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Δημοσιεύτηκε: **Δευ Αύγ 07, 2017 11:47 pm**

1. $\displaystyle{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\arcsin \left( \sqrt{1-s}\sqrt{y} \right)}{\sqrt{1-y}\sqrt{sy-y+1}}dsdy}}=2\pi \left( 1-\ln 2 \right)}$

2. $\displaystyle{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\arcsin \left( \sqrt{1-s}\sqrt{y} \right)}{\sqrt{1-s}\sqrt{y}\sqrt{sy-y+1}}dsdy}}=-\frac{7}{4}\zeta \left( 3 \right)+\frac{\pi ^{2}}{2}\ln 2}$

3. $\displaystyle{\int\limits_{0}^{1}{\frac{Li_{2}\left( x \right)}{\sqrt{1-x^{2}}}dx}=\frac{5\pi ^{3}}{24}-4\Im Li_{3}\left( 1+i \right)}$

Δημοσιεύτηκε: **Τετ Αύγ 16, 2017 2:32 pm**

pprime έγραψε:2. $\displaystyle{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\arcsin \left( \sqrt{1-s}\sqrt{y} \right)}{\sqrt{1-s}\sqrt{y}\sqrt{sy-y+1}}dsdy}}=-\frac{7}{4}\zeta \left( 3 \right)+\frac{\pi ^{2}}{2}\ln 2}$

$\displaystyle{I = \int\limits_0^1 {\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {1 - s} \sqrt y } \right)}}{{\sqrt {1 - s} \sqrt y \sqrt {sy - y + 1} }}ds\;dy} } = \int\limits_0^1 {\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {1 - s} \sqrt y } \right)}}{{\sqrt {1 - s} \sqrt y \sqrt {1 - y\left( {1 - s} \right)} }}ds\;dy} } }$ $\displaystyle{\mathop { = = = = = }\limits^{1 - s \to s} \int\limits_0^1 {\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {ys} } \right)}}{{\sqrt {1 - s} \sqrt y \sqrt {1 - ys} }}ds\;dy} } }$

Κάνουμε τον μετασχηματισμό $\displaystyle{ys = u}$ και $\displaystyle{y = v}$ , οπότε $\displaystyle{s = \frac{u}{v}}$ , $\displaystyle{y = v}$ και $\displaystyle{\left| {\frac{{\partial \left( {y,s} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \frac{1}{v}}$
[attachment=0]pprime-2.jpg[/attachment]
Τότε $\displaystyle{I = \int\limits_0^1 {\int\limits_u^1 {\frac{{\arcsin \left( {\sqrt u } \right)}}{{v\sqrt u \sqrt {1 - u} }}dv\;du} } = \int\limits_0^1 {\frac{{\arcsin \left( {\sqrt u } \right)}}{{\sqrt u \sqrt {1 - u} }}\int\limits_u^1 {\frac{1}{v}dv\;du} } = }$ $\displaystyle{ - \int\limits_0^1 {\frac{{\arcsin \left( {\sqrt u } \right)\log u}}{{\sqrt u \sqrt {1 - u} }}du} \mathop { = = = = }\limits^{u = {w^2}} - 4\int\limits_0^1 {\frac{{\arcsin \left( w \right)\log w}}{{\sqrt {1 - {w^2}} }}dw} = }$

$\displaystyle{ = - 2\int\limits_0^1 {{{\left( {{{\arcsin }^2}\left( w \right)} \right)}{'}}\log wdw} = - 2\left[ {{{\arcsin }^2}\left( w \right)\log w} \right]_0^1 + }$ $\displaystyle{2\int\limits_0^1 {\frac{{{{\arcsin }^2}\left( w \right)}}{w}dw} = 2\int\limits_0^1 {\frac{{{{\arcsin }^2}\left( w \right)}}{w}dw} = \mathop {\mathop { = = = = = = }\limits_{w = \sin x} }\limits^{\arcsin \left( w \right) = x} }$

$\displaystyle{ = 2\int\limits_0^{\pi /2} {\frac{{{x^2}}}{{\sin x}}\cos x\;dx} = 2\int\limits_0^{\pi /2} {{x^2}\left( {\log \sin x} \right)'\;dx} = 2\left[ {{x^2}\log \sin x} \right]_0^{\pi /2}}$ $\displaystyle{ - 4\int\limits_0^{\pi /2} {x\log \left( {\sin x} \right)\;dx} = - 4\int\limits_0^{\pi /2} {x\log \left( {\sin x} \right)\;dx} = }$

$\displaystyle{ = 4\int\limits_0^{\pi /2} {x\left( {\log 2 + \sum\limits_{n = 1}^\infty {\frac{{\cos 2nx}}{n}} } \right)\;dx} = = 4\log 2\int\limits_0^{\pi /2} {x\;dx} + }$ $\displaystyle{4\sum\limits_{n = 1}^\infty {\frac{1}{n}\int\limits_0^{\pi /2} {x\cos 2nx\;dx} } = \frac{{{\pi ^2}\log 2}}{2} + 4\sum\limits_{n = 1}^\infty {\frac{1}{n}\int\limits_0^{\pi /2} {x\cos 2nx\;dx} } = }$

$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{2} + \sum\limits_{n = 1}^\infty {\frac{{ - 1 + \cos \left( {n\pi } \right)}}{{{n^3}}}} = \frac{{{\pi ^2}\log 2}}{2} + \sum\limits_{n = 1}^\infty {\frac{{ - 1 + {{\left( { - 1} \right)}^n}}}{{{n^3}}}} }$ $\displaystyle{ = \frac{{{\pi ^2}\log 2}}{2} - \zeta \left( 3 \right) + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} = \frac{{{\pi ^2}\log 2}}{2} - \frac{7}{4}\zeta \left( 3 \right)}$

Χρησιμοποιήθηκε η σχέση $\displaystyle{\log \left( {\sin x} \right) = - \log 2 - \sum\limits_{n = 1}^\infty {\frac{{\cos 2nx}}{n}} }$ (έχει αποδειχθεί αρκετές φορές εδώ μέσα).

Δημοσιεύτηκε: **Τετ Αύγ 16, 2017 7:02 pm**

pprime έγραψε:1. $\displaystyle{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{\arcsin \left( \sqrt{1-s}\sqrt{y} \right)}{\sqrt{1-y}\sqrt{sy-y+1}}dsdy}}=2\pi \left( 1-\ln 2 \right)}$

Λήμμα 1: $\displaystyle{\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {ys} } \right)}}{{\sqrt {1 - ys} }}ds} = \frac{{2\left( {\sqrt y - \sqrt {1 - y} \arcsin \left( {\sqrt y } \right)} \right)}}{y}}$ διότι $\displaystyle{\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {ys} } \right)}}{{\sqrt {1 - ys} }}ds} \mathop { = = = }\limits^{ys = w} \frac{1}{y}\int\limits_0^y {\frac{{\arcsin \left( {\sqrt w } \right)}}{{\sqrt {1 - w} }}dw} }$

$\displaystyle{ = \frac{1}{y}\left[ {2\sqrt x - 2\sqrt {1 - x} \arcsin \left( {\sqrt x } \right)} \right]_0^y = \frac{{2\left( {\sqrt y - \sqrt {1 - y} \arcsin \left( {\sqrt y } \right)} \right)}}{y}}$

Λήμμα 2: $\displaystyle{\int\limits_0^1 {\frac{1}{{\sqrt y \sqrt {1 - y} }}dy} = \pi }$ διότι $\displaystyle{\int\limits_0^1 {\frac{1}{{\sqrt y \sqrt {1 - y} }}dy} \mathop { = = = }\limits^{y = {x^2}} \int\limits_0^1 {\frac{{2x}}{{x\sqrt {1 - {x^2}} }}dx} = 2\left[ {\arcsin \left( x \right)} \right]_0^1 = \pi }$

Λήμμα 3: $\displaystyle{\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt y } \right)}}{y}dy} = \pi \log 2}$ διότι $\displaystyle{\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt y } \right)}}{y}dy} \mathop { = = = }\limits^{y = {x^2}} 2\int\limits_0^1 {\frac{{\arcsin \left( x \right)}}{x}dx} = 2\int\limits_0^1 {\arcsin \left( x \right)\left( {\log x} \right)'dx} = }$

$\displaystyle{ = - 2\int\limits_0^1 {\frac{{\log x}}{{\sqrt {1 - {x^2}} }}dx} \mathop { = = = }\limits^{x = \sin w} - 2\int\limits_0^{\pi /2} {\log \left( {\sin w} \right)dw} = - 2\int\limits_0^{\pi /2} {\left( { - \log 2 - \sum\limits_{n = 1}^\infty {\frac{{\cos 2nw}}{n}} } \right)dw} = 2\pi \log 2}$

Τότε

$\displaystyle{I = \int\limits_0^1 {\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {1 - s} \sqrt y } \right)}}{{\sqrt {1 - y} \sqrt {sy - y + 1} }}ds\;dy} } = \int\limits_0^1 {\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {1 - s} \sqrt y } \right)}}{{\sqrt {1 - y} \sqrt {1 - y\left( {1 - s} \right)} }}ds\;dy} } }$ $\displaystyle{\mathop { = = = = = }\limits^{1 - s \to s} \int\limits_0^1 {\frac{1}{{\sqrt {1 - y} }}\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {ys} } \right)}}{{\sqrt {1 - ys} }}ds} } \;dy = }$

$\displaystyle{ = 2\int\limits_0^1 {\left( {\frac{{\sqrt y - \sqrt {1 - y} \arcsin \left( {\sqrt y } \right)}}{{y\sqrt {1 - y} }}} \right)} \;dy = 2\int\limits_0^1 {\left( {\frac{1}{{\sqrt y \sqrt {1 - y} }} - \frac{{\arcsin \left( {\sqrt y } \right)}}{y}} \right)dy} }$ $\displaystyle{ = 2\int\limits_0^1 {\frac{1}{{\sqrt y \sqrt {1 - y} }}dy} - 2\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt y } \right)}}{y}dy} \Rightarrow }$

$\displaystyle{ \Rightarrow \int\limits_0^1 {\int\limits_0^1 {\frac{{\arcsin \left( {\sqrt {1 - s} \sqrt y } \right)}}{{\sqrt {1 - y} \sqrt {sy - y + 1} }}ds\;dy} } = 2\pi - 2\pi \log 2}$

Δημοσιεύτηκε: **Δευ Αύγ 21, 2017 3:39 am**

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