Editing Fubini's theorem (section)

=== 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>