Ολοκλήρωμα 3

china university · #1

Να αποδείξετε ότι: $\int_{0}^{2\pi}\;x^{2}\cdot\ln^{2}{\left[2\cos\left(\frac{x}{4}\right)\right]}\;\textbf dx\;=\;\boxed{\frac{22}{45}\;\pi^{5}}$

#2

china university έγραψε:Να αποδείξετε ότι: $\int_{0}^{2\pi}\;x^{2}\cdot\ln^{2}{\left[2\cos\left(\frac{x}{4}\right)\right]}\;\textbf dx\;=\;\boxed{\frac{22}{45}\;\pi^{5}}$

$\displaystyle{\int\limits_0^{2\pi } {{x^2}{{\log }^2}\left( {2\cos \frac{x}{4}} \right)dx} = \mathop = \limits^{x/4 \to x} = 64\int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right)dx} }$ . Με τον συνήθη συμβολισμό $\displaystyle{{H_n} = 1 + \frac{1}{2} + \frac{1}{3} + .. + \frac{1}{n}}$

Λήμμα 0: $\displaystyle{\frac{{{H_n}}}{n} = - \int\limits_0^1 {{x^{n - 1}}\log \left( {1 - x} \right)dx} }$ διότι
$\displaystyle{{H_n} = 1 + \frac{1}{2} + .. + \frac{1}{n} = \int\limits_0^1 {\left( {1 + x + .. + {x^{n - 1}}} \right)dx} = \int\limits_0^1 {\frac{{1 - {x^n}}}{{1 - x}}dx} = - \int\limits_0^1 {\left( {1 - {x^n}} \right){{\left( {\log \left( {1 - x} \right)} \right)}{'}}dx} = - n\int\limits_0^1 {{x^{n - 1}}\log \left( {1 - x} \right)dx} }$

Λήμμα 1: $\displaystyle{ - \frac{{\log \left( {1 - x} \right)}}{{1 - x}} = \sum\limits_{n = 1}^\infty {{H_n}{x^n}} }$ διότι

$\displaystyle{ - \frac{{\log \left( {1 - x} \right)}}{{1 - x}} = \left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \frac{{{x^4}}}{4} + ..} \right)\left( {1 + x + {x^2} + {x^3} + ..} \right) = x + \left( {1 + \frac{1}{2}} \right){x^2} + \left( {1 + \frac{1}{2} + \frac{1}{3}} \right){x^2} + .. = \sum\limits_{n = 1}^\infty {{H_n}{x^n}} }$

Λήμμα 2: $\displaystyle{\frac{1}{2}{\log ^2}\left( {1 - x} \right) = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}{x^{n + 1}}} }$ διότι

$\displaystyle{\left[ {L:1} \right] \Rightarrow - \frac{{\log \left( {1 - x} \right)}}{{1 - x}} = \sum\limits_{n = 1}^\infty {{H_n}{x^n}} \Rightarrow - \int\limits_0^x {\frac{{\log \left( {1 - y} \right)}}{{1 - y}}dy} = \sum\limits_{n = 1}^\infty {{H_n}\int\limits_0^x {{y^n}dy} } \Rightarrow \frac{1}{2}{\log ^2}\left( {1 - x} \right) = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}{x^{n + 1}}} }$

Λήμμα 3: $\displaystyle{\frac{{{H_n}}}{{n + 1}} = - \frac{1}{{{{\left( {n + 1} \right)}^2}}} - \int\limits_0^1 {{x^n}\log \left( {1 - x} \right)dx} }$ διότι

$\displaystyle{\frac{{{H_n}}}{{n + 1}} = \frac{{{H_{n + 1}} - 1/\left( {n + 1} \right)}}{{n + 1}} = \frac{{{H_{n + 1}}}}{{n + 1}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}} = \mathop = \limits^{\left[ {L:0} \right]} = - \frac{1}{{{{\left( {n + 1} \right)}^2}}} - \int\limits_0^1 {{x^n}\log \left( {1 - x} \right)dx} }$

Λήμμα 4: $\displaystyle{\int\limits_0^1 {\frac{{{{\log }^3}\left( {1 - x} \right)}}{x}dx} = - \frac{{{\pi ^4}}}{{15}}}$ διότι

$\displaystyle{\int\limits_0^1 {\frac{{{{\log }^3}\left( {1 - x} \right)}}{x}dx} = \mathop = \limits^{x = 1 - y} = \int\limits_0^1 {\frac{{{{\log }^3}y}}{{1 - y}}dy} = \int\limits_0^1 {{{\log }^3}y\sum\limits_{n = 0}^\infty {{y^n}} dy} = \sum\limits_{n = 0}^\infty {\int\limits_0^1 {{y^n}{{\log }^3}ydy} } = \mathop = \limits^{\pi \alpha \rho \alpha \gamma o\nu \tau \iota \kappa \varepsilon \varsigma } = - 6\zeta \left( 4 \right) = - \frac{{{\pi ^4}}}{{15}}}$

Λήμμα 5: $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}} = \frac{{{\pi ^4}}}{{30}} - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} }$ διότι

$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}} = \mathop = \limits^{\left[ {L:3} \right]} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}\left( { - \frac{1}{{{{\left( {n + 1} \right)}^2}}} - \int\limits_0^1 {{x^n}\log \left( {1 - x} \right)dx} } \right)} = - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} - \int\limits_0^1 {\log \left( {1 - x} \right)\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}{x^n}} dx} = }$

$\displaystyle{ = \mathop = \limits^{\left[ {L:2} \right]} = - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} - \int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{x}\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}{x^{n + 1}}} dx} = - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} - \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^3}\left( {1 - x} \right)}}{x}dx} = \mathop = \limits^{\left[ {L:4} \right]} = \frac{{{\pi ^4}}}{{30}} - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} }$

Λήμμα 6: $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} = \frac{{{\pi ^4}}}{{72}}}$ διότι

$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}\int\limits_0^1 {\frac{{1 - {x^n}}}{{1 - x}}dx} } = \int\limits_0^1 {\frac{1}{{1 - x}}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}\left( {1 - {x^n}} \right)\,} dx} = \int\limits_0^1 {\frac{1}{{1 - x}}\left( {\zeta \left( 3 \right) - L{i_3}\left( x \right)} \right)dx} = }$

$\displaystyle{ = - \int\limits_0^1 {\left( {\zeta \left( 3 \right) - L{i_3}\left( x \right)} \right){{\left( {\log \left( {1 - x} \right)} \right)}{'}}dx} = - \left[ {\left( {\zeta \left( 3 \right) - L{i_3}\left( x \right)} \right)\log \left( {1 - x} \right)} \right]_0^1 - \int\limits_0^1 {\log \left( {1 - x} \right){{\left( {L{i_3}\left( x \right)} \right)}{'}}dx} = }$

$\displaystyle{ = - \int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{x}L{i_2}\left( x \right)dx} = \int\limits_0^1 {L{i_2}\left( x \right){{\left( {L{i_2}\left( x \right)} \right)}{'}}dx} = \frac{1}{2}\left[ {{{\left( {L{i_2}\left( x \right)} \right)}^2}} \right]_0^1 = \frac{1}{2}{\zeta ^2}\left( 2 \right) = \frac{{{\pi ^4}}}{{72}}}$

Λήμμα 7: $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}} = \frac{{11{\pi ^4}}}{{360}}}$ διότι

$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}} = \mathop = \limits^{\left[ {L:5} \right]} = \frac{{{\pi ^4}}}{{30}} - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} = \frac{{{\pi ^4}}}{{30}} - \sum\limits_{n = 1}^\infty {\frac{{{H_{n + 1}} - 1/\left( {n + 1} \right)}}{{{{\left( {n + 1} \right)}^3}}}} = \frac{{{\pi ^4}}}{{30}} - \sum\limits_{n = 1}^\infty {\frac{{{H_{n + 1}}}}{{{{\left( {n + 1} \right)}^3}}}} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^4}}}} = \mathop = \limits^{\left[ {L:6} \right]} = \frac{{11{\pi ^4}}}{{360}}}$

Τότε με χρήση των παραπάνω έχουμε

$\displaystyle{\left[ {L:2} \right] \Rightarrow \frac{1}{2}{\log ^2}\left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{n + 1}}{x^{n + 1}}} \Rightarrow \frac{1}{2}{\log ^2}\left( {1 + {e^{2ix}}} \right) = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{n + 1}}{e^{2i\left( {n + 1} \right)x}}} \Rightarrow }$

$\displaystyle{ \Rightarrow \frac{1}{2}{\left( {\log \left( {2\cos x} \right) + ix} \right)^2} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{n + 1}}\cos 2\left( {n + 1} \right)x} + i\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{n + 1}}\sin 2\left( {n + 1} \right)x} \Rightarrow }$

$\displaystyle{ \Rightarrow x\log \left( {2\cos x} \right) = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{n + 1}}\sin 2\left( {n + 1} \right)x} \Rightarrow {x^2}{\log ^2}\left( {2\cos x} \right) = \sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}{{\sin }^2}2\left( {n + 1} \right)x} + }$

$\displaystyle{ + 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{H_n}}}{{n + 1}}\sin 2\left( {n + 1} \right)x\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)}^{k - 1}}\frac{{{H_k}}}{{k + 1}}\sin 2\left( {k + 1} \right)x} } = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}\left( {1 - \cos 4\left( {n + 1} \right)x} \right)} + }$

$\displaystyle{ + \sum\limits_{n = 1}^\infty {\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)}^{n + k}}\frac{{{H_n}{H_k}}}{{\left( {n + 1} \right)\left( {k + 1} \right)}}\left( {\cos 2\left( {n - k} \right)x - \cos 2\left( {n + k} \right)} \right)} } \Rightarrow \int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right)dx} = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}\int\limits_0^{\pi /2} {1dx} } - }$

$\displaystyle{ - \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}\int\limits_0^{\pi /2} {\cos 4\left( {n + 1} \right)xdx} } + \sum\limits_{n = 1}^\infty {\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)}^{n + k}}\frac{{{H_n}{H_k}}}{{\left( {n + 1} \right)\left( {k + 1} \right)}}\left( {\int\limits_0^{\pi /2} {\cos 2\left( {n - k} \right)xdx} - \int\limits_0^{\pi /2} {\cos 2\left( {n + k} \right)dx} } \right)} } \Rightarrow }$

$\displaystyle{ \Rightarrow \int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right)dx} = \frac{\pi }{4}\sum\limits_{n = 1}^\infty {\frac{{H_n^2}}{{{{\left( {n + 1} \right)}^2}}}} = \frac{\pi }{4} \cdot \frac{{11{\pi ^4}}}{{360}} \Rightarrow \boxed{\int\limits_0^{2\pi } {{x^2}{{\log }^2}\left( {2\cos \frac{x}{4}} \right)dx} = 64\int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right)dx} = \frac{{22{\pi ^5}}}{{45}}}}$

Ολοκλήρωμα 3

Ολοκλήρωμα 3

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