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Δημοσιεύτηκε: **Πέμ Σεπ 15, 2011 6:50 pm**

Έστω $\displaystyle{G\left( {n,s} \right) = - \int\limits_0^\infty {{x^{2n - 1}}\frac{{{e^{ - sx}}}}{{1 + {e^{ - sx}}}}dx} }$ , τότε $\displaystyle{\frac{d}{{ds}}\left( {G\left( {n,s} \right)} \right) = - \int\limits_0^\infty {{x^{2n - 1}}\frac{\partial }{{\partial s}}\left( {\frac{{{e^{ - sx}}}}{{1 + {e^{ - sx}}}}} \right)dx} = \int\limits_0^\infty {{x^{2n}}\frac{{{e^{ - sx}}}}{{{{\left( {1 + {e^{ - sx}}} \right)}^2}}}dx} = \int\limits_0^\infty {\frac{{{x^{2n}}}}{{\left( {1 + {e^{sx}}} \right)\left( {1 + {e^{ - sx}}} \right)}}dx} }$ .

Επομένως $\displaystyle{F\left( n \right) = {G{'}}\left( {n,1} \right)}$ .

Όμως $\displaystyle{G\left( {n,s} \right) = - \int\limits_0^\infty {{x^{2n - 1}}\frac{{{e^{ - sx}}}}{{1 + {e^{ - sx}}}}dx} = - \int\limits_0^\infty {{x^{2n - 1}} \cdot {e^{ - sx}}\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - skx}}} dx} = - \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\int\limits_0^\infty {{x^{2n - 1}}{e^{ - s\left( {k + 1} \right)x}}dx} } = }$

$\displaystyle{ = - \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{\Gamma \left( {2n} \right)}}{{{s^{2n}}{{\left( {k + 1} \right)}^{2n}}}}} = \frac{{\Gamma \left( {2n} \right)}}{{{s^{2n}}}}\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{k^{2n}}}}} }$

και $\displaystyle{\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{k^{2n}}}}} = - 1 + \frac{1}{{{2^{2n}}}} - \frac{1}{{{3^{2n}}}} + .. = \left( { - 1 - \frac{1}{{{2^{2n}}}} - \frac{1}{{{3^{2n}}}} - ..} \right) + 2\left( {\frac{1}{{{2^{2n}}}} + \frac{1}{{{4^{2n}}}} + \frac{1}{{{6^{2n}}}} + ..} \right) = }$

$\displaystyle{ = - \zeta \left( {2n} \right) + {2^{1 - 2n}}\zeta \left( {2n} \right) = \left( { - 1 + {2^{1 - 2n}}} \right)\zeta \left( {2n} \right)}$

Άρα $\displaystyle{G\left( {n,s} \right) = \frac{{\Gamma \left( {2n} \right)}}{{{s^{2n}}}}\left( { - 1 + {2^{1 - 2n}}} \right)\zeta \left( {2n} \right)}$ , οπότε $\displaystyle{{G{'}}\left( {n,s} \right) = 2n\frac{{\Gamma \left( {2n} \right)}}{{{s^{2n + 1}}}}\left( {1 - {2^{1 - 2n}}} \right)\zeta \left( {2n} \right) \Rightarrow \boxed{F\left( n \right) = {G{'}}\left( {n,1} \right) = 2n\Gamma \left( {2n} \right)\left( {1 - {2^{1 - 2n}}} \right)\zeta \left( {2n} \right)}}$ .

Δημοσιεύτηκε: **Σάβ Σεπ 17, 2011 8:23 pm**

Σεραφείμ έγραψε:Έστω $\displaystyle{G\left( {n,s} \right) = - \int\limits_0^\infty {{x^{2n - 1}}\frac{{{e^{ - sx}}}}{{1 + {e^{ - sx}}}}dx} }$ , τότε $\displaystyle{\frac{d}{{ds}}\left( {G\left( {n,s} \right)} \right) = - \int\limits_0^\infty {{x^{2n - 1}}\frac{\partial }{{\partial s}}\left( {\frac{{{e^{ - sx}}}}{{1 + {e^{ - sx}}}}} \right)dx} = \int\limits_0^\infty {{x^{2n}}\frac{{{e^{ - sx}}}}{{{{\left( {1 + {e^{ - sx}}} \right)}^2}}}dx} = \int\limits_0^\infty {\frac{{{x^{2n}}}}{{\left( {1 + {e^{sx}}} \right)\left( {1 + {e^{ - sx}}} \right)}}dx} }$ .

Επομένως $\displaystyle{F\left( n \right) = {G{'}}\left( {n,1} \right)}$ .

Όμως $\displaystyle{G\left( {n,s} \right) = - \int\limits_0^\infty {{x^{2n - 1}}\frac{{{e^{ - sx}}}}{{1 + {e^{ - sx}}}}dx} = - \int\limits_0^\infty {{x^{2n - 1}} \cdot {e^{ - sx}}\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - skx}}} dx} = - \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\int\limits_0^\infty {{x^{2n - 1}}{e^{ - s\left( {k + 1} \right)x}}dx} } = }$

$\displaystyle{ = - \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{\Gamma \left( {2n} \right)}}{{{s^{2n}}{{\left( {k + 1} \right)}^{2n}}}}} = \frac{{\Gamma \left( {2n} \right)}}{{{s^{2n}}}}\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{k^{2n}}}}} }$

και $\displaystyle{\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{k^{2n}}}}} = - 1 + \frac{1}{{{2^{2n}}}} - \frac{1}{{{3^{2n}}}} + .. = \left( { - 1 - \frac{1}{{{2^{2n}}}} - \frac{1}{{{3^{2n}}}} - ..} \right) + 2\left( {\frac{1}{{{2^{2n}}}} + \frac{1}{{{4^{2n}}}} + \frac{1}{{{6^{2n}}}} + ..} \right) = }$

$\displaystyle{ = - \zeta \left( {2n} \right) + {2^{1 - 2n}}\zeta \left( {2n} \right) = \left( { - 1 + {2^{1 - 2n}}} \right)\zeta \left( {2n} \right)}$

Άρα $\displaystyle{G\left( {n,s} \right) = \frac{{\Gamma \left( {2n} \right)}}{{{s^{2n}}}}\left( { - 1 + {2^{1 - 2n}}} \right)\zeta \left( {2n} \right)}$ , οπότε $\displaystyle{{G{'}}\left( {n,s} \right) = 2n\frac{{\Gamma \left( {2n} \right)}}{{{s^{2n + 1}}}}\left( {1 - {2^{1 - 2n}}} \right)\zeta \left( {2n} \right) \Rightarrow \boxed{F\left( n \right) = {G{'}}\left( {n,1} \right) = 2n\Gamma \left( {2n} \right)\left( {1 - {2^{1 - 2n}}} \right)\zeta \left( {2n} \right)}}$ .

ωραία λύση Σεραφείμ

μια άλλη λύση εδώ
$\displaystyle{ F\left( n \right) = \int_0^{ + \infty } {\frac{{x^{2n} }} {{\left( {e^x + 1} \right)\left( {e^{ - x} + 1} \right)}}} dx }$
θεωρώ
$\displaystyle{ K\left( s \right) = - \int_0^{ + \infty } {\frac{{x^{2n - 1} }} {{e^{sx} + 1}}} dx \Rightarrow K'\left( 1 \right) = \int_0^{ + \infty } {\frac{{x^{2n} }} {{\left( {e^x + 1} \right)\left( {e^{ - x} + 1} \right)}}} dx }$
$\displaystyle{ K\left( s \right) = \int_0^{ + \infty } {x^{2n - 1} \sum\limits_{k = 1}^{ + \infty } {\left( { - 1} \right)^k e^{ - ksx} } } dx = \sum\limits_{k = 1}^{ + \infty } {\left( { - 1} \right)^k \int_0^{ + \infty } {x^{2n - 1} } e^{ - ksx} dx} \underbrace = _{x \mapsto \frac{y} {{ks}}}\sum\limits_{k = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {ks} \right)^{2n} }}\int_0^{ + \infty } {y^{2n - 1} } e^{ - y} dy} }$
$\displaystyle{ = \Gamma \left( {2n} \right)\frac{1} {{s^{2n} }}\sum\limits_{k = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{k^{2n} }}} = - \Gamma \left( {2n} \right)\eta \left( {2n} \right)\frac{1} {{s^{2n} }} }$

Ισχύει $\eta(2n) = \zeta( {2n})\left( {1 - 2^{1 - 2n} } \right)$ http://es.wikipedia.org/wiki/Funci%C3%B ... _Dirichlet
και με την αντικατάσταση s=1

προκύπτει
$\displaystyle{F( n) = K^{\prime}( 1) = 2n\,\Gamma( {2n} )\,\zeta ( {2n} )\left( {1 - 2^{1 - 2n} } \right) }$

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