Fubini's theorem

For the Fubini theorem forcategory,seeKuratowski–Ulam theorem.

Inmathematical analysis,Fubini's theoremcharacterizes the conditions under which it is possible to compute adouble integralby using aniterated integral.It was introduced byGuido Fubiniin 1907. The theorem states that if a function isLebesgue integrableon a rectangle$X\times Y$,then one can evaluate the double integral as an iterated integral:$$\,\iint \limits _{X\times Y}f(x,y)\,{\text{d}}(x,y)=\int _{X}\left(\int _{Y}f(x,y)\,{\text{d}}y\right){\text{d}}x=\int _{Y}\left(\int _{X}f(x,y)\,{\text{d}}x\right){\text{d}}y.$$ This formula is generally not true for theRiemann integral,but it is true if the function is continuous on the rectangle. Inmultivariable calculus,this weaker result is sometimes also called Fubini's theorem, although it was already known byLeonhard Euler.

Tonelli's theorem,introduced byLeonida Tonelliin 1909, is similar but is applied to a non-negativemeasurable functionrather than to an integrable function over its domain. The Fubini and Tonelli theorems are usually combined and form the Fubini-Tonelli theorem, which gives the conditions under which it is possible to switch theorder of integrationin an iterated integral.

A related theorem is often calledFubini's theorem for infinite series,although it is due toAlfred Pringsheim.It states that if${\textstyle \{a_{m,n}\}_{m=1,n=1}^{\infty }}$is a double-indexed sequence of real numbers, and if${\textstyle \displaystyle \sum _{(m,n)\in \mathbb {N} \times \mathbb {N} }a_{m,n}}$is absolutely convergent, then

$\sum _{(m,n)\in \mathbb {N} \times \mathbb {N} }a_{m,n}=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{m,n}=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }a_{m,n}.$

Although Fubini's theorem for infinite series is a special case of the more general Fubini's theorem, it is not necessarily appropriate to characterize the former as being proven by the latter because the properties of measures needed to prove Fubini's theorem proper, in particular subadditivity of measure, may be proven using Fubini's theorem for infinite series.

History

A special case of Fubini's theorem for continuous functions on the product of closed, bounded subsets of real vector spaces was known toLeonhard Eulerin the 18th century. In 1904,Henri Lebesgueextended this result to bounded measurable functions on a product of intervals.Levi conjectured that the theorem could be extended to functions that are integrable rather than boundedand this was proven by Fubini in 1907.In 1909, Leonida Tonelli gave a variation of the Fubini theorem that applies to non-negative functions rather than integrable functions.

Product measures

If$X$and$Y$aremeasure spaces,there are several natural ways to define aproduct measureon the product$X\times Y$.

Inthe sense of category theory,measurable sets in the product$X\times Y$of measure spaces are the elements of theσ-algebragenerated by the products$A\times B$,where$A$is measurable in$X$and$B$is measurable in$Y$.

A measureμonX×Yis called aproduct measureifμ(A×B) =μ₁(A)μ₂(B) for measurable subsetsA⊂XandB⊂Yand measuresμ₁onXandμ₂onY.In general, there may be many different product measures onX×Y.Fubini's theorem and Tonelli's theorem both require technical conditions to avoid this complication; the most common approach is to assume that all measure spaces areσ-finite,in which case there is a unique product measure onX×Y.There is always a unique maximal product measure onX×Y,where the measure of a measurable set is theinfof the measures of sets containing it that are countable unions of products of measurable sets. The maximal product measure can be constructed by applyingCarathéodory's extension theoremto the additive function μ such thatμ(A×B) =μ₁(A)μ₂(B) on the ring of sets generated by products of measurable sets. (Carathéodory's extension theorem gives a measure on a measure space that in general contains more measurable sets than the measure spaceX×Y,so strictly speaking, the measure should be restricted to theσ-algebragenerated by the productsA×Bof measurable subsets ofXandY.)

The product of twocomplete measure spacesis not usually complete. For example, the product of theLebesgue measureon the unit intervalIwith itself is not the Lebesgue measure on the squareI×I.There is a variation of Fubini's theorem for complete measures, which uses the completion of the product of measures rather than the uncompleted product.

For integrable functions

SupposeXandYareσ-finitemeasure spaces and suppose thatX×Yis given the product measure (which is unique asXandYare σ-finite). Fubini's theorem states that iffisX×Yintegrable, meaning thatfis ameasurable functionand $$\int _{X\times Y}|f(x,y)|\,{\text{d}}(x,y)<\infty,$$ then $$\int _{X}\left(\int _{Y}f(x,y)\,{\text{d}}y\right)\,{\text{d}}x=\int _{Y}\left(\int _{X}f(x,y)\,{\text{d}}x\right)\,{\text{d}}y=\int _{X\times Y}f(x,y)\,{\text{d}}(x,y).$$

The first two integrals are iterated integrals with respect to two measures, respectively, and the third is an integral with respect to the product measure. The partial integrals${\textstyle \int _{Y}f(x,y)\,{\text{d}}y}$and${\textstyle \int _{X}f(x,y)\,{\text{d}}x}$need not be defined everywhere, but this does not matter as the points where they are not defined form a set of measure 0.

If the above integral of the absolute value is not finite, then the two iterated integrals may have different values. Seebelowfor an illustration of this possibility.

The condition thatXandYare σ-finite is usually harmless because almost all measure spaces for which one wishes to use Fubini's theorem are σ-finite. Fubini's theorem has some rather technical extensions to the case whenXandYare not assumed to be σ-finite (Fremlin 2003). The main extra complication in this case is that there may be more than one product measure onX×Y.Fubini's theorem continues to hold for the maximal product measure but can fail for other product measures. For example, there is a product measure and a non-negative measurable functionffor which the double integral of |f| is zero but the two iterated integrals have different values; see the section on counterexamples below for an example of this. Tonelli's theorem and the Fubini–Tonelli theorem (stated below) can fail on non σ-finite spaces, even for the maximal product measure.

Tonelli's theorem for non-negative measurable functions

Tonelli's theorem,named afterLeonida Tonelli,is a successor of Fubini's theorem. The conclusion of Tonelli's theorem is identical to that of Fubini's theorem, but the assumption that$|f|$has a finite integral is replaced by the assumption that$f$is a non-negative measurable function.

Tonelli's theorem states that if$(X,A,\mu )$and$(Y,B,\nu )$areσ-finite measure spaces,while$f:X\times Y\to [0,\infty ]$is non-negative measurable function, then $$\int _{X}\left(\int _{Y}f(x,y)\,{\text{d}}y\right)\,{\text{d}}x=\int _{Y}\left(\int _{X}f(x,y)\,{\text{d}}x\right)\,{\text{d}}y=\int _{X\times Y}f(x,y)\,{\text{d}}(x,y).$$

A special case of Tonelli's theorem is in the interchange of the summations, as in${\textstyle \sum _{x}\sum _{y}a_{xy}=\sum _{y}\sum _{x}a_{xy}}$,where$a_{xy}$are non-negative for allxandy.The crux of the theorem is that the interchange of order of summation holds even if the series diverges. In effect, the only way a change in order of summation can change the sum is when there exist some subsequences that diverge to$+\infty$and others diverging to$-\infty$.With all elements non-negative, this does not happen in the stated example.

Without the condition that the measure spaces are σ-finite, all three of these integrals can have different values. Some authors give generalizations of Tonelli's theorem to some measure spaces that are not σ-finite, but these generalizations often add conditions that immediately reduce the problem to the σ-finite case. For example, one could take the σ-algebra onA×Bto be that generated by the product of subsets of finite measure, rather than that generated by all products of measurable subsets, though this has the undesirable consequence that the projections from the product to its factorsAandBare not measurable. Another way is to add the condition that the support offis contained in a countable union of products of sets of finite measures.Fremlin (2003)gives some rather technical extensions of Tonelli's theorem to some non σ-finite spaces. None of these generalizations have found any significant applications outside of abstract measure theory, largely because almost all measure spaces of practical interest are σ-finite.

Fubini–Tonelli theorem

Combining Fubini's theorem with Tonelli's theorem gives the Fubini–Tonelli theorem. Often just called Fubini's theorem, it states that if$X$and$Y$areσ-finite measurespaces, and if$f$is a measurable function, then $$\int _{X}\left(\int _{Y}|f(x,y)|\,{\text{d}}y\right)\,{\text{d}}x=\int _{Y}\left(\int _{X}|f(x,y)|\,{\text{d}}x\right)\,{\text{d}}y=\int _{X\times Y}|f(x,y)|\,{\text{d}}(x,y)$$ Furthermore, if any one of these integrals is finite, then $$\int _{X}\left(\int _{Y}f(x,y)\,{\text{d}}y\right)\,{\text{d}}x=\int _{Y}\left(\int _{X}f(x,y)\,{\text{d}}x\right)\,{\text{d}}y=\int _{X\times Y}f(x,y)\,{\text{d}}(x,y).$$

The absolute value of$f$in the conditions above can be replaced by either the positive or the negative part of$f$;these forms include Tonelli's theorem as a special case as the negative part of a non-negative function is zero and so has finite integral. Informally, all these conditions say that the double integral of$f$is well defined, though possibly infinite.

The advantage of the Fubini–Tonelli over Fubini's theorem is that the repeated integrals of$|f|$may be easier to study than the double integral. As in Fubini's theorem, the single integrals may fail to be defined on a measure 0 set.

For complete measures

The versions of Fubini's and Tonelli's theorems above do not apply to integration on the product of the real line$\mathbb {R}$with itself with Lebesgue measure. The problem is that Lebesgue measure on$\mathbb {R} \times \mathbb {R}$is not the product of Lebesgue measure on$\mathbb {R}$with itself, but rather the completion of this: a product of two complete measure spaces$X$and$Y$is not in general complete. For this reason, one sometimes uses versions of Fubini's theorem for complete measures: roughly speaking, one replaces all measures with their completions. The various versions of Fubini's theorem are similar to the versions above, with the following minor differences:

• Instead of taking a product$X\times Y$of two measure spaces, one takes the completion of some product.
• If$f$is measurable on the completion of$X\times Y$then its restrictions to vertical or horizontal lines may be non-measurable for a measure zero subset of lines, so one has to allow for the possibility that the vertical or horizontal integrals are undefined on a set of measure 0 because they involve integrating non-measurable functions. This makes little difference, because they can already be undefined due to the functions not being integrable.
• One generally also assumes that the measures on$X$and$Y$are complete, otherwise the two partial integrals along vertical or horizontal lines may be well-defined but not measurable. For example, if$f$is the characteristic function of a product of a measurable set and a non-measurable set contained in a measure 0 set then its single integral is well defined everywhere but non-measurable.

Proofs

Proofs of the Fubini and Tonelli theorems are necessarily somewhat technical, as they have to use ahypothesisrelated toσ-finiteness.Most proofs involve building up to the full theorems by proving them for increasingly complicated functions, with the steps as follows.

1. Use the fact that the measure on the product is multiplicative for rectangles to prove the theorems for the characteristic functions of rectangles.
2. Use the condition that the spaces are σ-finite (or some related condition) to prove the theorem for the characteristic functions of measurable sets. This also covers the case of simple measurable functions (measurable functions taking only a finite number of values).
3. Use the condition that the functions are measurable to prove the theorems for positive measurable functions by approximating them by simple measurable functions. This proves Tonelli's theorem.
4. Use the condition that the functions are integrable to write them as the difference of two positive integrable functions and apply Tonelli's theorem to each of these. This proves Fubini's theorem.

Riemann integrals

ForRiemann integrals,Fubini's theorem is proven by refining the partitions along the x-axis and y-axis as to create a joint partition of the form$[x_{i},x_{i+1}]\times [y_{j},y_{j+1}]$,which is a partition over$X\times Y$.This is used to show that the double integrals of either order are equal to the integral over$X\times Y$.

Counterexamples

The following examples show how Fubini's theorem and Tonelli's theorem can fail if any of their hypotheses are omitted.

Failure of Tonelli's theorem for non σ-finite spaces

Suppose thatXis the unit interval with the Lebesgue measurable sets and Lebesgue measure, andYis the unit interval with all the subsets measurable and thecounting measure,so thatYis not σ-finite. Iffis the characteristic function of the diagonal ofX×Y,then integratingfalongXgives the 0 function onY,but integratingfalongYgives the function 1 onX.So, the two iterated integrals are different. This shows that Tonelli's theorem can fail for spaces that are not σ-finite no matter which product measure is chosen. The measures are bothdecomposable,showing that Tonelli's theorem fails for decomposable measures (which are slightly more general than σ-finite measures).

Failure of Fubini's theorem for non-maximal product measures

Fubini's theorem holds for spaces even if they are not assumed to be σ-finite provided one uses the maximal product measure. In the example above, for the maximal product measure, the diagonal has infinite measure so the double integral of |f| is infinite, and Fubini's theorem holds vacuously. However, if we giveX×Ythe product measure such that the measure of a set is the sum of the Lebesgue measures of its horizontal sections, then the double integral of |f| is zero, but the two iterated integrals still have different values. This gives an example of a product measure where Fubini's theorem fails.

This gives an example of two different product measures on the same product of two measure spaces. For products of two σ-finite measure spaces, there is only one product measure.

Failure of Tonelli's theorem for non-measurable functions

Suppose thatXis the first uncountable ordinal, with the finite measure where the measurable sets are either countable (with measure 0) or the sets of countable complement (with measure 1). The (non-measurable) subsetEofX×Xgiven by pairs (x,y) withx<yis countable on every horizontal line and has countable complement on every vertical line. Iffis the characteristic function ofEthen the two iterated integrals offare defined and have different values 1 and 0. The functionfis not measurable. This shows that Tonelli's theorem can fail for non-measurable functions.

Failure of Fubini's theorem for non-measurable functions

A variation of the example above shows that Fubini's theorem can fail for non-measurable functions even if |f| is integrable and both repeated integrals are well defined: if we takefto be 1 onEand –1 on the complement ofE,then |f| is integrable on the product with integral 1, and both repeated integrals are well defined, but have different values 1 and –1.

Assuming the continuum hypothesis, one can identifyXwith the unit intervalI,so there is a bounded non-negative function onI×Iwhose two iterated integrals (using Lebesgue measure) are both defined but unequal. This example was found byWacław Sierpiński(1920). The stronger versions of Fubini's theorem on a product of two unit intervals with Lebesgue measure, where the function is no longer assumed to be measurable but merely that the two iterated integrals are well defined and exist, are independent of the standardZermelo–Fraenkel axiomsofset theory.The continuum hypothesis andMartin's axiomboth imply that there exists a function on the unit square whose iterated integrals are not equal, whileHarvey Friedman(1980) showed that it is consistent with ZFC that a strong Fubini-type theorem for [0,1] does hold, and whenever the two iterated integrals exist they are equal.SeeList of statements undecidable in ZFC.

Failure of Fubini's theorem for non-integrable functions

Fubini's theorem tells us that (for measurable functions on a product of σ-finite measure spaces) if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect toxand then with respect toy,we get the same result as if we integrate first with respect toyand then with respect tox.The assumption that the integral of the absolute value is finite is "Lebesgue integrability",and without it the two repeated integrals can have different values.

A simple example to show that the repeated integrals can be different in general is to take the two measure spaces to be the positive integers, and to take the functionf(x,y) to be 1 ifx=y,−1 ifx=y+ 1, and 0 otherwise. Then the two repeated integrals have different values 0 and 1.

Another example is as follows for the function $${\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}=-{\frac {\partial ^{2}}{\partial x\,\partial y}}\arctan(y/x).$$ Theiterated integrals

$$\int _{x=0}^{1}\left(\int _{y=0}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,{\text{d}}y\right)\,{\text{d}}x={\frac {\pi }{4}}$$ and $$\int _{y=0}^{1}\left(\int _{x=0}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,{\text{d}}x\right)\,{\text{d}}y=-{\frac {\pi }{4}}$$ have different values. The corresponding double integral does notconverge absolutely(in other words the integral of theabsolute valueis not finite): $$\int _{0}^{1}\int _{0}^{1}\left|{\frac {x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}}}\right|\,{\text{d}}y\,{\text{d}}x=\infty.$$

Fubini's theorem in multiplications of integrals

Product of two integrals

For the product of two integrals with lower limit zero and a common upper limit we have the following formula:

${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{1}\int _{0}^{u}x\,v(xy)\,w(x)+x\,v(x)\,w(xy)\,\mathrm {d} x\,\mathrm {d} y$

Proof

Let$V(x)$and$W(x)$are primitive functions of the functions$v(x)$and$w(x)$respectively, which pass through the origin:

$\int _{0}^{u}v(x)\,\mathrm {d} x=V(u),\quad \quad \int _{0}^{u}w(x)\,\mathrm {d} x=W(u)$

Therefore, we have

$${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=V(u)W(u)$$

By theproduct rule,the derivative of the right-hand side is

$${\frac {\mathrm {d} }{\mathrm {d} x}}{\bigl [}V(x)W(x){\bigr ]}=V(x)w(x)+v(x)W(x)$$

and by integrating we have:

$$\int _{0}^{u}V(x)w(x)+v(x)W(x)\,\mathrm {d} x=V(u)W(u)$$

Thus, the equation from the beginning we get:

$${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{u}V(x)w(x)+v(x)W(x)\,\mathrm {d} x$$

Now, we introduce a second integration parameter$y$for the description of the antiderivatives$V(x)$and$W(x)$:

$\int _{0}^{1}x\,v(xy)\,\mathrm {d} y={\biggl [}V(xy){\biggr ]}_{y=0}^{y=1}=V(x)$

$\int _{0}^{1}x\,w(xy)\,\mathrm {d} y={\biggl [}W(xy){\biggr ]}_{y=0}^{y=1}=W(x)$

By insertion, a double integral appears:

$${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{u}{\biggl [}\int _{0}^{1}x\,v(xy)\,\mathrm {d} y{\biggr ]}w(x)+v(x){\biggl [}\int _{0}^{1}x\,w(xy)\,\mathrm {d} y{\biggr ]}\,\mathrm {d} x$$

Functions that are foreign to the concerned integration parameter can be imported into the inner function as a factor:

$${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{u}{\biggl [}\int _{0}^{1}x\,v(xy)\,w(x)\,\mathrm {d} y{\biggr ]}+{\biggl [}\int _{0}^{1}x\,v(x)\,w(xy)\,\mathrm {d} y{\biggr ]}\,\mathrm {d} x$$

In the next step, thesum ruleis applied to the integrals:

$${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{u}\int _{0}^{1}x\,v(xy)\,w(x)+x\,v(x)\,w(xy)\,\mathrm {d} y\,\mathrm {d} x$$

And finally, we use theFubini theorem

$${\biggl [}\int _{0}^{u}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{u}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{1}\int _{0}^{u}x\,v(xy)\,w(x)+x\,v(x)\,w(xy)\,\mathrm {d} x\,\mathrm {d} y$$

Calculation examples

Arcsine Integral

The Arcsine Integral, also called the Inverse Sine Integral, is a function that cannot be represented byelementary functions.However, the Arcsine Integral does have some elementary function values. These values can be determined by integrating the derivative of the arcsine integral, which is the quotient of the Arcsine divided by theIdentity Function-the Cardinalized Arcsine. The Arcsine Integral is exactly the original antiderivative of the Cardinalized Arcsine. To integrate this function, Fubini's theorem serves as a key, which unlocks the integral by exchanging the order of the integration parameters. When applied correctly, Fubini's theorem leads directly to an antiderivative function that can be integrated in an elementary way, which is shown in cyan in the following equation chain:

$$\operatorname {Si} _{2}(1)=\int _{0}^{1}{\frac {1}{x}}\arcsin(x)\,\mathrm {d} x={\color {blue}\int _{0}^{1}}{\color {green}\int _{0}^{1}}{\frac {{\sqrt {1-x^{2}}}\,y}{(1-x^{2}y^{2}){\sqrt {1-y^{2}}}}}\,{\color {green}\mathrm {d} y}\,{\color {blue}\mathrm {d} x}=$$ $$={\color {green}\int _{0}^{1}}{\color {blue}\int _{0}^{1}}{\frac {{\sqrt {1-x^{2}}}\,y}{(1-x^{2}y^{2}){\sqrt {1-y^{2}}}}}{\color {blue}\,\mathrm {d} x}{\color {green}\,\mathrm {d} y}=\int _{0}^{1}{\frac {\pi \,y}{2{\sqrt {1-y^{2}}}(1+{\sqrt {1-y^{2}}}\,)}}\,\mathrm {d} y=$$ $$={\color {RoyalBlue}{\biggl \{}{\frac {\pi }{2}}\ln {\bigl [}2{\bigl (}1+{\sqrt {1-y^{2}}}\,{\bigr )}^{-1}{\bigr ]}{\biggr \}}_{y=0}^{y=1}}={\frac {\pi }{2}}\ln(2)$$

Dirichlet Eta Function

TheDirichlet seriesdefines theDirichlet Eta Functionas follows:

$$\eta (s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}}=1-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+{\frac {1}{5^{s}}}-{\frac {1}{6^{s}}}\pm \cdots$$

The value η(2) is equal to π²/12 and this can be proven with Fubini's theoremin this way:

$$\eta (2)=\sum _{n=1}^{\infty }(-1)^{n-1}{\frac {1}{n^{2}}}=\sum _{n=1}^{\infty }\int _{0}^{1}(-1)^{n-1}{\frac {1}{n}}{x}^{n-1}\,\mathrm {d} x=\int _{0}^{1}\sum _{n=1}^{\infty }(-1)^{n-1}{\frac {1}{n}}{x}^{n-1}\,\mathrm {d} x=\int _{0}^{1}{\frac {1}{x}}\ln(x+1)\,\mathrm {d} x$$

The integral of the product of theReciprocal Functionand theNatural Logarithmof theSuccessor Functionis aPolylogarithmic Integraland it cannot be represented by elementary function expressions. Fubini's theorem again unlocks this integral in a combinatorial way. This works by carrying out double integration on the basis of Fubini's theorem used on an additive combination of fractionally rational functions with fractions of linear and square denominators:

$$\int _{0}^{1}{\frac {1}{x}}\ln(x+1)\,\mathrm {d} x={\color {blue}\int _{0}^{1}}{\color {green}\int _{0}^{1}}{\frac {4}{3(x^{2}+2xy+1)}}+{\frac {2x}{3(x^{2}y+1)}}-{\frac {1}{3(xy+1)}}\,{\color {green}\mathrm {d} y}\,{\color {blue}\mathrm {d} x}=$$

$$={\color {green}\int _{0}^{1}}{\color {blue}\int _{0}^{1}}{\frac {4}{3(x^{2}+2xy+1)}}+{\frac {2x}{3(x^{2}y+1)}}-{\frac {1}{3(xy+1)}}\,{\color {blue}\mathrm {d} x}\,{\color {green}\mathrm {d} y}=\int _{0}^{1}{\frac {2\arccos(y)}{3{\sqrt {1-y^{2}}}}}\,\mathrm {d} y=$$

$$={\color {RoyalBlue}{\biggl [}{\frac {\pi ^{2}}{12}}-{\frac {1}{3}}\arccos(y)^{2}{\biggr ]}_{y=0}^{y=1}}={\frac {\pi ^{2}}{12}}$$

This way of working out the integral of theCardinalizednatural logarithm of the successor function was discovered by James Harper and it is described in his workAnother simple proof of 1 + 1/2² + 1/3² +... = π²/6accurately.

The original antiderivative, shown here in cyan, leads directly to the value of η(2):

$\eta (2)={\frac {\pi ^{2}}{12}}$

Integrals of Complete Elliptic Integrals

Theimproper integralof theComplete Elliptic Integral of first kindKtakes the value of twice theCatalan constantaccurately. The antiderivative of that K-integral belongs to the so-calledElliptic Polylogarithms.The Catalan constant can only be obtained via theArctangent Integral,which results from the application of Fubini's theorem:

$$\int _{0}^{1}K(x)\,\mathrm {d} x={\color {blue}\int _{0}^{1}}{\color {green}\int _{0}^{1}}{\frac {1}{\sqrt {(1-x^{2}y^{2})(1-y^{2})}}}\,{\color {green}\mathrm {d} y}\,{\color {blue}\mathrm {d} x}={\color {green}\int _{0}^{1}}{\color {blue}\int _{0}^{1}}{\frac {1}{\sqrt {(1-x^{2}y^{2})(1-y^{2})}}}\,{\color {blue}\mathrm {d} x}\,{\color {green}\mathrm {d} y}=$$

$$=\int _{0}^{1}{\frac {\arcsin(y)}{y{\sqrt {1-y^{2}}}}}\,\mathrm {d} y={\color {RoyalBlue}{\biggl \{}2\,\mathrm {Ti} _{2}{\bigl [}y{\bigl (}1+{\sqrt {1-y^{2}}}\,{\bigr )}^{-1}{\bigr ]}{\biggr \}}_{y=0}^{y=1}}=2\,\mathrm {Ti} _{2}(1)=2\beta (2)=2\,C$$

This time, the expression now in royal cyan color tone is not elementary, but it leads directly to the equally non-elementary value of the "Catalan constant" using the Arctangent Integral, also called Inverse Tangent Integral.

The same procedure also works for theComplete Elliptic Integral of the second kindEin the following way:

$$\int _{0}^{1}E(x)\,\mathrm {d} x={\color {blue}\int _{0}^{1}}{\color {green}\int _{0}^{1}}{\frac {\sqrt {1-x^{2}y^{2}}}{\sqrt {1-y^{2}}}}\,{\color {green}\mathrm {d} y}\,{\color {blue}\mathrm {d} x}={\color {green}\int _{0}^{1}}{\color {blue}\int _{0}^{1}}{\frac {\sqrt {1-x^{2}y^{2}}}{\sqrt {1-y^{2}}}}\,{\color {blue}\mathrm {d} x}\,{\color {green}\mathrm {d} y}=$$ $$=\int _{0}^{1}{\biggl [}{\frac {\arcsin(y)}{2y{\sqrt {1-y^{2}}}}}+{\frac {1}{2}}{\biggr ]}\,\mathrm {d} y={\color {RoyalBlue}{\biggl \{}\mathrm {Ti} _{2}{\bigl [}y{\bigl (}1+{\sqrt {1-y^{2}}}\,{\bigr )}^{-1}{\bigr ]}+{\frac {1}{2}}y{\biggr \}}_{y=0}^{y=1}}=\mathrm {Ti} _{2}(1)+{\frac {1}{2}}=\beta (2)+{\frac {1}{2}}=C+{\frac {1}{2}}$$

Double execution for the Exponential Integral Function

TheMascheroni constantemerges as theImproper Integralfrom zero to infinity at the integration on the product of negativeNatural Logarithmand theExponential reciprocal.But it is also the improper integral within the same limits on theCardinalized Differenceof the reciprocal of the Successor Function and theExponential Reciprocal:

$$\gamma ={\color {WildStrawberry}\int _{0}^{\infty }{\frac {-\ln(x)}{\exp(x)}}\,\mathrm {d} x}={\color {cornflowerblue}\int _{0}^{\infty }{\frac {1}{x}}{\biggl [}{\frac {1}{x+1}}-\exp(-x){\biggr ]}\,\mathrm {d} x}$$

The concord of these two integrals can be shown by successively executing theFubini's Theoremtwice and by leading this double execution of that theorem over the identity to an integral of the complementaryExponential Integral Function:

This is how the complementary integral exponential function is defined:

$\mathrm {E} _{1}(x)=\exp(-x)\int _{0}^{\infty }{\frac {\exp(-xy)}{y+1}}\,\mathrm {d} y$

This is the derivative of that function:

${\frac {\mathrm {d} }{\mathrm {d} x}}\,\mathrm {E} _{1}(x)=-{\frac {1}{x}}\exp(-x)$

First implementation of Fubini's theorem:

This integral from a construction of the integral exponential function leads to the integral from the negative Natural Logarithm and the Exponential Reciprocal:

$$\gamma ={\color {WildStrawberry}\int _{0}^{\infty }-\exp(-y)\ln(y)\,\mathrm {d} y}={\color {green}\int _{0}^{\infty }}{\color {blue}\int _{0}^{\infty }}\exp(-y){\bigl (}{\frac {1}{x+y}}-{\frac {1}{x+1}}{\bigr )}\,{\color {blue}\mathrm {d} x}\,{\color {green}\mathrm {d} y}=$$ $$={\color {blue}\int _{0}^{\infty }}{\color {green}\int _{0}^{\infty }}\exp(-y){\bigl (}{\frac {1}{x+y}}-{\frac {1}{x+1}}{\bigr )}\,{\color {green}\mathrm {d} y}\,{\color {blue}\mathrm {d} x}={\color {Dandelion}\int _{0}^{\infty }{\biggl [}\exp(x)\,\mathrm {E} _{1}(x)-{\frac {1}{x+1}}{\biggr ]}\,\mathrm {d} x}$$

Second implementation of Fubini's theorem:

The previously described integral from the described cardinalized difference leads to the previously mentioned integral from the Exponential Integral function:

$$\gamma ={\color {Dandelion}\int _{0}^{\infty }{\biggl [}\exp(x)\,\mathrm {E} _{1}(x)-{\frac {1}{x+1}}{\biggr ]}\,\mathrm {d} x}={\color {blue}\int _{0}^{\infty }}{\color {blueviolet}\int _{0}^{\infty }}\exp(-xz){\biggl [}{\frac {1}{z+1}}-\exp(-z){\biggr ]}\,{\color {blueviolet}\mathrm {d} z}\,{\color {blue}\mathrm {d} x}=$$ $$={\color {blueviolet}\int _{0}^{\infty }}{\color {blue}\int _{0}^{\infty }}\exp(-xz){\biggl [}{\frac {1}{z+1}}-\exp(-z){\biggr ]}\,{\color {blue}\mathrm {d} x}\,{\color {blueviolet}\mathrm {d} z}={\color {cornflowerblue}\int _{0}^{\infty }{\frac {1}{z}}{\biggl [}{\frac {1}{z+1}}-\exp(-z){\biggr ]}\,\mathrm {d} z}$$

In principle, products from exponential functions and fractionally rational functions can be integrated like this:

$${\frac {\exp(-ax)}{bx+c}}={\frac {\mathrm {d} }{\mathrm {d} x}}{\biggl \{}-{\frac {1}{b}}\exp {\bigl (}{\frac {ac}{b}}{\bigr )}\,\mathrm {E} _{1}{\bigl [}{\frac {a}{b}}{\bigl (}bx+c{\bigr )}{\bigr ]}{\biggr \}}$$ $$\int _{0}^{\infty }{\frac {\exp(-ax)}{bx+c}}\,\mathrm {d} x={\biggl \{}-{\frac {1}{b}}\exp {\bigl (}{\frac {ac}{b}}{\bigr )}\,\mathrm {E} _{1}{\bigl [}{\frac {a}{b}}{\bigl (}bx+c{\bigr )}{\bigr ]}{\biggr \}}_{x=0}^{x=\infty }={\frac {1}{b}}\exp {\bigl (}{\frac {ac}{b}}{\bigr )}\,\mathrm {E} _{1}{\bigl (}{\frac {ac}{b}}{\bigr )}$$

In this way it is shown accurately by using theFubini's Theoremtwice that these integrals are indeed identical to each other.

Gauss curve integral

Now this formula for the squaring of an integral is set up:

${\biggl [}\int _{0}^{\infty }f(x)\,\mathrm {d} x{\biggr ]}^{2}=\int _{0}^{1}\int _{0}^{\infty }2x\,f(x)\,f(xy)\,\mathrm {d} x\,\mathrm {d} y$

This chain of equations can then be generated accordingly:

$${\biggl [}\int _{0}^{\infty }\exp(-x^{2})\,\mathrm {d} x{\biggr ]}^{2}=\int _{0}^{1}\int _{0}^{\infty }2x\exp(-x^{2})\exp(-x^{2}y^{2})\,\mathrm {d} x\,\mathrm {d} y=\int _{0}^{1}\int _{0}^{\infty }2x\exp {\bigl [}-x^{2}(y^{2}+1){\bigr ]}\,\mathrm {d} x\,\mathrm {d} y=$$

$$=\int _{0}^{1}{\biggl \{}{\frac {1}{y^{2}+1}}-{\frac {1}{y^{2}+1}}\exp {\biggl [}-x^{2}(y^{2}+1){\biggr ]}{\biggr \}}_{x=0}^{x=\infty }\,\mathrm {d} y=\int _{0}^{1}{\frac {1}{y^{2}+1}}\,\mathrm {d} y=\arctan(1)={\frac {\pi }{4}}$$

For the integral of theGauss curvethis value can be generated:

$\int _{0}^{\infty }\exp(-x^{2})\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi }}$

Dilogarithm of one

Now another formula for the squaring of an integral is set up again:

${\biggl [}\int _{0}^{1}g(x)\,\mathrm {d} x{\biggr ]}^{2}=\int _{0}^{1}\int _{0}^{1}2x\,g(x)\,g(xy)\,\mathrm {d} x\,\mathrm {d} y$

So this chain of equations applies as a new example:

$${\frac {\pi ^{2}}{4}}=\arcsin(1)^{2}={\biggl [}\int _{0}^{1}{\frac {1}{\sqrt {1-x^{2}}}}\,\mathrm {d} x{\biggr ]}^{2}=\int _{0}^{1}\int _{0}^{1}{\frac {2x}{\sqrt {(1-x^{2})(1-x^{2}y^{2})}}}\,\mathrm {d} x\,\mathrm {d} y=$$

$$=\int _{0}^{1}{\biggl [}{\frac {2}{y}}\operatorname {artanh} (y)-{\frac {2}{y}}\operatorname {artanh} {\biggl (}{\frac {{\sqrt {1-x^{2}}}\,y}{\sqrt {1-x^{2}y^{2}}}}{\biggr )}{\biggr ]}_{x=0}^{x=1}\,\mathrm {d} y=\int _{0}^{1}{\frac {2}{y}}\operatorname {artanh} (y)\,\mathrm {d} y=$$

$$={\biggl [}2\,\mathrm {Li} _{2}(y)-{\frac {1}{2}}\,\mathrm {Li} _{2}(y^{2}){\biggr ]}_{y=0}^{y=1}={\frac {3}{2}}\,\mathrm {Li} _{2}(1)$$

For theDilogarithmof one this value appears:

$\mathrm {Li} _{2}(1)={\frac {\pi ^{2}}{6}}$

In this way theBasel problemcan be solved.

Legendre's relation

In this next example, the more generalized form of the equation is used again as a mold:

${\biggl [}\int _{0}^{1}v(x)\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{1}w(x)\,\mathrm {d} x{\biggr ]}=\int _{0}^{1}\int _{0}^{1}\left(x\,v(xy)\,w(x)+x\,v(x)\,w(xy)\right)\,\mathrm {d} x\,\mathrm {d} y$

The following integrals can be computed by using the incompleteElliptic Integralsof the first and second kind as antiderivatives and these integrals have values that can be represented withComplete Elliptic Integrals:

$\int _{0}^{1}{\frac {1}{\sqrt {1-x^{4}}}}\,\mathrm {d} x=$

$${\biggl \{}-{\frac {1}{2}}{\sqrt {2}}\,F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}{\biggr \}}_{x=0}^{x=1}={\frac {1}{2}}{\sqrt {2}}\,K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}$$

$\int _{0}^{1}{\frac {x^{2}}{\sqrt {1-x^{4}}}}\,\mathrm {d} x=$

$${\biggl \{}{\frac {1}{2}}{\sqrt {2}}\,F{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}-{\sqrt {2}}\,E{\biggl [}\arccos(x);{\frac {1}{2}}{\sqrt {2}}{\biggr ]}{\biggr \}}_{x=0}^{x=1}={\frac {1}{2}}{\sqrt {2}}{\biggl [}2\,E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\biggr ]}$$

Inserting these two integrals into the above form gives:

$${\biggl [}\int _{0}^{1}{\frac {1}{\sqrt {1-x^{4}}}}\,\mathrm {d} x{\biggr ]}{\biggl [}\int _{0}^{1}{\frac {x^{2}}{\sqrt {1-x^{4}}}}\,\mathrm {d} x{\biggr ]}=\int _{0}^{1}\int _{0}^{1}{\frac {x^{3}(y^{2}+1)}{\sqrt {(1-x^{4})(1-x^{4}y^{4})}}}\,\mathrm {d} x\,\mathrm {d} y=$$

$$=\int _{0}^{1}{\biggl \{}{\frac {y^{2}+1}{2\,y^{2}}}{\biggl [}{\text{artanh}}{\bigl (}y^{2}{\bigr )}-{\text{artanh}}{\biggl (}{\frac {{\sqrt {1-x^{4}}}\,y^{2}}{\sqrt {1-x^{4}y^{4}}}}{\biggr )}{\biggr ]}{\biggr \}}_{x=0}^{x=1}\,\mathrm {d} y=\int _{0}^{1}{\frac {y^{2}+1}{2\,y^{2}}}\,\mathrm {artanh} {\bigl (}y^{2}{\bigr )}\,\mathrm {d} y=$$

$$={\biggl [}\arctan(y)-{\frac {1-y^{2}}{2\,y}}\,\mathrm {artanh} {\bigl (}y^{2}{\bigr )}{\biggr ]}_{y=0}^{y=1}=\arctan(1)={\frac {\pi }{4}}$$

For theLemniscaticspecial case ofLegendre's relation,this result emerges:

$K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\biggl [}2\,E{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\biggr ]}={\frac {\pi }{2}}$

See also

• Cavalieri's principle– Geometrical concept − an early particular case
• Coarea formula− generalization to geometric measure theory
• Disintegration theorem– theorem in measure theory − a restricted converse to Fubini's theorem
• Fubini's theorem for distributions– Mathematical analysis term similar to generalized functionPages displaying short descriptions of redirect targets
• Kuratowski–Ulam theorem– analog of Fubini's theorem for arbitrary second countable Baire spaces
• Symmetry of second derivatives− analogue for differentiation
• Fubini's nightmare– Apparent violation of Fubini's theorem

Further reading

• Billingsley, Patrick(1995), "Product Measure and Fubini's Theorem",Probability and Measure,New York: Wiley, pp. 231–240,ISBN0-471-00710-2
• DiBenedetto, Emmanuele (2002),Real Analysis,Birkhäuser Advanced Texts: Basler Lehrbücher, Boston: Birkhäuser,doi:10.1007/978-1-4612-0117-5,ISBN0-8176-4231-5,MR1897317
• Fremlin, D. H. (2003),Measure theory,vol. 2, Colchester: Torres Fremlin,ISBN0-9538129-2-8,MR2462280
• Weir, Alan J. (1973), "Fubini's Theorem",Lebesgue Integration and Measure,Cambridge: Cambridge University Press, pp. 83–92,ISBN0-521-08728-7

External links

• Kudryavtsev, L.D. (2001) [1994],"Fubini theorem",Encyclopedia of Mathematics,EMS Press