Fubini's theorem

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Template:Short description Template:For Fubini's theorem is a theorem in measure theory that gives conditions under which a double integral can be computed as an iterated integral, i.e. by integrating in one variable at a time. Intuitively, just as the volume of a loaf of bread is the same whether one sums over standard slices or over long thin slices, the value of a double integral does not depend on the order of integration when the hypotheses of the theorem are satisfied. The theorem is named after Guido Fubini, who proved a general result in 1907; special cases were known earlier through results such as Cavalieri's principle, which was used by Leonhard Euler.

More formally, the theorem states that if a function is Lebesgue integrable on a rectangle <math>X\times Y</math>, then one can evaluate the double integral as an iterated integral:<math display="block">\, \iint\limits_{X\times Y} f(x,y)\,\mathrm{d}(x,y) = \int_X\left(\int_Y f(x,y)\,\mathrm{d}y\right)\mathrm{d}x=\int_Y\left(\int_X f(x,y) \, \mathrm{d}x \right) \mathrm{d}y.</math> This formula is generally not true for the Riemann integral (however, it is true if the function is continuous on the rectangle; in multivariable calculus, this weaker result is sometimes also called Fubini's theorem).

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

A related theorem is often called Fubini's theorem for infinite series,<ref>Template:Citation</ref> although it is due to Alfred Pringsheim.<ref>Template:Citation</ref> It states that if <math display="inline">\{a_{m,n}\}_{m=1,n=1}^{\infty}</math> is a double-indexed sequence of real numbers, and if <math display="inline">\displaystyle \sum_{(m,n)\in \N\times \N} a_{m,n} </math> is absolutely convergent, then <math display=block> \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}. </math>

History

A special case of Fubini's theorem for continuous functions on the 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.<ref>Template:Citation</ref> Levi conjectured that the theorem could be extended to functions that are integrable rather than boundedTemplate:Citation needed and this was proven by Fubini in 1907.<ref>Template:Citation; reprinted in Template:Citation</ref> In 1909, Leonida Tonelli gave a variation of the Fubini theorem that applies to non-negative functions rather than integrable functions.<ref>Template:Citation</ref>

Product measures

If Template:Mvar and Template:Mvar are measure spaces, there are several natural ways to define a product measure on the product Template:Math.

In the sense of category theory, measurable sets in the product Template:Math of measure spaces are the elements of the σ-algebra generated by the products Template:Math, where Template:Mvar is measurable in Template:Mvar, and Template:Mvar is measurable in Template:Mvar.

A measure Template:Mvar on Template:Math is called a Template:Dfn if Template:Math for measurable subsets Template:Math and Template:Math and measures Template:Math on Template:Mvar and Template:Math on Template:Mvar. In general, there may be many different product measures on Template:Math. 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 are σ-finite, in which case there is a unique product measure on Template:Math. There is always a unique maximal product measure on Template:Math, 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 Template:Mvar such that Template:Math 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 Template:Math, so strictly speaking, the measure should be restricted to the σ-algebra generated by the products Template:Math of measurable subsets of Template:Mvar and Template:Mvar.)

The product of two complete measure spaces is not usually complete. For example, the product of the Lebesgue measure on the unit interval Template:Mvar with itself is not the Lebesgue measure on the square Template:Math. 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 Template:Mvar and Template:Mvar are σ-finite measure spaces and suppose that Template:Math is given the product measure (which is unique as Template:Mvar and Template:Mvar are σ-finite). Fubini's theorem states that if Template:Mvar is Template:Math integrable, meaning that Template:Mvar is a measurable function and <math display="block">\int_{X\times Y} |f(x,y)|\,\mathrm{d}(x,y) < \infty,</math> then <math display="block">\int_X\left(\int_Y f(x,y)\,\mathrm{d}y\right)\,\mathrm{d}x = \int_Y\left(\int_X f(x,y)\,\mathrm{d}x\right)\,\mathrm{d}y = \int_{X\times Y} f(x,y)\,\mathrm{d}(x,y).</math>

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 <math display="inline">\int_Y f(x,y)\,\mathrm{d}y</math> and <math display="inline">\int_X f(x,y)\,\mathrm{d}x</math> 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 Template:See below.

The condition that Template:Mvar and Template:Mvar are 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 when Template:Mvar and Template:Mvar are not assumed to be σ-finite.Template:Sfnp The main extra complication in this case is that there may be more than one product measure on Template:Math. 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 Template:Mvar for which the double integral of Template:Math is zero but the two iterated integrals have different values Template:See below. Tonelli's theorem and the Fubini–Tonelli theorem (Template:Xref-print) can fail on non σ-finite spaces, even for the maximal product measure.

Tonelli's theorem for non-negative measurable functions

Template:Visible anchor, 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 Template:Math has a finite integral is replaced by the assumption that Template:Mvar is a non-negative measurable function.

Tonelli's theorem states that if Template:Math and Template:Math are σ-finite measure spaces, while <math display="inline">f:X\times Y \to [0,\infty]</math> is a non-negative measurable function, then <math display="block">\int_X\left(\int_Y f(x,y)\,\mathrm{d}y\right)\,\mathrm{d}x = \int_Y\left(\int_X f(x,y)\,\mathrm{d}x\right)\,\mathrm{d}y = \int_{X\times Y} f(x,y)\,\mathrm{d}(x,y).</math>

A special case of Tonelli's theorem is in the interchange of the summations, as in Template:Nowrap where Template:Mvar are non-negative for all Template:Mvar and Template:Mvar. 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 Template:Math and others diverging to Template:Math. 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 Template:Math 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 Template:Mvar and Template:Mvar are not measurable. Another way is to add the condition that the support of Template:Mvar is contained in a countable union of products of sets of finite measures. D. H. Fremlin 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.Template:Sfnp

Fubini–Tonelli theorem

Combining Fubini's theorem with Tonelli's theorem gives the Template:Dfn. Often just called Fubini's theorem, it states that if Template:Mvar and Template:Mvar are σ-finite measure spaces, and if Template:Mvar is a measurable function, then <math display="block">\int_X\left(\int_Y |f(x,y)|\,\mathrm{d}y\right)\,\mathrm{d}x=\int_Y\left(\int_X |f(x,y)|\,\mathrm{d}x\right)\,\mathrm{d}y=\int_{X\times Y} |f(x,y)|\,\mathrm{d}(x,y)</math> Furthermore, if any one of these integrals is finite, then <math display="block">\int_X\left(\int_Y f(x,y)\,\mathrm{d}y\right)\,\mathrm{d}x=\int_Y\left(\int_X f(x,y)\,\mathrm{d}x\right)\,\mathrm{d}y=\int_{X\times Y} f(x,y)\,\mathrm{d}(x,y).</math>

The absolute value of Template:Mvar in the conditions above can be replaced by either the positive or the negative part of Template:Mvar; 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 Template:Mvar is well defined, though possibly infinite.

The advantage of the Fubini–Tonelli over Fubini's theorem is that the repeated integrals of Template:Math 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 Template:Math with itself with Lebesgue measure. The problem is that Lebesgue measure on Template:Math is not the product of Lebesgue measure on Template:Math with itself, but rather the completion of this: a product of two complete measure spaces Template:Mvar and Template:Mvar 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 Template:Math of two measure spaces, one takes the completion of the product.
  • If Template:Mvar is measurable on the completion of Template:Math 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 Template:Mvar and Template:Mvar are complete, otherwise the two partial integrals along vertical or horizontal lines may be well-defined but not measurable. For example, if Template:Mvar 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 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

For Riemann integrals, Fubini's theorem is proven by refining the partitions along the Template:Mvar-axis and Template:Mvar-axis as to create a joint partition of the form Template:Math, which is a partition over Template:Math. This is used to show that the double integrals of either order are equal to the integral over Template:Math.

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 Template:Mvar is the unit interval with the Lebesgue measurable sets and Lebesgue measure, and Template:Mvar is the unit interval with all the subsets measurable and the counting measure, so that Template:Mvar is not σ-finite. If Template:Mvar is the characteristic function of the diagonal of Template:Math, then integrating Template:Mvar along Template:Mvar gives the Template:Math function on Template:Mvar, but integrating Template:Mvar along Template:Mvar gives the function Template:Math on Template:Mvar. 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 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 Template:Math is infinite, and Fubini's theorem holds vacuously.

However, if we give Template:Math 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 Template:Math 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 Template:Mvar 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 Template:Mvar of Template:Math given by pairs Template:Math with Template:Math is countable on every horizontal line and has countable complement on every vertical line. If Template:Mvar is the characteristic function of Template:Mvar then the two iterated integrals of Template:Mvar are defined and have different values 1 and 0. The function Template:Mvar 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 Template:Math is integrable and both repeated integrals are well defined: if we take Template:Mvar to be 1 on Template:Mvar and –1 on the complement of Template:Mvar, then Template:Math is integrable on the product with integral 1, and both repeated integrals are well defined, but have different values 1 and –1.

Wacław Sierpiński demonstrated that assuming the continuum hypothesis, one can identify Template:Mvar with the unit interval Template:Mvar, so there is a bounded non-negative function on Template:Math whose two iterated integrals (using Lebesgue measure) are both defined but unequal.<ref>Template:Citation</ref> 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 with the axiom of choice (ZFC). 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 Harvey Friedman showed that it is consistent with Template:Abbr that a strong Fubini-type theorem for Template:Closed-closed does hold, and whenever the two iterated integrals exist they are equal.<ref>Template:Citation</ref> Template:Hatnote inline

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 Template:Mvar and then with respect to Template:Mvar, we get the same result as if we integrate first with respect to Template:Mvar and then with respect to Template:Mvar. 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 Template:Math to be Template:Math if Template:Math, Template:Math if Template:Math, and Template:Math otherwise. Then the two repeated integrals have different values Template:Math and Template:Math.

Another example is as follows for the function <math display="block">\frac{x^2-y^2}{(x^2+y^2)^2} = -\frac{\partial^2}{\partial x\,\partial y} \arctan(y/x).</math>

The iterated integrals <math display="block">\begin{align} \int_{x=0}^1\left(\int_{y=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\mathrm{d}y\right)\,\mathrm{d}x &= \frac{\pi}{4} \\ \int_{y=0}^1\left(\int_{x=0}^1\frac{x^2-y^2}{(x^2+y^2)^2}\,\mathrm{d}x\right)\,\mathrm{d}y &=-\frac{\pi}{4} \end{align}</math> have different values. The corresponding double integral does not converge absolutely (in other words the integral of the absolute value is not finite): <math display="block">\int_0^1\int_0^1 \left|\frac{x^2-y^2}{\left(x^2 + y^2\right)^2}\right|\,\mathrm{d}y\,\mathrm{d}x=\infty.</math>

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: Template:Equation box 1

Proof

Let Template:Math and Template:Math are primitive functions of the functions Template:Math and Template:Math respectively, which pass through the origin: <math display="block">\int_{0}^{u} v(x) \,\mathrm{d}x = V(u), \quad \quad \int_{0}^{u} w(x) \,\mathrm{d}x = W(u).</math>

Therefore, we have <math display="block">\left[\int_{0}^{u} v(x) \,\mathrm{d}x\right]\left[\int_{0}^{u} w(x) \,\mathrm{d}x\right] = V(u) W(u) </math>

By the product rule, the derivative of the right-hand side is <math display="block">\frac{\mathrm{d}}{\mathrm{d}x} \left[V(x) W(x)\right] = V(x)w(x) + v(x)W(x)</math>

and by integrating we have: <math display="block">\int_{0}^{u} V(x)w(x) + v(x)W(x) \,\mathrm{d}x=V(u) W(u) </math>

Thus, the equation from the beginning we get: <math display="block">\left[\int_{0}^{u} v(x) \,\mathrm{d}x\right]\left[\int_{0}^{u} w(x) \,\mathrm{d}x\right]= \int_{0}^{u} V(x)w(x) + v(x)W(x) \,\mathrm{d}x </math>

Now, we introduce a second integration parameter Template:Mvar for the description of the antiderivatives Template:Math and Template:Math: <math display="block">\begin{alignat}{3} \int_{0}^{1} x\,v(xy) \,\mathrm{d}y &= \left[V(xy)\right]_{y = 0}^{y = 1} &&= V(x) \\ \int_{0}^{1} x\,w(xy) \,\mathrm{d}y &= \left[W(xy)\right]_{y = 0}^{y = 1} &&= W(x) \end{alignat}</math>

By insertion, a double integral appears: <math display="block">\left[\int_{0}^{u} v(x) \,\mathrm{d}x\right]\left[\int_{0}^{u} w(x) \,\mathrm{d}x\right]= \int_{0}^{u} \left[\int_{0}^{1} x\,v(xy) \,\mathrm{d}y\right]w(x) + v(x)\left[\int_{0}^{1} x\,w(xy) \,\mathrm{d}y\right] \,\mathrm{d}x </math>

Functions that are foreign to the concerned integration parameter can be imported into the inner function as a factor: <math display="block">\left[\int_{0}^{u} v(x) \,\mathrm{d}x\right]\left[\int_{0}^{u} w(x) \,\mathrm{d}x\right]= \int_{0}^{u} \left[\int_{0}^{1} x\,v(xy) \,w(x) \,\mathrm{d}y\right] + \left[\int_{0}^{1} x\,v(x)\,w(xy) \,\mathrm{d}y\right] \,\mathrm{d}x </math>

In the next step, the sum rule is applied to the integrals: <math display="block">\left[\int_{0}^{u} v(x) \,\mathrm{d}x\right]\left[\int_{0}^{u} w(x) \,\mathrm{d}x\right]= \int_{0}^{u} \int_{0}^{1} x\,v(xy) \,w(x) + x\,v(x)\,w(xy) \,\mathrm{d}y \,\mathrm{d}x </math>

And finally, we use Fubini's theorem <math display="block">\left[\int_{0}^{u} v(x) \,\mathrm{d}x\right]\left[\int_{0}^{u} w(x) \,\mathrm{d}x\right]= \int_{0}^{1} \int_{0}^{u} x\,v(xy) \,w(x) + x\,v(x)\,w(xy) \,\mathrm{d}x \,\mathrm{d}y </math>

Calculation examples

Template:Tone

Arcsine integral

The arcsine integral, also called the inverse sine integral, is a function that cannot be represented by elementary functions. However, some of the values of the arcsine integral can be expressed with elementary functions. These values can be determined by integrating the derivative of the arcsine integral, which is the quotient of the arcsine divided by the identity functionTemplate:Snd the cardinalized arcsine.Template:Clarify 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):

<math display="block">\begin{align} \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}\left\{ \frac{\pi}{2} \ln\left[2 \left(1 + \sqrt{1 - y^2}\,\right)^ {-1}\right] \right\}_{y = 0}^{y = 1}} = \frac{\pi}{2}\ln(2) \end{align}</math>

Dirichlet eta Function

The Dirichlet series defines the Dirichlet eta function as follows: <math display="block"> \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 </math>

The value Template:Math is equal to Template:Math and this can be proven with Fubini's theorem{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B= Template:Fix }} in this way: <math display="block"> \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 </math>

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: <math display="block">\begin{align} &\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} \\ &\hphantom{\quad\quad\quad} = {\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} \\ &\hphantom{\quad\quad\quad} = \int_{0}^{1} \frac{2\arccos(y)}{3\sqrt{1-y^2}} \,\mathrm {d}y = {\color{RoyalBlue}\left[\frac{\pi^2}{12}-\frac{1}{3}\arccos(y)^2 \right]_{y = 0} ^{y = 1}} \\ &\hphantom{\quad\quad\quad} = \frac{\pi^2}{12} \end{align}</math>

This way of working out the integral of the cardinalized natural logarithmTemplate:Definition needed of the successor function was discovered by James Harper.<ref>Template:Cite journal + \frac{1}{3^{2}} + \cdot \cdot \cdot = \frac{\pi^{2}}{6}</math> |journal= The American Mathematical Monthly |volume= 110 |number= 6 |pages= 540–541 |doi= 10.2307/3647912 |jstor= 3647912 }}</ref>

The original antiderivative, shown here in cyan, leads directly to the value of Template:Math: <math display="block"> \eta(2) = \frac{\pi^2}{12}</math>

Integrals of Complete Elliptic Integrals

The improper integral of the complete elliptic integral of first kind, Template:Math, takes the value of twice the Catalan constant accurately. The antiderivative of that K-integral belongs to the so-called elliptic polylogarithms.Template:Definition needed The Catalan constant can only be obtained via the arctangent integral, which results from the application of Fubini's theorem: <math display="block">\begin{align} \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} \left\{ 2\,\mathrm{Ti}_{2} \left[ y\left(1 + \sqrt{1 - y^2}\,\right)^{-1} \right] \right\}_{y = 0}^{y = 1}} \\ &= 2\,\mathrm{Ti}_{2}(1) =2\beta(2) =2\,C \end{align}</math>

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 second kind, Template:Math, in the following way: <math display="block">\begin{align} \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} \left[\frac{\arcsin(y)}{2y\sqrt{1 - y^2}} + \frac{1}{2}\right] \,\mathrm{d}y \\ &= {\color{RoyalBlue}\left\{ \mathrm{Ti}_{2} \left[ y\left(1 + \sqrt{1 - y^2}\,\right )^{-1} \right] + \frac{1}{2} y \right\}_{y = 0}^{y = 1}} \\ &= \mathrm{Ti}_{2}(1) + \frac{1}{2} =\beta(2) + \frac{1}{2} =C + \frac{1}{2} \end{align}</math>

Double execution for the Exponential Integral Function

The Euler-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: <math display="block"> \definecolor{cerulean}{rgb}{0.0, 0.48, 0.65} \gamma = {\color{WildStrawberry}\int_0^\infty \frac{-\ln(x)}{\exp(x)}\,\mathrm{d}x} = {\color{cerulean}\int_{0}^{\infty} \frac{1}{x}\left[\frac{1}{x + 1}-\exp(-x)\right] \,\mathrm{d}x } </math>

The concord of these two integrals can be shown by successively executing 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.

The complementary integral exponential function is defined as: <math display="block">\mathrm{E}_{1}(x) = \exp(-x)\int_{0}^{\infty} \frac{\exp(-xy)}{y+1} \, \mathrm{d}y </math>

The derivative of the complementary integral exponential function is: <math display="block">\frac{\mathrm{d}}{\mathrm{d}x} \,\mathrm{E}_{1}(x) = -\frac{1}{x}\exp(-x) </math>

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: <math display="block">\begin{align} \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)\left(\frac{1}{x + y} - \frac{1}{x + 1}\right) \,{\color{blue}\mathrm{d}x} \,{\color{green}\mathrm{d}y} \\ &= {\color{blue}\int_{0}^{\infty}}{\color{green}\int_{0}^{\infty}} \exp(-y)\left(\frac{1}{x + y} - \frac{1}{x + 1}\right) \,{\color{green}\mathrm{d}y} \,{\color{blue}\mathrm{d}x} \\ &= {\color{brown}\int_{0}^{\infty} \left[\exp(x)\,\mathrm{E}_{1}(x) - \frac{1}{x + 1}\right] \,\mathrm{d}x} \end{align}</math>

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: <math display="block">\definecolor{cerulean}{rgb}{0.0, 0.48, 0.65} \begin{align} \gamma &= {\color{brown}\int_{0}^{\infty} \left[\exp(x)\,\mathrm{E}_{1}(x) - \frac{1}{x + 1}\right] \,\mathrm{d}x} \\ &= {\color{blue}\int_{0}^{\infty}}{\color{blueviolet}\int_{0}^{\infty}} \exp(-xz)\left[\frac{1}{z + 1}-\exp(-z)\right] \,{\color{blueviolet}\mathrm{d}z} \,{\color{blue}\mathrm{d}x} \\ &= {\color{blueviolet}\int_{0}^{\infty}}{\color{blue}\int_{0}^{\infty}} \exp(-xz)\left[\frac{1}{z + 1}-\exp(-z)\right] \,{\color{blue}\mathrm{d}x} \,{\color{blueviolet}\mathrm{d}z} \\ &= {\color{cerulean}\int_{0}^{\infty} \frac{1}{z}\left[\frac{1}{z + 1}-\exp(-z)\right] \,\mathrm{d}z} \end{align}</math>

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

<math display="block">\frac{\exp(-ax)}{bx + c} = \frac{\mathrm{d}}{\mathrm{d}x} \left\{-\frac{1}{b} \exp\left(\frac{ac}{b}\right)\,\mathrm{E}_{1}\left[\frac{a}{b}\left(bx + c\right)\right] \right\} </math> <math display="block">\int_0^{\infty} \frac{\exp(-ax)}{bx + c} \,\mathrm{d}x = \left\{-\frac{1}{b}\exp \left(\frac{ac}{b}\right)\,\mathrm{E}_{1}\left[\frac{a}{b}\left(bx + c\right)\right]\right \}_{x = 0}^{x = \infty} = \frac{1}{b}\exp\left(\frac{ac}{b}\right) \,\mathrm{E}_{1 }\left(\frac{ac}{b}\right) </math>

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

Normal distribution integral

Now this formula for the squaring of an integral is set up: Template:Equation box 1

This chain of equations can then be generated accordingly: <math display="block">\begin{align} \left[\int_{0}^{\infty} \exp(-x^2) \,\mathrm{d}x\right]^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\left[-x^2 (y^2 + 1)\right] \,\mathrm{d}x\,\mathrm{d}y \\ &= \int_{0}^{1} \left\{- \frac{1}{y^2 + 1}\exp\left[-x^2(y^2 + 1)\right]\right\}_{x = 0}^{x = \infty} \,\mathrm{d}y \\ &= \int_{0}^{1} \frac{1}{y^2 + 1} \,\mathrm{d}y \\ &= \arctan(1) \\ &= \frac{\pi}{4} \end{align}</math>

For the integral of the normal distribution this value can be generated: <math display="block">\int_{0}^{\infty} \exp(-x^2) \,\mathrm{d}x = \frac{1}{2}\sqrt{\pi} </math>

Dilogarithm of one

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

Template:Equation box 1

So this chain of equations applies as a new example: <math display="block">\begin{align} \frac{\pi^2}{4} &= \left(\arcsin(1)\right)^2 = \left[\int_{0}^{1} \frac{1}{\sqrt{1 - x^2}} \,\mathrm{d}x\right]^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} \left[\frac{2}{y}\operatorname{artanh}(y) - \frac{2}{y}\operatorname{artanh}\left(\frac{\sqrt{1 - x^2}\,y}{\sqrt{1 - x^2 y^2}}\right)\right]_{x = 0}^{x = 1}\,\mathrm{d}y \\ &= \int_{0}^{1} \frac{2}{y}\operatorname{artanh}(y) \,\mathrm{d}y \\ &= \left[2\,\mathrm{Li}_{2}(y) - \frac{1}{2}\,\mathrm{Li}_{2}(y^2)\right]_{y = 0}^{y = 1} \\ &= \frac{3}{2}\,\mathrm{Li}_{2}(1) \end{align}</math>

For the Dilogarithm of one this value appears: <math display="block">\mathrm{Li}_{2}(1) = \frac{\pi^2}{6} </math>

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: Template:Equation box 1

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: <math display="block">\begin{align} \int_{0}^{1} \frac{1}{\sqrt{1 - x^4}} \,\mathrm{d}x &= \left\{ - \frac{1}{2}\sqrt{2}\,F\left[\arccos(x);\frac{1}{2}\sqrt{2}\right] \right\}_{x = 0}^{x = 1} \\ &= \frac{1}{2}\sqrt{2}\,K\left(\frac{1}{2}\sqrt{2}\right) \\ \int_{0}^{1} \frac{x^2}{\sqrt{1 - x^4}} \,\mathrm{d}x &= \left\{\frac{1}{2}\sqrt{2}\,F\left[\arccos(x);\frac{1}{2}\sqrt{2}\right] - \sqrt{2}\,E\left[\arccos(x);\frac{1}{2}\sqrt{2}\right] \right\}_{x = 0}^{x = 1} \\ &= \frac{1}{2}\sqrt{2}\left[2\,E\left(\frac{1}{2}\sqrt{2}\right) - K\left(\frac{1}{2}\sqrt{2}\right)\right] \end{align}</math>

Inserting these two integrals into the above form gives: <math display="block">\begin{align} &\left[\int_{0}^{1} \frac{1}{\sqrt{1 - x^4}} \,\mathrm{d}x\right]\left[\int_{0}^{1} \frac{x^2}{\sqrt{1 - x^4}} \,\mathrm{d}x\right] = \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 \\ &\hphantom{\quad\quad\quad}= \int_{0}^{1} \left\{\frac{y^2 + 1}{2\,y^2}\left[\text{artanh}\left(y^2\right) - \text{artanh}\left(\frac{\sqrt{1 - x^4}\,y^2}{\sqrt{1 - x^4 y^4}}\right)\right]\right\}_{x = 0}^{x = 1}\,\mathrm{d}y \\ &\hphantom{\quad\quad\quad}= \int_{0}^{1} \frac{y^2 + 1}{2\,y^2}\,\mathrm{artanh}\left(y^2\right) \,\mathrm{d}y \\ &\hphantom{\quad\quad\quad}= \left[\arctan(y) - \frac{1 - y^2}{2\,y}\,\mathrm{artanh}\left(y^2\right)\right]_{y = 0}^{y = 1} \\ &\hphantom{\quad\quad\quad}= \arctan(1) \\ &\hphantom{\quad\quad\quad}= \frac{\pi}{4} \end{align}</math>

For the lemniscatic special case of Legendre's relation, this result emerges: <math display="block">K\left(\frac{1}{2}\sqrt{2}\right)\left[2\,E\left(\frac{1}{2}\sqrt{2}\right) - K\left(\frac{1}{2}\sqrt{2}\right)\right] = \frac{\pi}{2} </math>

See also

References

Template:Reflist

Further reading


Template:Measure theory

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