Fubini's theorem

In mathematical analysis, the Fubini theorem gives the conditions under which it is possible to compute a double integral by using an iterated integral. It was introduced by Guido Fubini in 1907. It states that if a function is Lebesgue integrable on a rectangle $$X\times Y$$, then one can evaluate the double integral as an iterated integral:$$\, \iint\limits_{X\times Y} f(x,y)\,\text{d}(x,y) = \int_X\left(\int_Y f(x,y)\,\text{d}y\right)\text{d}x=\int_Y\left(\int_X f(x,y) \, \text{d}x \right) \text{d}y.$$ The formula is not true in general for Riemann integrals, but it is true if the function is continuous on the rectangle. In multivariable calculus, this weaker result is sometimes also called Fubini's theorem, although it was already known by Leonhard Euler.

The Tonelli theorem, introduced by Leonida Tonelli in 1909, is similar but applied to a non-negative measurable function rather than 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 the order of integration in an iterated integral.

A related theorem is often called Fubini's theorem for infinite series, which states that: if $\{a_{m,n}\}_{m=1,n=1}^{\infty}$ is a double-indexed sequence of real numbers, and if $\displaystyle \sum_{(m,n)\in \N\times \N} a_{m,n} $  is absolutely convergent, then


 * $$ \sum_{(m,n)\in\N\times \N}a_{m,n} = \sum_{m=1}^\infty\sum_{n=1}^{\infty} a_{m,n} = \sum_{n=1}^\infty \sum_{m=1}^\infty a_{m,n} $$

Although the Fubini theorem for infinite series is a special case of the more general Fubini's theorem, it is not appropriate to characterize the former as a logical consequence of the latter. This is because some properties of measures, in particular subadditivity, are often proved using the Fubini theorem for infinite series. In this case, the Fubini theorem for integrals is a logical consequence of the Fubini's theorem for infinite series.

History
The special case of the Fubini theorem for continuous functions on a product of closed bounded subsets of real vector spaces was known to Leonhard Euler in the 18th century. In 1904, Henri Lebesgue extended 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 bounded and 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 $$Y$$ are measure spaces, there are several natural ways to define a product measure on the product $$X\times Y$$.

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

A measure μ on X × Y is called a product measure if μ(A × B) = μ1(A)μ2(B) for measurable subsets A ⊂ X and B ⊂ Y and measures μ1 on X and μ2 on Y. In general, there may be many different product measures on X × Y. Fubini's theorem and Tonelli's theorem both need technical conditions to avoid this complication; the most common way is to assume all measure spaces are σ-finite, in which case there is a unique product measure on X×Y. There is always a unique maximal product measure on X × Y, where the measure of a measurable set is the inf of the measures of sets containing it that are countable unions of products of measurable sets. The maximal product measure can be constructed by applying Carathéodory's extension theorem to the additive function μ such that μ(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 space X × Y, so strictly speaking, the measure should be restricted to the σ-algebra generated by the products A × B of measurable subsets of X and Y.)

The product of two complete measure spaces is not usually complete. For example, the product of the Lebesgue measure on the unit interval I with itself is not the Lebesgue measure on the square I × 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
Suppose X and Y are σ-finite measure spaces and suppose that X × Y is given the product measure (which is unique as X and Y are σ-finite). Fubini's theorem states that if f is X × Y integrable, meaning that f is a measurable function and $$\int_{X\times Y} |f(x,y)|\,\text{d}(x,y) < \infty,$$ 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 $\int_Y f(x,y)\,\text{d}y$ and $\int_X f(x,y)\,\text{d}x$  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. See below for an illustration of this possibility.

The condition that X and Y are σ-finite is usually harmless because, in practice 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 when X and Y are not assumed to be σ-finite. The main extra complication in this case is that there may be more than one product measure on X×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 function f for 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
, named after Leonida 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 $$ (Y, B, \nu) $$ are σ-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 $\sum_x \sum_y a_{xy} = \sum_y \sum_x a_{xy}$, where $$a_{xy}$$ are non-negative for all x and y. 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 on A×B to 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 factors A and B are not measurable. Another way is to add the condition that the support of f is contained in a countable union of products of sets of finite measures. 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 $$Y$$ are σ-finite measure spaces, 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 $$\R$$ with itself with Lebesgue measure. The problem is that Lebesgue measure on $$\R\times\R$$ is not the product of Lebesgue measure on $$\R$$ 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 a hypothesis related to σ-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 a product measure 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
For Riemann 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 that X is the unit interval with the Lebesgue measurable sets and Lebesgue measure, and Y is the unit interval with all subsets measurable and the counting measure, so that Y is not σ-finite. If f is the characteristic function of the diagonal of X×Y, then integrating f along X gives the 0 function on Y, but integrating f along Y gives the function 1 on X. So, the two iterated integrals are different. This shows that Tonelli's theorem can fail for spaces that are not σ-finite no matter what product measure is chosen. The measures are both decomposable, 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 give X×Y the 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 that X is 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) subset E of X×X given by pairs (x, y) with x<y is countable on every horizontal line and has countable complement on every vertical line. If f is the characteristic function of E then the two iterated integrals of f are defined and have different values 1 and 0. The function f is 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 take f to be 1 on E and –1 on the complement of E, 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 identify X with the unit interval I, so there is a bounded non-negative function on I×I whose two iterated integrals (using Lebesgue measure) are both defined but unequal. This example was found by. 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 standard Zermelo–Fraenkel axioms of set theory. The continuum hypothesis and Martin's axiom both imply that there exists a function on the unit square whose iterated integrals are not equal, while 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. See List 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 to x and then with respect to y, we get the same result as if we integrate first with respect to y and then with respect to x. 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 function f(x,y) to be 1 if x = y, −1 if x = y + 1, and 0 otherwise. Then the two repeated integrals have different values 0 and 1.

Another example is as follows for the function $$\frac{x^2-y^2}{(x^2+y^2)^2} = -\frac{\partial^2}{\partial x\,\partial y} \arctan(y/x).$$ The iterated 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 not converge absolutely (in other words the integral of the absolute value is not finite): $$\int_0^1\int_0^1 \left|\frac{x^2-y^2}{\left(x^2 + y^2\right)^2}\right|\,\text{d}y\,\text{d}x=\infty.$$

Product of two integrals
For the product of two integrals with lower limit zero and a common upper limit we have the following formula:
 * {| class = "wikitable"


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

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 the product 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, the sum rule is 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 the Fubini theorem

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

Arcsine Integral
The Arcsine Integral, also called the Inverse Sine Integral, is a function that cannot be represented by elementary functions. The Arcsine Integral does, however, have some elementary function values. These values may be determined by integrating the derivative of the arcsine integral, which is the quotient of the Arcsine divided by the Identity Function-the Cardinalized Arcsine. The Arcsine Integral is exactly the original antiderivative of the Cardinalized Arcsine. For the integration of 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
The Dirichlet series defines the Dirichlet Eta Function as 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 theorem in 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 the Reciprocal Function and the Natural Logarithm of the Successor Function is a Polylogarithmic Integral and 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:

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

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

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

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

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
The improper integral of the Complete Elliptic Integral of first kind K takes the value of twice the Catalan constant accurately. The antiderivative of that K-integral belongs to the so-called Elliptic Polylogarithms. The Catalan constant can only be obtained via the Arctangent 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 the Complete Elliptic Integral of the second kind E in 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
The Mascheroni constant emerges as the Improper Integral from zero to infinity at the integration on the product of negative Natural Logarithm and the Exponential reciprocal. But it is also the improper integral within the same limits on the Cardinalized Difference of the reciprocal of the Successor Function and the Exponential 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 the Fubini's Theorem twice and by leading this double execution of that theorem over the identity to an integral of the complementary Exponential 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 the Fubini's Theorem twice that these integrals are indeed identical to each other.

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


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

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 the Gauss curve this 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:


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

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 the Dilogarithm of one this value appears:


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

In this way the Basel problem can be solved.

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


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

The following integrals can be computed by using the incomplete Elliptic Integrals of the first and second kind as antiderivatives and these integrals have values that can be represented with Complete Elliptic Integrals:


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

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


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

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

By Insertion of these two integrals into the mentioned mold, this chain of equations results:

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

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

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

For the Lemniscatic special case of Legendre's relation, this result emerges:


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