirely reduce the "Lebesgue theory of geometric integrals" to the Lebesgue theory of (classical) integrals. In other words, because continuous functions like ${\displaystyle \exp }$ and ${\displaystyle \ln }$ can be interchanged with limits, and the product integral of any product-integrable function ${\displaystyle f}$ is equal to the limit of some increasing sequence of product integrals of simple functions, it follows that the relationship

${\displaystyle \prod _{X}f(x)^{d\mu (x)}=\exp \left(\int _{X}\ln f(x)\,d\mu (x)\right)}$

holds generally for any product-integrable ${\displaystyle f}$. This generalizes the property of geometric integrals mentioned above.