Ολοκλήρωμα με διλογάριθμο

Tolaso J Kos · #1

Έστω $\mathbb{N} \ni s \geq 2$ και έστω

άρτιος. και ας δηλώσουμε τον διλογάριθμο με ${\rm Li}_2$ και το τριλογάριθμο με ${\rm Li}_3$ . Τότε δείξατε ότι
$\displaystyle{\int_0^{\infty} \frac{x^{s/2-1}{\rm Li}_2(-x)}{1+x^s}\, {\rm d}x}=- \frac{\pi^3}{4} \left( \frac1{3 s}+ \frac1{ s^3}\right)$ Ερώτηση: Τι μπορούμε να πούμε για το παρόμοιο
$\displaystyle{\mathcal{J}=\int_0^{\infty} \frac{x^{s/2-1}{\rm Li}_3(-x)}{1+x^s}\, {\rm d}x}}$

#2

Tolaso J Kos έγραψε:Έστω $\mathbb{N} \ni s \geq 2$ και έστω άρτιος. και ας δηλώσουμε τον διλογάριθμο με ${\rm Li}_2$ και το τριλογάριθμο με ${\rm Li}_3$ . Τότε δείξατε ότι

$\displaystyle{\int_0^{\infty} \frac{x^{s/2-1}{\rm Li}_2(-x)}{1+x^s}\, {\rm d}x}=- \frac{\pi^3}{4} \left( \frac1{3 s}+ \frac1{ s^3}\right)$

$\displaystyle{I = \int\limits_0^\infty {\frac{{{x^{s/2 - 1}}L{i_2}\left( { - x} \right)}}{{1 + {x^s}}}dx} \mathop { = = = = = = = = }\limits^{x = 1/y} \int\limits_0^\infty {\frac{{{x^{s/2 - 1}}L{i_2}\left( { - \frac{1}{x}} \right)}}{{1 + {x^s}}}dx} }$

Τότε $\displaystyle{2 \cdot I = \int\limits_0^\infty {\frac{{{x^{s/2 - 1}}\left( {L{i_2}\left( { - x} \right) + L{i_2}\left( { - \frac{1}{x}} \right)} \right)}}{{1 + {x^s}}}dx} = \int\limits_0^\infty {\frac{{{x^{s/2 - 1}}\left( { - \frac{1}{2}{{\log }^2}x - \frac{{{\pi ^2}}}{6}} \right)}}{{1 + {x^s}}}dx} = }$

$\displaystyle{ = - \frac{1}{2}\int\limits_0^\infty {\frac{{{x^{s/2 - 1}}{{\log }^2}x}}{{1 + {x^s}}}dx} - \frac{{{\pi ^2}}}{6}\int\limits_0^\infty {\frac{{{x^{s/2 - 1}}}}{{1 + {x^s}}}dx} \mathop { = = = }\limits^{s = 2n} - \frac{{{\pi ^2}}}{6}\underbrace {\int\limits_0^\infty {\frac{{{x^{n - 1}}}}{{1 + {x^{2n}}}}dx} }_{{I_1}} - \frac{1}{2}\underbrace {\int\limits_0^\infty {\frac{{{x^{n - 1}}{{\log }^2}x}}{{1 + {x^{2n}}}}dx} }_{{I_2}}}$

Όμως $\displaystyle{{I_1} = \int\limits_0^\infty {\frac{{{x^{n - 1}}}}{{1 + {x^{2n}}}}dx} = \int\limits_0^1 {\frac{{{x^{n - 1}}}}{{1 + {x^{2n}}}}dx} + \int\limits_1^\infty {\frac{{{y^{n - 1}}}}{{1 + {y^{2n}}}}dy} \mathop { = = = = = = = }\limits^{y = 1/x} 2\int\limits_0^1 {\frac{{{x^{n - 1}}}}{{1 + {x^{2n}}}}dx} = }$

$\displaystyle{ = 2\sum\limits_{m = 0}^\infty {{{\left( { - 1} \right)}^m}\int\limits_0^1 {{x^{2nm + n - 1}}dx} } = \frac{2}{n}\sum\limits_{m = 0}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{2m + 1}}} = \frac{2}{n} \cdot \frac{\pi }{4} = \frac{\pi }{{2n}} = \frac{\pi }{s}}$

και $\displaystyle{{I_2} = \int\limits_0^\infty {\frac{{{x^{n - 1}}{{\log }^2}x}}{{1 + {x^{2n}}}}dx} = \int\limits_0^1 {\frac{{{x^{n - 1}}{{\log }^2}x}}{{1 + {x^{2n}}}}dx} + \int\limits_1^\infty {\frac{{{x^{n - 1}}{{\log }^2}x}}{{1 + {x^{2n}}}}dx} = 2\int\limits_0^1 {\frac{{{x^{n - 1}}{{\log }^2}x}}{{1 + {x^{2n}}}}dx} = }$

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

$\displaystyle{ = \frac{4}{{{n^3}}}\sum\limits_{m = 0}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{{{\left( {2m + 1} \right)}^3}}}} = \frac{4}{{{n^3}}}\frac{{{\pi ^3}}}{{32}} = \frac{{{\pi ^3}}}{{{s^3}}}}$

Τελικά $\displaystyle{2I = - \frac{{{\pi ^3}}}{{6 \cdot s}} - \frac{{{\pi ^3}}}{{2 \cdot {s^3}}} \Rightarrow I = - \frac{{{\pi ^3}}}{{12 \cdot s}} - \frac{{{\pi ^3}}}{{4 \cdot {s^3}}} = - \frac{{{\pi ^3}}}{4}\left( {\frac{1}{{3 \cdot s}} + \frac{{{1}}}{{{s^3}}}} \right)}$

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Χρησιμοποιήθηκαν οι σχέσεις

1) $\displaystyle{L{i_2}\left( { - x} \right) + L{i_2}\left( { - \frac{1}{x}} \right) = - \frac{1}{2}{\log ^2}x - \frac{{{\pi ^2}}}{6}}$ από εδώ http://functions.wolfram.com/ZetaFuncti ... owAll.html

2) $\displaystyle{\int\limits_0^1 {{{\log }^2}x \cdot {x^{2nm + n - 1}}dx} = \frac{2}{{{n^3} \cdot {{\left( {2m + 1} \right)}^3}}}}$ (στοιχειώδης παραγοντική ολοκλήρωση) και

3) $\displaystyle{\sum\limits_{m = 0}^\infty {\frac{{{{\left( { - 1} \right)}^m}}}{{{{\left( {2m + 1} \right)}^3}}}} = \frac{{{\pi ^3}}}{{32}}}$ αποδείχθηκε εδώ viewtopic.php?f=59&t=3614

Τόλη .. δεν δουλεύει ο reflection τύπος στους τριλογάριθμους, οπότε θα δούμε ..

pprime · #3

Tolaso J Kos έγραψε:Έστω $\mathbb{N} \ni s \geq 2$ και έστω άρτιος. και ας δηλώσουμε τον διλογάριθμο με ${\rm Li}_2$ και το τριλογάριθμο με ${\rm Li}_3$ . Τότε δείξατε ότι $\displaystyle{\int_0^{\infty} \frac{x^{s/2-1}{\rm Li}_2(-x)}{1+x^s}\, {\rm d}x}=- \frac{\pi^3}{4} \left( \frac1{3 s}+ \frac1{ s^3}\right)$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\cdot Li_{2}\left( -x \right)dx}\underbrace{=}_{x=\frac{1}{y}}\int\limits_{0}^{+\infty }{\frac{\left( \frac{1}{y} \right)^{\frac{s}{2}-1}}{1+\frac{1}{y^{s}}}\cdot Li_{2}\left( -\frac{1}{y} \right)\frac{dy}{y^{2}}}=\int\limits_{0}^{+\infty }{\frac{y^{\frac{s}{2}-1}}{1+y^{s}}\cdot Li_{2}\left( -\frac{1}{y} \right)dy}}$

$\displaystyle{=\frac{1}{2}\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\cdot \left( Li_{2}\left( -x \right)+Li_{2}\left( -\frac{1}{x} \right) \right)dx}=\frac{1}{2}\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\cdot \left( -\frac{\pi ^{2}}{6}-\frac{1}{2}\ln ^{2}x \right)dx}}$

$\displaystyle{=-\frac{\pi ^{2}}{12}\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}dx}-\frac{1}{4}\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\ln ^{2}xdx}}$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}dx}\underbrace{=}_{x=u^{\frac{1}{s}}}\frac{1}{s}\int\limits_{0}^{+\infty }{\frac{\left( u^{\frac{1}{s}} \right)^{\frac{s}{2}-1}}{1+u}u^{\frac{1}{s}-1}du}=\frac{1}{s}\int\limits_{0}^{+\infty }{\frac{u^{-\frac{1}{2}}}{1+u}du}=\frac{1}{s}\int\limits_{0}^{+\infty }{\frac{u^{\frac{1}{2}-1}}{\left( 1+u \right)^{\frac{1}{2}+\frac{1}{2}}}du}=\frac{1}{s}\beta \left( \frac{1}{2},\frac{1}{2} \right)=\frac{\pi }{s}}$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\ln ^{2}xdx}\underbrace{=}_{x=u^{\frac{1}{s}}}\frac{1}{s}\int\limits_{0}^{+\infty }{\frac{\left( u^{\frac{1}{s}} \right)^{\frac{s}{2}-1}}{1+u}\left( \frac{1}{s}\ln u \right)^{2}u^{\frac{1}{s}-1}du}=\frac{1}{s^{3}}\int\limits_{0}^{+\infty }{\frac{u^{\frac{1}{2}-\frac{1}{s}}\cdot u^{\frac{1}{s}-1}}{1+u}\ln ^{2}udu}}$

$\displaystyle{=\frac{1}{s^{3}}\int\limits_{0}^{+\infty }{\frac{u^{-\frac{1}{2}}}{1+u}\ln ^{2}udu}\underbrace{=}_{u=\frac{t}{1-t}}\frac{1}{s^{3}}\int\limits_{0}^{1}{\frac{\left( \frac{t}{1-t} \right)^{-\frac{1}{2}}}{1+\frac{t}{1-t}}\left( \ln \frac{t}{1-t} \right)^{2}\frac{dt}{\left( 1-t \right)^{2}}}}$

$\displaystyle{=\frac{1}{s^{3}}\int\limits_{0}^{1}{t^{-\frac{1}{2}}\left( 1-t \right)^{-\frac{1}{2}}\left( \ln ^{2}t-2\ln t\ln \left( 1-t \right)+\ln ^{2}\left( 1-t \right) \right)dt}}$

$\displaystyle{\beta \left( x,y \right)=\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}dt}}$
$\displaystyle{\frac{\partial ^{2}}{\partial x^{2}}\beta \left( x,y \right)=\frac{\partial ^{2}}{\partial x^{2}}\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}dt}=\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}\ln ^{2}tdt}}$

$\displaystyle{=\left( \left( \psi ^{\left( 0 \right)}\left( x \right)-\psi ^{\left( 0 \right)}\left( x+y \right) \right)^{2}+\psi ^{\left( 1 \right)}\left( x \right)-\psi ^{\left( 1 \right)}\left( x+y \right) \right)\beta \left( x,y \right)}$

$\displaystyle{\frac{\partial ^{2}}{\partial y^{2}}\beta \left( x,y \right)=\frac{\partial ^{2}}{\partial y^{2}}\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}dt}=\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}\ln ^{2}\left( 1-t \right)dt}}$

$\displaystyle{=\left( \left( \psi ^{\left( 0 \right)}\left( y \right)-\psi ^{\left( 0 \right)}\left( x+y \right) \right)^{2}+\psi ^{\left( 1 \right)}\left( y \right)-\psi ^{\left( 1 \right)}\left( x+y \right) \right)\beta \left( x,y \right)}$

$\displaystyle{\frac{\partial ^{2}}{\partial y\partial x}\beta \left( x,y \right)=\frac{\partial ^{2}}{\partial y\partial x}\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}dt}=\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}\ln t\ln \left( 1-t \right)dt}}$

$\displaystyle{=\left( \left( \psi ^{\left( 0 \right)}\left( x \right)-\psi ^{\left( 0 \right)}\left( x+y \right) \right)\left( \psi ^{\left( 0 \right)}\left( y \right)-\psi ^{\left( 0 \right)}\left( x+y \right) \right)-\psi ^{\left( 1 \right)}\left( x+y \right) \right)\beta \left( x,y \right)}$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\ln ^{2}xdx}=\frac{1}{s^{3}}\left( \frac{\partial ^{2}}{\partial x^{2}}\beta \left( x,y \right)\left| _{\left( \frac{1}{2},\frac{1}{2} \right)} \right.+\frac{\partial ^{2}}{\partial y^{2}}\beta \left( x,y \right)\left| _{\left( \frac{1}{2},\frac{1}{2} \right)} \right.-\frac{2\partial ^{2}}{\partial y\partial x}\beta \left( x,y \right)\left| _{\left( \frac{1}{2},\frac{1}{2} \right)} \right. \right)}$

$\displaystyle{=\frac{1}{s^{3}}\left( 2\left( \psi ^{\left( 0 \right)}\left( \frac{1}{2} \right)-\psi ^{\left( 0 \right)}\left( 1 \right) \right)^{2}+2\psi ^{\left( 1 \right)}\left( \frac{1}{2} \right)-2\psi ^{\left( 1 \right)}\left( 1 \right)+2\psi ^{\left( 1 \right)}\left( 1 \right)-2\left( \psi ^{\left( 0 \right)}\left( \frac{1}{2} \right)-\psi ^{\left( 0 \right)}\left( 1 \right) \right)^{2} \right)\beta \left( \frac{1}{2},\frac{1}{2} \right)}$

$\displaystyle{=\frac{2\pi }{s^{3}}\psi ^{\left( 1 \right)}\left( \frac{1}{2} \right)=\frac{2\pi }{s^{3}}\psi ^{\left( 1 \right)}\left( \frac{1}{2} \right)=\frac{\pi ^{3}}{s^{3}}}$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}Li_{2}\left( -x \right)dx}=-\frac{\pi ^{2}}{12}\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}dx}-\frac{1}{4}\int\limits_{0}^{+\infty }{\frac{x^{\frac{s}{2}-1}}{1+x^{s}}\ln ^{2}xdx}}$

$\displaystyle{=-\frac{\pi ^{3}}{12s}-\frac{\pi ^{3}}{4s^{3}}=-\frac{\pi ^{3}}{4s}\left( \frac{1}{3}+\frac{1}{s^{2}} \right)}$

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