Γενικευμενα Ολοκληρωματα

coheN · #1

Θα ήθελα λίγη βοήθεια στα παρακάτω:
Να υπολογίσετε τα παρακάτω :

1) $\int_{0}^{+\infty}{({\frac{\arctan{t}}{t})}^{2}}\,dt$
2) $\int_{0}^{+\infty}{\frac{\arctan{x}}{x(1+{a}^2{x}^{2})}}\,dx$

#2

Για το πρώτο

$\displaystyle{\int\limits_0^\infty {\frac{{{{\arctan }^2}\left( x \right)}}{{{x^2}}}dx} = \mathop = \limits^{\arctan \left( x \right) = y} = \int\limits_0^{\pi /2} {\frac{{{y^2}}}{{{{\sin }^2}\left( y \right)}}dy} = - \int\limits_0^{\pi /2} {{y^2} \cdot {{\left( {\sigma \varphi (y)} \right)}^\prime }dy} = 2\int\limits_0^{\pi /2} {y \cdot \sigma \varphi (y)dy} = 2\int\limits_0^{\pi /2} {y \cdot {{\left( {\ln (\sin \left( y \right))} \right)}^\prime }dy} = }$

$\displaystyle{ = - 2\int\limits_0^{\pi /2} {\ln \left( {\sin \left( y \right)} \right)dy} = \pi \ln \left( 2 \right)}$

Διότι $\displaystyle{I = \int\limits_0^{\pi /2} {\ln (\sin (2x))dx} = \mathop = \limits^{2 \cdot x = u} = \frac{1}{2} \cdot \int\limits_0^\pi {\ln (\sin (u))du} = \frac{1}{2} \cdot \int\limits_0^{\pi /2} {\ln (\sin (x))dx} + \frac{1}{2} \cdot \int\limits_{\pi /2}^\pi {\ln (\sin (u))du} = }$

$\displaystyle{ = \mathop = \limits^{u = \pi - x} = \int\limits_0^{\pi /2} {\ln (\sin (x))dx} = \mathop = \limits^{x \to \pi /2 - x} = \int\limits_0^{\pi /2} {\ln (\cos (x))dx} }$

Τότε $\displaystyle{I = \int\limits_0^{\pi /2} {\ln (\sin (2 \cdot x))dx} = \int\limits_0^{\pi /2} {\left( {\ln (\sin (x)) + \ln (\cos (x)) + \ln (2)} \right)dx} = 2I + \frac{\pi }{2}\ln (2) \Rightarrow I = - \frac{\pi }{2}\ln (2)}$

#3

Και το δεύτερο

$\displaystyle{\int\limits_0^\infty {\frac{{\arctan \left( x \right)}}{{x\left( {1 + {a^2}{x^2}} \right)}}dx} = \mathop = \limits^{x = 1/y} = \int\limits_0^\infty {\frac{{\left( {\frac{\pi }{2} - \arctan \left( y \right)} \right) \cdot y}}{{{a^2} + {y^2}}}dy} = \frac{1}{2}\int\limits_0^\infty {\left( {\frac{\pi }{2} - \arctan \left( y \right)} \right) \cdot \ln {{\left( {{a^2} + {y^2}} \right)}^\prime }dy} =}$

$\displaystyle{= \frac{1}{2}\left[ {\left( {\frac{\pi }{2} - \arctan \left( y \right)} \right) \cdot \ln \left( {{a^2} + {y^2}} \right)} \right]_0^\infty + \frac{1}{2}\int\limits_0^\infty {\frac{{\ln \left( {{a^2} + {y^2}} \right)}}{{1 + {y^2}}}dy} = - \frac{{\pi \ln \left( {{a^2}} \right)}}{4} + \frac{1}{2}\int\limits_0^\infty {\frac{{\ln \left( {{a^2} + {y^2}} \right)}}{{1 + {y^2}}}dy}}$

Έστω $\displaystyle{f\left( a \right) = \int\limits_0^\infty {\frac{{\ln \left( {{a^2} + {y^2}} \right)}}{{1 + {y^2}}}dy} \Rightarrow f'\left( a \right) = 2a\int\limits_0^\infty {\frac{1}{{\left( {1 + {y^2}} \right)\left( {{a^2} + {y^2}} \right)}}dy} = \frac{{2a}}{{{a^2} - 1}}\int\limits_0^\infty {\left( {\frac{1}{{1 + {y^2}}} - \frac{1}{{{a^2} + {y^2}}}} \right)dy} =}$

$\displaystyle{= \frac{{2a}}{{{a^2} - 1}}\int\limits_0^\infty {\frac{1}{{1 + {y^2}}}dy} - \frac{{2a}}{{{a^2} - 1}}\int\limits_0^\infty {\frac{1}{{{a^2} + {u^2}}}du} = \mathop = \limits^{u = ax} = \frac{{\pi a}}{{{a^2} - 1}} - \frac{{2a}}{{{a^2} - 1}}\int\limits_0^\infty {\frac{a}{{{a^2} + {a^2}{x^2}}}dx} =}$

$\displaystyle{= \frac{{\pi a}}{{{a^2} - 1}} - \frac{2}{{{a^2} - 1}}\int\limits_0^\infty {\frac{1}{{1 + {x^2}}}dx} = \frac{{\pi a}}{{{a^2} - 1}} - \frac{\pi }{{{a^2} - 1}} = \frac{\pi }{{a + 1}} \Rightarrow f\left( a \right) = \pi \ln \left( {a + 1} \right) + c}$

Για $\displaystyle{a = 1:\int\limits_0^\infty {\frac{{\ln \left( {1 + {y^2}} \right)}}{{1 + {y^2}}}dy} = \mathop = \limits^{y = \tan \left( x \right)} = - 2\int\limits_0^\infty {\ln \left( {\cos \left( x \right)} \right)dx} = - 2\left( { - \frac{\pi }{2}\ln \left( 2 \right)} \right) = \pi \ln \left( 2 \right) + c \Rightarrow c = 0}$

Και τελικά $\displaystyle{\int\limits_0^\infty {\frac{{\arctan \left( x \right)}}{{x\left( {1 + {a^2}{x^2}} \right)}}dx} = - \frac{{\pi \ln \left( {{a^2}} \right)}}{4} + \frac{1}{2}\int\limits_0^\infty {\frac{{\ln \left( {{a^2} + {y^2}} \right)}}{{1 + {y^2}}}dy} = \frac{\pi }{2}\ln \left( {\frac{{a + 1}}{a}} \right)}$

Θεωρήσαμε $\displaystyle{a>0}$ , σε κάθε περίπτωση πρέπει $\displaystyle{a \ne 0}$ και $\displaystyle{\int\limits_0^\infty {\frac{{\arctan \left( x \right)}}{{x\left( {1 + {a^2}{x^2}} \right)}}dx} = \frac{\pi }{2}\ln \left( {\frac{{\left| a \right| + 1}}{{\left| a \right|}}} \right)}$

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