ολοκλήρωμα 2

Συντονιστές: grigkost, Κοτρώνης Αναστάσιος

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thepathofresistance
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Εγγραφή: Τρί Ιουν 14, 2011 2:56 am

ολοκλήρωμα 2

#1

Μη αναγνωσμένη δημοσίευση από thepathofresistance » Σάβ Σεπ 17, 2011 4:50 pm

Να βρεθεί το \displaystyle{ \int\limits_0^\pi {\sin \left( {nx } \right)\arctan \left( {\frac{{\tan \frac{x } {2}}} {{\tan \frac{\theta } {2}}}} \right)} dx }

\displaystyle{ n \in {\Bbb Z}^ + \wedge \theta \in \left( {0,\frac{\pi } {2}} \right) }


είναι
\displaystyle{ u = \arctan \left( {\frac{{\tan \frac{\alpha } {2}}} {{\tan \frac{\theta } {2}}}} \right) \wedge dv = \sin \left( {n\alpha } \right)d\alpha }
\displaystyle{ du = \frac{1} {{1 + \left( {\frac{{\tan \frac{\alpha } {2}}} {{\tan \frac{\theta } {2}}}} \right)^2 }}\frac{1} {2}\frac{1} {{\tan \frac{\theta } {2}}}\sec ^2 \frac{\alpha } {2}d\alpha \wedge v = - \frac{1} {n}\cos \left( {n\alpha } \right) }

\displaystyle{ K\left( \theta \right) = - \frac{1} {n}\cos \left( {n\alpha } \right)\arctan \left( {\frac{{\tan \frac{\alpha } {2}}} {{\tan \frac{\theta } {2}}}} \right)\left| {_0^\pi } \right. + \frac{1} {n}\int\limits_0^\pi {\cos \left( {n\alpha } \right)} \frac{1} {{1 + \left( {\frac{{\tan \frac{\alpha } {2}}} {{\tan \frac{\theta } {2}}}} \right)^2 }}\frac{1} {2}\frac{1} {{\tan \frac{\theta } {2}}}\sec ^2 \frac{\alpha } {2}d\alpha }
\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n + \frac{1} {n}\int\limits_0^\pi {\cos \left( {n\alpha } \right)} \frac{1} {{1 + \left( {\frac{{\tan \frac{\alpha } {2}}} {{\tan \frac{\theta } {2}}}} \right)^2 }}\frac{1} {2}\frac{1} {{\tan \frac{\theta } {2}}}\sec ^2 \frac{\alpha } {2}d\alpha }

αφ 'ετέρου
\displaystyle{ \frac{1} {{1 + \left( {\frac{{\tan \frac{\alpha } {2}}} {{\tan \frac{\theta } {2}}}} \right)^2 }}\frac{1} {2}\frac{1} {{\tan \frac{\theta } {2}}}\sec ^2 \frac{\alpha } {2} = \frac{1} {2}\frac{{\tan \frac{\theta } {2}}} {{\tan ^2 \frac{\theta } {2} + \tan ^2 \frac{\alpha } {2}}}\frac{1} {{\cos ^2 \frac{\alpha } {2}}} }
\displaystyle{ = \frac{{\sin \frac{\theta } {2}}} {{2\cos \frac{\theta } {2}}}\frac{{\cos ^2 \frac{\theta } {2}}} {{\cos ^2 \frac{\alpha } {2}\sin ^2 \frac{\theta } {2} + \sin ^2 \frac{\alpha } {2}\cos ^2 \frac{\theta } {2}}} }
\displaystyle{ = \frac{{\sin \theta }} {4}\frac{1} {{\cos ^2 \frac{\alpha } {2}\sin ^2 \frac{\theta } {2} + \sin ^2 \frac{\alpha } {2}\sin ^2 \frac{\theta } {2} + \sin ^2 \frac{\alpha } {2}\cos ^2 \frac{\theta } {2} - \sin ^2 \frac{\alpha } {2}\sin ^2 \frac{\theta } {2}}} }
\displaystyle{ = \frac{{\sin \theta }} {4}\frac{1} {{\sin ^2 \frac{\alpha } {2} + \sin ^2 \frac{\theta } {2}\cos \alpha }} = \frac{{\sin \theta }} {4}\frac{1} {{\frac{{1 - \cos \alpha }} {2} + \frac{{1 - \cos \theta }} {2}\cos \alpha }} }
\displaystyle{ = \frac{{\sin \theta }} {2}\frac{1} {{1 - \cos \theta \cos \alpha }} }

\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n + \frac{1} {{2n}}\tan \theta \int\limits_0^\pi {\cos \left( {n\alpha } \right)} \frac{1} {{\sec \theta - \cos \alpha }}d\alpha }
\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n + \frac{1} {{2n}}\tan \theta \int\limits_0^{2\pi } {\frac{{e^{in\alpha } }} {{2\sec \theta - e^{i\alpha } - e^{ - i\alpha } }}} d\alpha }
\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n + \frac{1} {{2n}}\tan \theta \int\limits_0^{2\pi } {\frac{{e^{in\alpha } e^{i\alpha } }} {{2\sec \theta e^{i\alpha } - e^{2i\alpha } - 1}}} d\alpha }
\displaystyle{ z = e^{i\alpha } \Rightarrow \frac{{dz}} {{d\alpha }} = ie^{i\alpha } = iz }

\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n - \frac{1} {{2in}}\tan \theta \int\limits_{\left| z \right| = 1} {\frac{{z^n }} {{z^2 - 2z\sec \theta + 1}}} dz }

\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n - \frac{1} {{2in}}\tan \theta \int\limits_{\left| z \right| = 1} {\frac{{z^n }} {{\left( {z - \left( {\sec \theta + \tan \theta } \right)} \right)\left( {z - \left( {\sec \theta - \tan \theta } \right)} \right)}}} dz }
αλλά
\displaystyle{{\theta \in \left( {0,\frac{\pi }{2}} \right)}}
\displaystyle{ {\sec \theta + \tan \theta > 1} }
\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n - \frac{{\tan \theta }} {{2in}}2\pi i\Re es\left( {F\left( z \right),z = \sec \theta - \tan \theta } \right) }
\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n - \frac{{\tan \theta }} {{2in}}2\pi i\mathop {\lim }\limits_{z \to \sec \theta - \tan \theta } \frac{{z^n \left( {z - \left( {\sec \theta - \tan \theta } \right)} \right)}} {{\left( {z - \left( {\sec \theta + \tan \theta } \right)} \right)\left( {z - \left( {\sec \theta - \tan \theta } \right)} \right)}} }
\displaystyle{ K\left( \theta \right) = - \frac{\pi } {{2n}}\left( { - 1} \right)^n - \frac{{\tan \theta }} {n}\pi \frac{{\left( {\sec \theta - \tan \theta } \right)^n }} {{ - 2\tan \theta }} = - \frac{\pi } {{2n}}\left( { - 1} \right)^n + \frac{\pi } {{2n}}\left( {\sec \theta - \tan \theta } \right)^n }


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