Κλειστός τύπος με γινόμενο λογαρίθμων

galactus · #1

συγχωρέστε αν τα ελληνικά μου δεν είναι πολύ καλή

Υπολογίστε:
$\displaystyle{\displaystyle{\int_{0}^{1}\log(1+x)\log(1-x^{3})dx}=}$

Δε ξερώ ακριβώς τη λύση.

Κατάφερα να βρω κάτι ισοδύναμο:

$\displaystyle{\displaystyle{-\gamma+2\gamma \log(2)-\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi\left(\frac{n+4}{3}\right)}{n(n+1)}}}$

χρησιμοποιώντας τη συνάρτηση Βeta. Έχω καταφέρει να βρω το παρόμοιο:

$\displaystyle{\displaystyle{\int_{0}^{1}\log(1-x)\log(1+x^{3})dx=-1/2\psi_{1}(1/3)+\log^{2}(2)-2\log(2)+\frac{5\pi^{2}}{36}-\frac{\pi}{\sqrt{3}}+6}}$

#2

galactus έγραψε:Υπολογίστε: $\displaystyle{\displaystyle{\int_{0}^{1}\log(1+x)\log(1-x^{3})dx}}$

Λήμμα : $\displaystyle{\int\limits_0^a {\log x \cdot \log \left( {z - x} \right)dx} = \left( {\log \left( {1 - \frac{a}{z}} \right)\left( {z - a} \right) - a\log z} \right)\left( {1 - \log a} \right) + 2a - z \cdot L{i_2}\left( {\frac{a}{z}} \right) - a\log a}$ διότι

$\displaystyle{\int\limits_0^a {\log x \cdot \log \left( {z - x} \right)dx} = \int\limits_0^a {\log x \cdot \log \left( {z \cdot \left( {1 - \frac{x}{z}} \right)} \right)dx} = \log z\int\limits_0^a {\log x\;dx} + \int\limits_0^a {\log x \cdot \log \left( {1 - \frac{x}{z}} \right)dx} = }$

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

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

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

$\displaystyle{ + \log a\sum\limits_{n = 1}^\infty {\frac{{{a^{n + 1}}}}{{{z^n}\left( {n + 1} \right)}}} = \log z \cdot a\left( { - 1 + \log a} \right) + \sum\limits_{n = 1}^\infty {\frac{{{a^{n + 1}}}}{{{z^n}}}\left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} - }$ $\displaystyle{z\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}{n^2}}}} - a\log a\sum\limits_{n = 1}^\infty {\frac{{{a^n}}}{{{z^n}n}}} + z\log a\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}n}}} = }$

$\displaystyle{ = \log z \cdot a\left( { - 1 + \log a} \right) + \sum\limits_{n = 1}^\infty {\frac{{{a^{n + 1}}}}{{{z^n}}}\left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} - z\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}{n^2}}}} - a\log a\sum\limits_{n = 1}^\infty {\frac{{{a^n}}}{{{z^n}n}}} + z\log a\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}n}}} = }$

$\displaystyle{ = \log z \cdot a\left( { - 1 + \log a} \right) + \sum\limits_{n = 1}^\infty {\frac{{{a^{n + 1}}}}{{{z^n}n}}} - \sum\limits_{n = 1}^\infty {\frac{{{a^{n + 1}}}}{{{z^n}\left( {n + 1} \right)}}} - z\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}{n^2}}}} - a\log a\sum\limits_{n = 1}^\infty {\frac{{{a^n}}}{{{z^n}n}}} + z\log a\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}n}}} = }$

$\displaystyle{ = \log z \cdot a\left( { - 1 + \log a} \right) + a\sum\limits_{n = 1}^\infty {\frac{{{a^n}}}{{{z^n}n}}} - z\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}n}}} - z\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}{n^2}}}} - a\log a\sum\limits_{n = 1}^\infty {\frac{{{a^n}}}{{{z^n}n}}} + z\log a\sum\limits_{n = 2}^\infty {\frac{{{a^n}}}{{{z^n}n}}} = }$

$\displaystyle{ = \log z\left( { - a + a\log a} \right) - a\log \left( {1 - \frac{a}{z}} \right) - z\left( { - \frac{a}{z} - \log \left( {1 - \frac{a}{z}} \right)} \right) - }$ $\displaystyle{z\left( { - \frac{a}{z} + L{i_2}\left( {\frac{a}{z}} \right)} \right) + a\log a\log \left( {1 - \frac{a}{z}} \right) + }$

$\displaystyle{ + z\log a\left( { - \frac{a}{z} - \log \left( {1 - \frac{a}{z}} \right)} \right) = \left( {\log \left( {1 - \frac{a}{z}} \right)\left( {z - a} \right) - a\log z} \right)\left( {1 - \log a} \right) + 2a - z \cdot L{i_2}\left( {\frac{a}{z}} \right) - a\log a}$

Τότε

$\displaystyle{\int\limits_0^1 {\log \left( {1 + x} \right)\log \left( {1 - {x^3}} \right)dx} = \int\limits_0^1 {\log \left( {1 + x} \right)\log \left( {1 - x} \right)dx} + \int\limits_0^1 {\log \left( {1 + x} \right)\log \left( {1 + x + {x^2}} \right)dx} = \mathop = \limits^{x + 1 \to x} = }$

$\displaystyle{ = \int\limits_1^2 {\log x\log \left( {2 - x} \right)dx} + \int\limits_1^2 {\log x\log \left( {1 - x + {x^2}} \right)dx} = }$ $\displaystyle{\int\limits_1^2 {\log x\log \left( {2 - x} \right)dx} + \int\limits_1^2 {\log x\log \left( {\frac{{1 + i\sqrt 3 }}{2} - x} \right)dx} + }$

$\displaystyle{ + \int\limits_1^2 {\log x\log \left( {\frac{{1 - i\sqrt 3 }}{2} - x} \right)dx} = \int\limits_0^2 {\log x\log \left( {2 - x} \right)dx} - }$ $\displaystyle{\int\limits_0^1 {\log x\log \left( {2 - x} \right)dx} + \int\limits_0^2 {\log x\log \left( {\frac{{1 + i\sqrt 3 }}{2} - x} \right)dx} - }$

$\displaystyle{ - \int\limits_0^1 {\log x\log \left( {\frac{{1 + i\sqrt 3 }}{2} - x} \right)dx} + \int\limits_0^2 {\log x\log \left( {\frac{{1 - i\sqrt 3 }}{2} - x} \right)dx} - \int\limits_0^1 {\log x\log \left( {\frac{{1 - i\sqrt 3 }}{2} - x} \right)dx} }$

Εφαρμόζοντας διαδοχικά το λήμμα για $\displaystyle{a = 2}$ , $\displaystyle{a = 1}$ και $\displaystyle{z = 2,{\rm{ }}\frac{{1 \pm i\sqrt 3 }}{2}}$ προκύπτει

$\displaystyle{\int\limits_0^1 {\log \left( {1 + x} \right) \cdot \log \left( {1 - {x^3}} \right)dx} = 2 + \frac{{\left( {1 - \log 2} \right)\left( { - 3\log 3 + 8 - \sqrt 3 \pi } \right)}}{2} + \frac{\pi }{{\sqrt 3 }} + {\log ^2}2 - }$

$\displaystyle{ - 2\log 2 - \frac{{{\pi ^2}}}{6} - 2{\mathop{\rm Re}\nolimits} \left( {\frac{{1 + i\sqrt 3 }}{2} \cdot \left( {L{i_2}\left( {1 - i\sqrt 3 } \right) - L{i_2}\left( {\frac{{1 - i\sqrt 3 }}{2}} \right)} \right)} \right)}$

Δεν μπορώ να φανταστώ πως μπορούν να φύγουν οι πολυλογάριθμοι .. αν και πιστεύω πως δεν μπορούν να φύγουν ..

galactus · #3

$\displaystyle{Li_{2}\left(\frac{1}{2}-\frac{\sqrt{3}i}{2}\right)=Li_{2}(e^{-\pi i/3})=\frac{\pi^{2}}{36}-Cl_{2}(\frac{\pi}{3})}i*$

$\displaystyle{*Cl_{2}(\frac{\pi}{3})=\sum_{n=1}^{\infty}\frac{\sin(\frac{\pi n}{3})}{n^{2}}=\frac{\sqrt{3}}{6}\psi_{1}(1/3)-\frac{\pi^{2}\sqrt{3}}{9}}$

Κλειστός τύπος με γινόμενο λογαρίθμων

Κλειστός τύπος με γινόμενο λογαρίθμων

Re: Κλειστός τύπος με γινόμενο λογαρίθμων

Re: Κλειστός τύπος με γινόμενο λογαρίθμων

Μέλη σε σύνδεση