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Fubini's theorem
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=== 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.
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