Editing Fubini's theorem (section)

== Calculation examples ==
{{tone|section|date=April 2024}}

=== Arcsine integral ===

The arcsine integral, also called the inverse sine integral, is a function that cannot be represented by [[elementary function]]s. 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 function]] - the cardinalized arcsine. {{Clarify|date=February 2025|reason= Isn't this just the definition of the "arcsine integral" (by analogy with the sine integral)?|text=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:

<math display="block">\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} =</math>
<math display="block">= {\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 =</math>
<math display="block">= {\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)</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 η(2) is equal to π²/12 and this can be proven with Fubini's theorem{{dubious|| the condition for using Fubini is not fulfilled since |f(x,n)|=\frac{1}{n} {x}^{n-1} is NOT (Lebesgue) integrable because the harmonic series diverges!|date=January 2024}} 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 [[Multiplicative inverse|Reciprocal Function]] and the [[Logarithm|Natural Logarithm]] of the ''Successor Function'' is a [[Polylogarithm|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"> \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} = </math>

<math display="block"> = {\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 = </math>

<math display="block"> = {\color{RoyalBlue}\biggl[\frac{\pi^2}{12}-\frac{1}{3}\arccos(y)^2 \biggr]_{y = 0} ^{y = 1}} = \frac{\pi^2}{12} </math>

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):

: <math> \eta(2) = \frac{\pi^2}{12} </math>

=== Integrals of Complete Elliptic Integrals ===

The [[improper integral]] of the [[Elliptic integral#Complete elliptic integral of the first kind|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 [[Inverse tangent integral|Arctangent Integral]], which results from the application of Fubini's theorem:

<math display="block"> \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}= </math>

<math display="block"> = \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 </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 the second kind'' '''E''' in the following way:

<math display="block"> \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}= </math>
<math display="block"> = \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} </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 [[Logarithm|Natural Logarithm]] and the [[Exponential function|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}\biggl[\frac{1}{x + 1}-\exp(-x)\biggr] \,\mathrm{d}x } </math>

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|Exponential Integral Function]]:

This is how the complementary integral exponential function is defined:

: <math>\mathrm{E}_{1}(x) = \exp(-x)\int_{0}^{\infty} \frac{\exp(-xy)}{y+1} \, \mathrm{d}y </math>

This is the derivative of that function:

: <math>\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"> \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} = </math>
<math display="block"> = {\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{brown}\int_{0}^{\infty} \biggl[\exp(x)\,\mathrm{E}_{1}(x) - \frac{1}{x + 1}\biggr] \,\mathrm{d}x} </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"> \gamma = {\color{brown}\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} = </math>
<math display="block">
\definecolor{cerulean}{rgb}{0.0, 0.48, 0.65}
= {\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{cerulean}\int_{0}^{\infty} \frac{1}{z}\biggl[\frac{1}{z + 1}-\exp(-z)\biggr] \,\mathrm{d}z} </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} \biggl\{-\frac{1}{b} \exp\bigl(\frac{ac}{b}\bigr)\,\mathrm{E}_{1}\bigl[\frac{a}{b}\bigl(bx + c\bigr)\bigr] \biggr\} </math>
<math display="block">\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) </math>

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:

:{| class = "wikitable"
|<math>\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 </math>
|}

This chain of equations can then be generated accordingly:

<math display="block">\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 = </math>

<math display="block">= \int_{0}^{1} \biggl\{- \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} </math>

For the integral of the [[Normal distribution|Gauss curve]] this value can be generated:

: <math>\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:

:{| class = "wikitable"
|<math>\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 </math>
|}

So this chain of equations applies as a new example:

<math display="block">\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 = </math>

<math display="block">= \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 = </math>

<math display="block">= \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) </math>

For the [[Dilogarithm]] of one this value appears:

: <math>\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:

:{| class = "wikitable"
| <math>\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 </math>
|}

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

: <math> \int_{0}^{1} \frac{1}{\sqrt{1 - x^4}} \,\mathrm{d}x = </math>

<math display="block"> \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) </math>

: <math>\int_{0}^{1} \frac{x^2}{\sqrt{1 - x^4}} \,\mathrm{d}x =</math>

<math display="block">\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] </math>

Inserting these two integrals into the above form gives:

<math display="block">\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 = </math>

<math display="block">= \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= </math>

<math display="block">= \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} </math>

For the [[Lemniscate elliptic functions|Lemniscatic]] special case of [[Legendre's relation]], this result emerges:

: <math>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} </math>