ολοκλήρωμα

#2

thepathofresistance έγραψε: $\displaystyle{\int_{0}^{1}\int_{0}^{1}\frac{ln\left ( 2-xy \right )}{1-xy}dxdy}$

Λήμμα 1 : Αν $\displaystyle{{H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} }$ (αρμονικός) τότε $\displaystyle{\frac{{{H_n}}}{n} = - \int\limits_0^1 {{x^{n - 1}}\log \left( {1 - x} \right)dx} }$ διότι
$\displaystyle{{H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} = \sum\limits_{k = 1}^n {\int\limits_0^1 {{x^{k - 1}}dx} } = \int\limits_0^1 {\sum\limits_{k = 1}^n {{x^{k - 1}}} dx} = \int\limits_0^1 {\frac{{1 - {x^n}}}{{1 - x}}dx} = - \int\limits_0^1 {\left( {1 - {x^n}} \right)\left( {\log \left( {1 - x} \right)} \right)'dx} = - n\int\limits_0^1 {{x^{n - 1}}\log \left( {1 - x} \right)dx} }$

Λήμμα 2 : $\displaystyle{\log \left( {1 - x} \right)\log \left( {1 + x} \right) = \frac{1}{2}\left( {{{\log }^2}\left( {1 - {x^2}} \right) - {{\log }^2}\left( {1 - x} \right) - {{\log }^2}\left( {1 + x} \right)} \right)}$ (φανερό).

Λήμμα 3 : $\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 - {x^2}} \right)}}{x}dx = \frac{1}{2}} \int\limits_0^1 {\frac{{{{\log }^2}\left( {1 - x} \right)}}{x}dx} }$ .. εύκολο με τον μετασχηματισμό $\displaystyle{x = \sqrt y }$

Στο θέμα μας.

$\displaystyle{I = \int\limits_0^1 {\int\limits_0^1 {\frac{{\log \left( {2 - xy} \right)}}{{1 - xy}}dx} dy} = \int\limits_0^1 {\int\limits_0^1 {\frac{{\log \left( {1 + 1 - xy} \right)}}{{1 - xy}}dx} dy} = \int\limits_0^1 {\left( {\int\limits_0^1 {\frac{1}{{1 - xy}}\sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\frac{{{{\left( {1 - xy} \right)}^{n + 1}}}}{{n + 1}}} \,dx} } \right)dy} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n + 1}}\int\limits_0^1 {\int\limits_0^1 {{{\left( {1 - xy} \right)}^n}dx} dy} } }$

$\displaystyle{ = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n + 1}}\int\limits_0^1 {\left( { - \frac{1}{{y\left( {n + 1} \right)}}\left[ {{{\left( {1 - xy} \right)}^{n + 1}}} \right]_{x = 0}^{x = 1}} \right)dy} } = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n + 1} \right)}^2}}}\int\limits_0^1 {\left( {\frac{1}{y}\left( {1 - {{\left( {1 - y} \right)}^{n + 1}}} \right)} \right)dy} } = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n + 1} \right)}^2}}}\int\limits_0^1 {\left( {\sum\limits_{k = 0}^n {{{\left( {1 - y} \right)}^k}} } \right)dy} } }$

$\displaystyle{ = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n + 1} \right)}^2}}}\sum\limits_{k = 0}^n {\int\limits_0^1 {{{\left( {1 - y} \right)}^k}dy} } } = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n + 1} \right)}^2}}}\sum\limits_{k = 0}^n {\int\limits_0^1 {{y^k}dy} } } = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n + 1} \right)}^2}}}\sum\limits_{k = 0}^n {\frac{1}{{k + 1}}} } = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{H_{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}}}{{{n^2}}}} = }$

$\displaystyle{ = \mathop = \limits^{Lemma\;1} = - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\int\limits_0^1 {{x^{n - 1}}\log \left( {1 - x} \right)dx} = } - \int\limits_0^1 {\log \left( {1 - x} \right)\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{x^{n - 1}}}}{n}\,} dx} = - \int\limits_0^1 {\frac{{\log \left( {1 - x} \right)\log \left( {1 + x} \right)}}{x}dx} \Rightarrow }$

$\displaystyle{ \Rightarrow \mathop \Rightarrow \limits^{Lemma\;2,3} \Rightarrow I = \frac{1}{4}\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 - x} \right)}}{x}dx} + \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + x} \right)}}{x}dx} }$

Όμως $\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 - x} \right)}}{x}dx} = \int\limits_0^1 {\frac{{{{\log }^2}\left( x \right)}}{{1 - x}}dx} = \int\limits_0^1 {{{\log }^2}\left( x \right)\sum\limits_{n = 0}^\infty {{x^n}} dx} = \sum\limits_{n = 0}^\infty {\int\limits_0^1 {{x^n}{{\log }^2}\left( x \right)dx} } = .. = 2\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^3}}}} = 2 \cdot \zeta \left( 3 \right)}$

και $\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + x} \right)}}{x}dx} = \mathop = \limits^{1 + x = y} = \int\limits_1^2 {\frac{{{{\log }^2}\left( y \right)}}{{y - 1}}dy} = \mathop = \limits^{y = 1/x} = \int\limits_{1/2}^1 {\frac{{{{\log }^2}\left( x \right)}}{{x\left( {1 - x} \right)}}dx} = \int\limits_{1/2}^1 {{{\log }^2}\left( x \right)\left( {\frac{1}{x} + \frac{1}{{1 - x}}} \right)dx} = \int\limits_{1/2}^1 {\frac{{{{\log }^2}\left( x \right)}}{x}dx} + }$

$\displaystyle{ + \int\limits_{1/2}^1 {\frac{{{{\log }^2}\left( x \right)}}{{1 - x}}dx} = \frac{{{{\log }^3}2}}{3} + \int\limits_{1/2}^1 {{{\log }^2}\left( x \right)\sum\limits_{n = 0}^\infty {{x^n}} dx} = \frac{{{{\log }^3}2}}{3} + \sum\limits_{n = 0}^\infty {\int\limits_{1/2}^1 {{{\log }^2}\left( x \right){x^n}dx} } }$
Με κλασσική παραγοντική ολοκλήρωση βρίσκουμε ότι $\displaystyle{\sum\limits_{n = 0}^\infty {\int\limits_{1/2}^1 {{{\log }^2}\left( x \right){x^n}dx} } = 2 \cdot \zeta \left( 3 \right) - {\log ^3}2 - 2\log 2 \cdot L{i_2}\left( {\frac{1}{2}} \right) - 2 \cdot L{i_3}\left( {\frac{1}{2}} \right)}$

όπου $\displaystyle{L{i_2}\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{{n^2}}}} }$ και $\displaystyle{L{i_3}\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{{n^3}}}} }$ . Μέσω των συναρτησιακών σχέσεων για τις $\displaystyle{L{i_2}}$ και $\displaystyle{L{i_3}}$ που αναλύθηκαν αποδείχθηκαν εδώ viewtopic.php?f=9&t=12954&p=72038&hilit ... kel#p72038 (Πρόβλημα Νο 9), βρίσκουμε ότι $\displaystyle{L{i_2}\left( {\frac{1}{2}} \right) = \frac{{{\pi ^2}}}{{12}} - \frac{1}{2} \cdot {\log ^2}2}$ και $\displaystyle{L{i_3}\left( {\frac{1}{2}} \right) = \frac{7}{8}\zeta \left( 3 \right) - \frac{{{\pi ^2}}}{{12}}\log 2 + \frac{1}{6} \cdot {\log ^3}2}$

Συμμαζεύοντας τα παραπάνω έχουμε $\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + x} \right)}}{x}dx} = \frac{1}{4} \cdot \zeta \left( 3 \right)}$

και τελικά $\displaystyle{I = \frac{1}{4}\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 - x} \right)}}{x}dx} + \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + x} \right)}}{x}dx} = \frac{1}{2}\zeta \left( 3 \right) + \frac{1}{8}\zeta \left( 3 \right) = \frac{5}{8}\zeta \left( 3 \right)}$

------------------------------------------------------------------------

Σχόλια :

1) Από τον Euler έχει αποδειχθεί ότι $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^2}}}} = 2\zeta \left( 3 \right)}$ . Στο παρόν προκύπτει ότι $\displaystyle{\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{{n^2}}}} = \frac{5}{8}\zeta \left( 3 \right)}$

2) Πολύ θα ήθελα να δω μια στοιχειωδέστερη απόδειξη του $\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + x} \right)}}{x}dx} = \frac{1}{4} \cdot \zeta \left( 3 \right)}$ (χωρίς πολυλογαριθμικές συναρτήσεις)

3) Πολύ δύσκολο θέμα εκτός κι αν πήρα λάθος δρόμο.

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