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Δημοσιεύτηκε: **Τετ Νοέμ 25, 2015 8:33 pm**

Για

θετικό ακέραιο έστω $\displaystyle{I_n:=\int_{0}^{+\infty}\frac{\arctan x}{(1+x^2)^n}\,dx}$ .

Ας δειχθεί ότι:

$\displaystyle{\sum_{n\geq1}\frac{I_n}{n}=\zeta(2)}$ και

$\displaystyle{\int_{0}^{+\infty}\arctan x\ln\left(1+\frac{1}{x^2}\right)\,dx=\zeta(2)}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

Δημοσιεύτηκε: **Σάβ Νοέμ 28, 2015 8:33 am**

Κοτρώνης Αναστάσιος έγραψε:Για θετικό ακέραιο έστω $\displaystyle{I_n:=\int_{0}^{+\infty}\frac{\arctan x}{(1+x^2)^n}\,dx}$ . Ας δειχθεί ότι: $\displaystyle{\sum_{n\geq1}\frac{I_n}{n}=\zeta(2)}$ και $\displaystyle{\int_{0}^{+\infty}\arctan x\ln\left(1+\frac{1}{x^2}\right)\,dx=\zeta(2)}$ .

Πρώτα το δεύτερο ολοκλήρωμα
$\displaystyle{I = \int\limits_0^\infty {\arctan x\log \left( {1 + \frac{1}{{{x^2}}}} \right)dx} = \int\limits_0^1 {\arctan x\log \left( {1 + \frac{1}{{{x^2}}}} \right)dx} + \underbrace {\int\limits_1^\infty {\arctan x\log \left( {1 + \frac{1}{{{x^2}}}} \right)dx} }_{x \to 1/x} = }$
$\displaystyle{ = \int\limits_0^1 {\arctan x\log \left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)dx} + \int\limits_0^1 {\left( {\frac{\pi }{2} - \arctan x} \right)\frac{{\log \left( {{x^2} + 1} \right)}}{{{x^2}}}dx} = }$

$\displaystyle{ = \underbrace {\int\limits_0^1 {\arctan x\log \left( {{x^2} + 1} \right)dx} }_{{I_1}} - \underbrace {\int\limits_0^1 {\arctan x\log \left( {{x^2}} \right)dx} }_{{I_2}} + }$ $\displaystyle{\frac{\pi }{2}\underbrace {\int\limits_0^1 {\frac{{\log \left( {{x^2} + 1} \right)}}{{{x^2}}}dx} }_{{I_3}} - \underbrace {\int\limits_0^1 {\frac{{\arctan x\log \left( {{x^2} + 1} \right)}}{{{x^2}}}dx} }_{{I_4}}}$

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Λήμμα $\displaystyle{\int\limits_0^1 {\arctan x \cdot \log \left( {1 + {x^2}} \right)dx} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_0^1 {{{\log }^2}\left( {1 + ix} \right)dx} } \right)}$ και $\displaystyle{\int\limits_0^1 {\frac{{\arctan x \cdot \log \left( {1 + {x^2}} \right)}}{{{x^2}}}dx} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}}{{{x^2}}}dx} } \right)}$

Διότι $\displaystyle{\int\limits_0^1 {{{\log }^2}\left( {1 + ix} \right)dx} = \int\limits_0^1 {{{\log }^2}\left( {\sqrt {1 + {x^2}} \cdot {e^{i \cdot arc\tan x}}} \right)dx} = \int\limits_0^1 {{{\left( {\frac{1}{2}\log \left( {1 + {x^2}} \right) + i \cdot \arctan x} \right)}^2}dx} = }$

$\displaystyle{ = A + i\int\limits_0^1 {\log \left( {1 + {x^2}} \right) \cdot \arctan x\;dx} \Rightarrow \int\limits_0^1 {\arctan x\log \left( {1 + {x^2}} \right)\;dx} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_0^1 {{{\log }^2}\left( {1 + ix} \right)dx} } \right)}$ .. όμοια και το δεύτερο.

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Υπολογισμοί:
$\displaystyle{\int\limits_0^1 {{{\log }^2}\left( {1 + ix} \right)dx} = \mathop = \limits^{1 + ix = y} = - i\int\limits_1^{1 + i} {{{\log }^2}\left( y \right)dy} = - i\left[ {2y - 2y\log y + y{{\log }^2}y} \right]_1^{1 + i} = }$

$\displaystyle{ = .. = \left( {2 - \log 2 - \frac{\pi }{2} + \frac{\pi }{4}\log 2 + \frac{1}{4}{{\log }^2}2 - \frac{{{\pi ^2}}}{{16}}} \right) + i\left( {\log 2 - \frac{\pi }{2} - \frac{1}{4}{{\log }^2}2 + \frac{{{\pi ^2}}}{{16}} + \frac{\pi }{4}\log 2} \right) \Rightarrow }$

$\displaystyle{ \Rightarrow {I_1} = \int\limits_0^1 {\log \left( {1 + {x^2}} \right) \cdot \arctan x\;dx} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_0^1 {{{\log }^2}\left( {1 + ix} \right)dx} } \right) = \log 2 - \frac{\pi }{2} - \frac{1}{4}{\log ^2}2 + \frac{{{\pi ^2}}}{{16}} + \frac{\pi }{4}\log 2}$

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$\displaystyle{\int\limits_0^1 {\arctan x\log \left( {{x^2}} \right)dx} = 2\int\limits_0^1 {\arctan x\log \left( x \right)dx} = }$ $\displaystyle{2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{2n - 1}}\int\limits_0^1 {\log \left( x \right){x^{2n - 1}}dx} } = - \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{\left( {2n - 1} \right){n^2}}}} = }$

$\displaystyle{ = - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {\frac{1}{{2n - 1}} - \frac{1}{{2n}}} \right)} = - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{n\left( {2n - 1} \right)}}} + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}} }$ $\displaystyle{ = - 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\left( {\frac{1}{{2n - 1}} - \frac{1}{{2n}}} \right)} + \frac{{{\pi ^2}}}{{24}} = }$

$\displaystyle{ = - 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{2n - 1}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}} + \frac{{{\pi ^2}}}{{24}} \Rightarrow }$ $\displaystyle{{I_2} = \int\limits_0^1 {\arctan x\log \left( {{x^2}} \right)dx} = - \frac{\pi }{2} + \log 2 + \frac{{{\pi ^2}}}{{24}}}$

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

$\displaystyle{ = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{2n - 1}}} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}} \Rightarrow {I_3} = \int\limits_0^1 {\frac{{\log \left( {1 + {x^2}} \right)}}{{{x^2}}}dx} = \frac{\pi }{2} - \log 2}$

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

$\displaystyle{ = i\sum\limits_{n = 1}^\infty {\left( {\frac{2}{{{n^2}}} + \frac{{ - 2}}{{{n^2}{{\left( {1 + i} \right)}^n}}} - \frac{{2\log \left( {1 + i} \right)}}{{n{{\left( {1 + i} \right)}^n}}} - \frac{{{{\log }^2}\left( {1 + i} \right)}}{{{{\left( {1 + i} \right)}^n}}}} \right)} = }$ $\displaystyle{i\frac{{{\pi ^2}}}{3} - 2i \cdot L{i_2}\left( {\frac{1}{{1 + i}}} \right) - 2i\log \left( {1 + i} \right)\log \left( {1 - \frac{1}{{1 + i}}} \right) - }$

$\displaystyle{ - {\log ^2}\left( {1 + i} \right) \Rightarrow .. \Rightarrow {I_4} = \int\limits_0^1 {\arctan x\frac{{\log \left( {1 + {x^2}} \right)}}{{{x^2}}}dx} = }$ $\displaystyle{{\mathop{\rm Im}\nolimits} \left( {\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}}{{{x^2}}}dx} } \right) = - \frac{\pi }{4}\log 2 + \frac{5}{{48}}{\pi ^2} - \frac{1}{4}{\log ^2}2}$

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Συμμαζεύοντας όλα τα παραπάνω έχουμε $\displaystyle{I = \int\limits_0^\infty {\arctan x\log \left( {1 + \frac{1}{{{x^2}}}} \right)dx} = {I_1} - {I_2} + \frac{\pi }{2}{I_3} - {I_4} = \frac{{{\pi ^2}}}{6} = \zeta \left( 2 \right)}$

Αναφορικά με το πρώτο έχουμε $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{I_n}}}{n}} = \int\limits_0^\infty {\arctan x\sum\limits_{n = 1}^\infty {\frac{1}{{n{{\left( {1 + {x^2}} \right)}^n}}}} \;dx} = - \int\limits_0^\infty {\arctan x\log \left( {1 - \frac{1}{{1 + {x^2}}}} \right)dx} }$ $\displaystyle{ = - \int\limits_0^\infty {\arctan x\log \frac{{{x^2}}}{{1 + {x^2}}}dx} = }$

$\displaystyle{ = \int\limits_0^\infty {\arctan x\log \frac{{{x^2} + 1}}{{{x^2}}}dx} = \int\limits_0^\infty {\arctan x\log \left( {1 + \frac{1}{{{x^2}}}} \right)dx} = I = \zeta \left( 2 \right)}$

Πήρα την ανάποδη σειρά γιατί το πρώτο ολοκλήρωμα με οδηγούσε μονοσήμαντα στο δεύτερο .. δεν βρήκα άλλο δρόμο υπολογισμού του πρώτου .. ας είναι ..

Δημοσιεύτηκε: **Κυρ Νοέμ 29, 2015 2:01 pm**

Γεια σου Σεραφείμ!

Εδώ μια άλλη λύση κι εδώ οι δημοσιευμένες. Είναι το 5360 το SSM.

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