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Δημοσιεύτηκε: **Παρ Ιούλ 23, 2010 7:20 pm**

Να υπολογίσετε το
$\displaystyle\int_{0}^{1}\int_{0}^{1}\ln(\sin^2(\pi x) + \sin^2(\pi y)) \,dx\,dy$

Δημοσιεύτηκε: **Σάβ Σεπ 04, 2010 9:35 pm**

Πολύ ζόρικο!!

Λήμμα 1

$\displaystyle{\mathbf{I = \int\limits_0^1 {\frac{1}{{{{\sin }^2}\left( {\pi \cdot x} \right) + a}} \cdot dx=} \mathop {.......} \limits^\big{{\pi \cdot x = u}} {\rm{ }}=\frac{1}{\pi } \cdot \int\limits_0^\pi {\frac{1}{{{{\sin }^2}\left( u \right) + a}} \cdot du} = \frac{2}{\pi } \cdot \int\limits_0^{\pi /2} {\frac{1}{{{{\sin }^2}\left( u \right) + a}} \cdot du=} \mathop {.......} \limits^{\big{\tan \left( u \right) = y}}= \frac{2}{\pi } \cdot \int\limits_0^\infty {\frac{1}{{\left( {1 + a} \right) \cdot {y^2} + a}} \cdot dy} = }}$

$\displaystyle{\mathbf{ = \dfrac{2}{{\pi \cdot \left( {1 + a} \right)}} \cdot \int\limits_0^\infty {\dfrac{1}{{{y^2} + \dfrac{a}{{1 + a}}}} \cdot dy} = \mathop {.......} \limits^\big{{y = \sqrt {\dfrac{a}{{1 + a}}}} \cdot w} =\dfrac{2}{{\pi \cdot \left( {1 + a} \right)}} \cdot \int\limits_0^\infty {\dfrac{1}{{\dfrac{a}{{1 + a}} \cdot \left( {1 + {w^2}} \right)}} \cdot \sqrt {\dfrac{a}{{1 + a}}} \cdot dw} = \dfrac{1}{{\sqrt {a \cdot \left( {1 + a} \right)} }}}}$

Τότε Λήμμα 2 $\displaystyle{\mathbf{\int\limits_0^1 {\frac{1}{{{{\sin }^2}\left( {\pi \cdot x} \right) + {{\sin }^2}\left( {\pi \cdot y} \right)}} \cdot dx} = \frac{1}{{\sin \left( {\pi \cdot y} \right)\sqrt {1 + {{\sin }^2}\left( {\pi \cdot y} \right)} }}}}$

Aν $\displaystyle{\mathbf{f\left( y \right) = \int\limits_0^1 {\ln \left( {{{\sin }^2}\left( {\pi \cdot x} \right) + {{\sin }^2}\left( {\pi \cdot y} \right)} \right) \cdot dx} }}$

τότε $\displaystyle{\mathbf{f'\left( y \right) = 2 \cdot \pi \cdot \sin \left( {\pi \cdot y} \right) \cdot \cos \left( {\pi \cdot y} \right) \cdot \int\limits_0^1 {\frac{1}{{{{\sin }^2}\left( {\pi \cdot x} \right) + {{\sin }^2}\left( {\pi \cdot y} \right)}} \cdot dx} = \frac{{2 \cdot \pi \cdot \sin \left( {\pi \cdot y} \right) \cdot \cos \left( {\pi \cdot y} \right)}}{{\sin \left( {\pi \cdot y} \right)\sqrt {1 + {{\sin }^2}\left( {\pi \cdot y} \right)} }} = \frac{{2 \cdot \pi \cdot \cos \left( {\pi \cdot y} \right)}}{{\sqrt {1 + {{\sin }^2}\left( {\pi \cdot y} \right)} }}}}$

Άρα $\displaystyle{\mathbf{f\left( y \right) = 2 \cdot \pi \cdot \int {\frac{{\cos \left( {\pi \cdot y} \right)}}{{\sqrt {1 + {{\sin }^2}\left( {\pi \cdot y} \right)} }}} \cdot dy\mathop = \limits^{\pi \cdot y = w} 2 \cdot \int {\frac{{\cos \left( w \right)}}{{\sqrt {1 + {{\sin }^2}\left( w \right)} }}} \cdot dw\mathop = \limits^{\sin \left( w \right) = v} 2 \cdot \int {\frac{1}{{\sqrt {1 + {v^2}} }}} \cdot dv = 2 \cdot \ln \left( {v + \sqrt {1 + {v^2}} } \right) + c \Rightarrow }}$

$\displaystyle{\mathbf{f\left( y \right) = 2 \cdot \ln \left( {\sin \left( {\pi \cdot y} \right) + \sqrt {1 + {{\sin }^2}\left( {\pi \cdot y} \right)} } \right) + c}}$ με $\displaystyle{\mathbf{c = f\left( 1 \right) }}$

Για $\displaystyle{\mathbf{y = 1:f\left( 1 \right) = 2 \cdot \int\limits_0^1 {\ln \left( {\sin \left( {\pi \cdot x} \right)} \right) \cdot dx} = - 2 \cdot \ln \left( 2 \right)}}$ (εύκολο) ..
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Άρα $\displaystyle{\mathbf{ \int\limits_0^1 {\ln \left( {{{\sin }^2}\left( {\pi \cdot x} \right) + {{\sin }^2}\left( {\pi \cdot y} \right)} \right) \cdot dx} = 2 \cdot \ln \left( {\sin \left( {\pi \cdot y} \right) + \sqrt {1 + {{\sin }^2}\left( {\pi \cdot y} \right)} } \right) - 2 \cdot \ln \left( 2 \right)}}$
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Τότε $\displaystyle{\mathbf{\int\limits_0^1 {\left( {\int\limits_0^1 {\ln \left( {{{\sin }^2}\left( {\pi \cdot x} \right) + {{\sin }^2}\left( {\pi \cdot y} \right)} \right) \cdot dx} } \right) \cdot dy} = \frac{2}{\pi } \cdot \int\limits_0^\pi {\ln \left( {\sin \left( y \right) + \sqrt {1 + {{\sin }^2}\left( y \right)} } \right) \cdot dy} - 2 \cdot \ln \left( 2 \right)}}$

Όμως $\displaystyle{\mathbf{\int\limits_0^\pi {\ln \left( {\sin \left( y \right) + \sqrt {1 + {{\sin }^2}\left( y \right)} } \right) \cdot dy} = 2 \cdot G}}$ όπου $\displaystyle{\mathbf{G \approx 0.915{\rm{ }}965{\rm{ }}594{\rm{ }}177{\rm{ }}219{\rm{ }}...}}$ είναι η σταθερά του Catalan.

Τελικά $\displaystyle{\mathbf{\int\limits_0^1 {\left( {\int\limits_0^1 {\ln \left( {{{\sin }^2}\left( {\pi \cdot x} \right) + {{\sin }^2}\left( {\pi \cdot y} \right)} \right) \cdot dx} } \right) \cdot dy} = - 2 \cdot \ln \left( 2 \right) + \frac{{4 \cdot G}}{\pi }}}$

Για την σταθερά του Catalan εδώ http://en.wikipedia.org/wiki/Catalan%27s_constant και εδώ http://www.cs.cmu.edu/~adamchik/article ... atalan.htm

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