Guía VII de Variable Compleja - Foro fmat.cl

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Guía VII de Variable Compleja, Prof. Dr. Carlos Lizama

ciunhaly Ver Perfil del Miembro Ver los mensajes de este miembro	Jul 19 2007, 01:16 AM Publicado: #1
Maestro Matemático Grupo: Colaborador Silver Mensajes: 128 Registrado: 30-June 06 Desde: En Rio de Janeiro, Brasil Miembro Nº: 1.477 Nacionalidad: Universidad: Sexo:	$TEX: \begin{center}<br />{\Huge Gu\'ia $N^o 7$}\\<br />{\large Variable Compleja}<br />\end{center}<br />\begin{enumerate}<br /> \item Sea $y > 0$. Demuestre que para cada $x\geq0$ se tiene<br /> \[ \displaystyle{\int_{-\infty}^{\infty}\frac{e^{-isx}}{s^2+t^2}ds=\frac{\pi}{t}e^{-tx}} \]<br /> \item Sean $a, b \in \mathbb{R}$ tales que $a > 0$ y $b > 0$. Demuestre que \[ \displaystyle{\int_{-\infty}^{\infty}\frac{x^2-b^2}{x^2+b^2}\;\;\frac{\sin(ax)}{x} dx=\pi\left( e^{-ab}-\frac{1}{2} \right)} \]<br /> \item Demuestre que<br /> \[ \displaystyle{\int_{0}^{\infty}\frac{dx}{x^6+b^1}=\frac{\pi}{3} } \]<br /> \item Demuestre que<br /> \[ \displaystyle{\int_{0}^{2\pi}\frac{\cos(3\theta)}{5-4\cos(\theta)}d\theta=\frac{\pi}{12} } \]<br /> \item Demuestre que<br /> \[ \displaystyle{\int_{0}^{2\pi}\frac{d\theta}{(5-3\sin(\theta))^2}=\frac{5\pi}{32} } \]<br /> \item Demuestre que<br /> \[ \displaystyle{\int_{0}^{\infty}\frac{\cos(mx)}{x^2+1}dx=\frac{\pi}{2}e^{-m},\;\;\;m>0 } \]<br /> \item Demuestre que<br /> \[ \displaystyle{\int_{0}^{\infty}\frac{\cosh(ax)}{\cosh(x)}dx=\frac{\pi}{2\cos\left( \dfrac{\pi a}{2}\right)},\;\;\;donde\;\;\;\|a\|<1 } \]<br />\end{enumerate}$ -------------------- _______________________________________________________________ "La primera regla de la enseñanza es saber lo que se debe enseñar. La segunda, es saber un poco más de aquello que se debe enseñar". George Polya Eu sou uma estudante da UFRJ.

master_c	May 25 2011, 11:52 PM Publicado: #2
Invitado	para el 5 $TEX: $\int_0^{2\pi } {\frac{{d\theta }}{{\left( {5 - 3\sin \theta } \right)^2 }} = } 4i\int_\gamma {\frac{{zdz}}{{\left( {3z^2 - 10iz - 3} \right)^2 }} = } \frac{{4i}}{9}\int_\gamma {\frac{{zdz}}{{\left( {\left( {z - 3i} \right)\left( {z - \frac{i}{3}} \right)} \right)^2 }}}$$ $TEX: $$<br />\frac{{4i}}<br />{9} \cdot 2\pi i\mathop {\lim }\limits_{z \to \frac{i}<br />{3}} \frac{d}<br />{{dz}}\left( {\left( {z - \frac{i}<br />{3}} \right)\frac{z}<br />{{\left( {\left( {z - 3i} \right)\left( {z - \frac{i}<br />{3}} \right)} \right)^2 }}} \right) = \frac{8}<br />{9}\pi \left( {\frac{{16 + 64}}<br />{9}} \right) \cdot \frac{{9^2 }}<br />{{8^4 }} = \pi \left( {\frac{{16 + 64}}<br />{{8^3 }}} \right) = \frac{5}<br />{{32}}\pi <br />$$$

master_c	May 26 2011, 11:06 AM Publicado: #3
Invitado	para el 6 $TEX: $$<br />\int_0^{ + \infty } {\frac{{\cos ax}}<br />{{x^2 + b^2 }}} dx = \frac{1}<br />{2}\int_{ - \infty }^{ + \infty } {\frac{{\cos ax}}<br />{{x^2 + b^2 }}} dx = \frac{1}<br />{2}\Re e\left( {2\pi i\Re es\left( {\frac{{e^{iaz} }}<br />{{z^2 + b^2 }},ib} \right)} \right) = \frac{1}<br />{2}\Re e\left( {2\pi i\frac{{e^{ - ab} }}<br />{{2ib}}} \right) = \frac{{\pi e^{ - ab} }}<br />{{2b}}<br />$$$ haciendo $TEX: $b = 1$$ se obtiene lo pedido

master_c	May 26 2011, 02:01 PM Publicado: #4
Invitado	algo mas general $TEX: $$<br />\int_{ - \infty }^{ + \infty } {\frac{{\cosh ax}}<br />{{\left( {\cosh x} \right)^b }}dx = 2^{b - 1} \frac{{\Gamma \left( {\frac{{b + a}}<br />{2}} \right)\Gamma \left( {\frac{{b - a}}<br />{2}} \right)}}<br />{{\Gamma \left( b \right)}}} <br />$$$

master_c	May 28 2011, 05:28 PM Publicado: #5
Invitado	para el 4) previo $TEX: $\cos 3x = \cos x\left( {\cos ^2 x - 3\sin ^2 x} \right) = \cos x\left( {1 - 4\sin ^2 x} \right)$$ entonces $TEX: $$<br />I = \int_0^{2\pi } {\frac{{\cos 3x}}<br />{{5 - 4\cos x}}dx} = \frac{i}<br />{2}\int_\gamma {\frac{{\left( {z^2 + 1} \right)\left( {z^2 + \left( {z^2 - 1} \right)^2 } \right)}}<br />{{z^3 \left( {2z^2 - 5z + 2} \right)}}} dz = \frac{i}<br />{4}\int_\gamma {\frac{{z^6 + 1}}<br />{{z^3 \left( {z - 2} \right)\left( {z - \frac{1}<br />{2}} \right)}}} dz<br />$$$ $TEX: $$<br /> = - \frac{\pi }<br />{2}\left( {\Re es\left( {f\left( z \right),z = 0} \right) + \Re es\left( {f\left( z \right),z = \frac{1}<br />{2}} \right)} \right) = - \frac{\pi }<br />{2}\left( {\frac{{63}}<br />{{12}} + - \frac{{65}}<br />{{12}}} \right) = \frac{\pi }<br />{{12}}<br />$$$ ya que $TEX: $$<br />\Re es\left( {f\left( z \right),z = 0} \right) = \frac{1}<br />{{2!}}\mathop {\lim }\limits_{z \to z_0 } \frac{{d^2 }}<br />{{dz^2 }}\left( {\frac{{\left( {z^2 + 1} \right)\left( {z^4 - \left( {z^2 - 1} \right)} \right)}}<br />{{\left( {z - 2} \right)\left( {z - \frac{1}<br />{2}} \right)}}} \right) = \mathop {\lim }\limits_{z \to z_0 } \frac{{d^2 }}<br />{{dz^2 }}\left( {\frac{{z^6 + 1}}<br />{{2z^2 - 5z + 2}}} \right) = \frac{{21}}<br />{4}<br />$$$ y $TEX: $$<br />\Re es\left( {f\left( z \right),z = \frac{1}<br />{2}} \right) = \frac{{\frac{1}<br />{{2^6 }} + 1}}<br />{{\frac{1}<br />{{2^3 }}\left( {\frac{1}<br />{2} - 2} \right)}} = \frac{{\frac{{65}}<br />{{8^2 }}}}<br />{{\frac{1}<br />{8} \cdot \frac{{ - 3}}<br />{2}}} = - \frac{{65}}<br />{{12}}<br />$$$

master_c	May 29 2011, 09:16 PM Publicado: #6
Invitado	para 3) el denominador vale $TEX: $\left( {x + i\root 6 \of b } \right)\left( {x - i\root 6 \of b } \right)\left( {x - \frac{{\root 6 \of b }}{{\sqrt 2 }}\sqrt {1 + i\sqrt 3 } } \right)\left( {x + \frac{{\root 6 \of b }}{{\sqrt 2 }}\sqrt {1 - i\sqrt 3 } } \right)\left( {x - \frac{{\root 6 \of b }}{{\sqrt 2 }}\sqrt {1 - i\sqrt 3 } } \right)\left( {x + \frac{{\root 6 \of b }}{{\sqrt 2 }}\sqrt {1 + i\sqrt 3 } } \right)$$ donde facilmente se desprende que $TEX: $\sqrt {1 \pm i\sqrt 3 } = \sqrt {\frac{3}{2}} \pm \sqrt {\frac{{ - 1}}{2}} = \frac{{\sqrt 2 }}{2}\left( {\sqrt 3 \pm i} \right)$$ aplicando teorema (ya que se cumple las propiedades no existen polos sobre el eje real y $TEX: $f\left( {\frac{1}{z}} \right)$$ tiene al menos un cero de orden dos en $TEX: $ z = 0 $$ ) consideramos la suma solo de polos en el semiplano superior $TEX: $\int_0^{ + \infty } {\frac{{dx}}{{b + x^6 }}} = \frac{\pi }{{3\root 6 \of {b^5 } }} \Rightarrow \int_0^{ + \infty } {\frac{{dx}}{{1 + x^6 }}} = \frac{\pi }{3}$$ yo creo que ese 'b' nunca existio sino que es un 1, saludos para el uno puedes dar un hint, solo se que $TEX: $\left\| {dz} \right\| = \left\| {dx + idy} \right\| = \sqrt {\left( {dx} \right)^2 + \left( {dy} \right)^2 } = ds$$ pero no logro expresar s como una longitud de arco... Mensaje modificado por master_c el May 30 2011, 06:17 PM

master_c	Jun 6 2011, 02:24 PM Publicado: #7
Invitado	la dos debes corregirla pues el verdadero resultado es $TEX: $$<br />\int_{ - \infty }^{ + \infty } {\frac{{x^2 - b^2 }}<br />{{x^2 + b^2 }}} \frac{{\sin ax}}<br />{x}dx = \pi \left( {2e^{ - ab} - 1} \right)<br />$$$ la hice por variable compleja y calculo avanzado y creo que te falto un dos Mensaje modificado por master_c el Jun 6 2011, 02:25 PM

master_c	Jan 10 2013, 11:38 AM Publicado: #8
Invitado	para 7) Manipulando la integral $TEX: $$<br />K\left( a \right) = \int\limits_0^{ + \infty } {\frac{{\cosh ax}}<br />{{\cosh x}}dx} = \int\limits_0^{ + \infty } {\frac{{e^{ax} + e^{ - ax} }}<br />{{e^x + e^{ - x} }}dx} = \int\limits_0^{ + \infty } {\frac{{\left( {e^{ax} + e^{ - ax} } \right)e^{ - x} }}<br />{{1 + e^{ - 2x} }}dx} = \int\limits_0^{ + \infty } {\frac{{e^{ - \left( {1 - a} \right)x} + e^{ - \left( {1 + a} \right)x} }}<br />{{1 + e^{ - 2x} }}dx} <br />$$$ $TEX: $$<br /> = \int\limits_0^{ + \infty } {\sum\limits_{n = 0}^{ + \infty } {\left( { - 1} \right)^n e^{ - 2nx} \left( {e^{ - \left( {1 - a} \right)x} + e^{ - \left( {1 + a} \right)x} } \right)dx} } = \sum\limits_{n = 0}^{ + \infty } {\left( { - 1} \right)^n \int\limits_0^{ + \infty } {\left( {e^{ - \left( {2n + 1 - a} \right)x} + e^{ - \left( {2n + 1 + a} \right)x} } \right)dx} } <br />$$$ $TEX: $$<br /> = \sum\limits_{n = 0}^{ + \infty } {\left( { - 1} \right)^n \left( {\frac{1}<br />{{2n + 1 - a}} + \frac{1}<br />{{2n + 1 + a}}} \right)} = \frac{1}<br />{2}\sum\limits_{n = 0}^{ + \infty } {\left( { - 1} \right)^n \left( {\frac{1}<br />{{n + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{1}<br />{{n + \frac{1}<br />{2} + \frac{a}<br />{2}}}} \right)} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\left( {\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{n + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{n + 1 - \left( {\frac{1}<br />{2} - \frac{a}<br />{2}} \right)}}} } } \right) = \frac{1}<br />{2}\left( {\frac{1}<br />{{\frac{1}<br />{2} - \frac{a}<br />{2}}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{n + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n - 1} }}<br />{{n - \left( {\frac{1}<br />{2} - \frac{a}<br />{2}} \right)}}} } } \right)<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\left( {\frac{1}<br />{{\frac{1}<br />{2} - \frac{a}<br />{2}}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{\frac{1}<br />{2} - \frac{a}<br />{2} + n}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{\frac{1}<br />{2} - \frac{a}<br />{2} - n}}} } } \right) = \frac{1}<br />{2}\sum\limits_{n = - \infty }^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{\frac{1}<br />{2} - \frac{a}<br />{2} + n}}} <br />$$$ Considere la identidad $TEX: $$<br />\sin \pi z = \pi z\prod\limits_{n = 1}^{ + \infty } {\left( {1 - \frac{{z^2 }}<br />{{n^2 }}} \right)} <br />$$$ $TEX: $$<br />\ln \sin \pi z = \ln \pi + \ln z + \sum\limits_{n = 1}^{ + \infty } {\ln \left( {1 - \frac{{z^2 }}<br />{{n^2 }}} \right)} <br />$$$ $TEX: $$<br />\frac{{\pi \cos \pi z}}<br />{{\sin \pi z}} = \frac{1}<br />{z} - \sum\limits_{n = 1}^{ + \infty } {\frac{{2z}}<br />{{n^2 - z^2 }}} = \frac{1}<br />{z} - \sum\limits_{n = 1}^{ + \infty } {\left( {\frac{1}<br />{{n - z}} - \frac{1}<br />{{n + z}}} \right)} = \frac{1}<br />{z} + \sum\limits_{n = 1}^{ + \infty } {\left( {\frac{1}<br />{{z + n}} + \frac{1}<br />{{z - n}}} \right)} <br />$$$ Entonces $TEX: $$<br />\sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{n + z}}} = \frac{\pi }<br />{{\tan \pi z}}<br />$$$ por otro lado expandiendo la suma $TEX: $$<br />\sum\limits_{n = - \infty }^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{n + \frac{1}<br />{2} - \frac{a}<br />{2}}}} = ... + \frac{{\left( { - 1} \right)^{ - 2} }}<br />{{ - 2 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{\left( { - 1} \right)^{ - 1} }}<br />{{ - 1 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{\left( { - 1} \right)^0 }}<br />{{0 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{\left( { - 1} \right)^1 }}<br />{{1 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{\left( { - 1} \right)^2 }}<br />{{2 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + ...<br />$$$ notese que $TEX: $$<br /> = \left( {... + \frac{1}<br />{{ - 6 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{1}<br />{{ - 4 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{1}<br />{{ - 2 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{1}<br />{{0 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{1}<br />{{2 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + ...} \right)<br />$$$ $TEX: $$<br /> + \left( {... + \frac{{ - 1}}<br />{{ - 5 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{ - 1}}<br />{{ - 3 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{ - 1}}<br />{{ - 1 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{ - 1}}<br />{{1 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + \frac{{ - 1}}<br />{{3 + \frac{1}<br />{2} - \frac{a}<br />{2}}} + ...} \right)<br />$$$ es decir; $TEX: $$<br /> = \sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{2n + \frac{1}<br />{2} - \frac{a}<br />{2}}}} - \sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{2n + 1 + \frac{1}<br />{2} - \frac{a}<br />{2}}}} <br />$$$ $TEX: $$<br />\int\limits_0^{ + \infty } {\frac{{\cosh ax}}<br />{{\cosh x}}dx} = \frac{1}<br />{2}\sum\limits_{n = - \infty }^{ + \infty } {\frac{{\left( { - 1} \right)^n }}<br />{{n + \frac{1}<br />{2} - \frac{a}<br />{2}}}} = \frac{1}<br />{2}\left( {\sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{2n + \frac{1}<br />{2} - \frac{a}<br />{2}}}} - \sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{2n + 1 + \frac{1}<br />{2} - \frac{a}<br />{2}}}} } \right)<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\left( {\frac{1}<br />{2}\sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{n + \frac{1}<br />{4} - \frac{a}<br />{4}}}} - \frac{1}<br />{2}\sum\limits_{n = - \infty }^{ + \infty } {\frac{1}<br />{{n + \frac{1}<br />{2} + \frac{1}<br />{4} - \frac{a}<br />{4}}}} } \right) = \frac{1}<br />{2}\left( {\frac{1}<br />{2}\frac{\pi }<br />{{\tan \pi \left( {\frac{1}<br />{4} - \frac{a}<br />{4}} \right)}} - \frac{1}<br />{2}\frac{\pi }<br />{{\tan \pi \left( {\frac{1}<br />{2} + \frac{1}<br />{4} - \frac{a}<br />{4}} \right)}}} \right)<br />$$$ $TEX: $$<br /> = \frac{\pi }<br />{4}\left( {\frac{1}<br />{{\tan \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}} - \frac{1}<br />{{\tan \left( {\frac{\pi }<br />{2} + \frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}}} \right) = \frac{\pi }<br />{4}\left( {\frac{1}<br />{{\tan \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}} + \tan \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)} \right)<br />$$$ $TEX: $$<br /> = \frac{\pi }<br />{4}\frac{{1 + \tan ^2 \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}}<br />{{\tan \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}} = \frac{\pi }<br />{4}\frac{{\sec ^2 \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}}<br />{{\tan \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}} = \frac{\pi }<br />{4}\frac{1}<br />{{\sin \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)\cos \left( {\frac{\pi }<br />{4} - \frac{{\pi a}}<br />{4}} \right)}} = \frac{\pi }<br />{2}\frac{1}<br />{{\sin \left( {\frac{\pi }<br />{2} - \frac{{\pi a}}<br />{2}} \right)}}<br />$$$ $TEX: $$<br /> = \frac{\pi }<br />{2}\frac{1}<br />{{\cos \left( { - \frac{{\pi a}}<br />{2}} \right)}} = \frac{\pi }<br />{2}\frac{1}<br />{{\cos \frac{{\pi a}}<br />{2}}} = \frac{\pi }<br />{2}\sec \frac{{\pi a}}<br />{2}<br />$$$ Saludos ! Mensaje modificado por master_c el Jan 10 2013, 12:31 PM

master_c	Jan 10 2013, 04:17 PM Publicado: #9
Invitado	para el 7) $TEX: $$<br />K\left( n \right) = \int\limits_0^{ + \infty } {\frac{{\cosh nx}}<br />{{\cosh x}}dx} = \frac{1}<br />{2}\int\limits_{ - \infty }^{ + \infty } {\frac{{e^{nx} + e^{ - nx} }}<br />{{e^x + e^{ - x} }}dx} = \frac{1}<br />{2}\int\limits_0^{ + \infty } {\frac{{y^n + y^{ - n} }}<br />{{1 + y^2 }}dy} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{4}\left( {2\int\limits_0^{\frac{\pi }<br />{2}} {\cos ^{2\left( {\frac{n}<br />{2} + \frac{1}<br />{2}} \right) - 1} \theta \sin ^{2\left( { - \frac{n}<br />{2} + \frac{1}<br />{2}} \right) - 1} \theta d\theta } + 2\int\limits_0^{\frac{\pi }<br />{2}} {\cos ^{2\left( { - \frac{n}<br />{2} + \frac{1}<br />{2}} \right) - 1} \theta \sin ^{2\left( {\frac{n}<br />{2} + \frac{1}<br />{2}} \right) - 1} \theta d\theta } } \right)<br />$$$ invocando a beta $TEX: $$<br /> = \frac{1}<br />{4}\left( {\beta \left( {\frac{n}<br />{2} + \frac{1}<br />{2}, - \frac{n}<br />{2} + \frac{1}<br />{2}} \right) + \beta \left( { - \frac{n}<br />{2} + \frac{1}<br />{2},\frac{n}<br />{2} + \frac{1}<br />{2}} \right)} \right)<br />$$$ haciendo el cambio a gamma $TEX: $$<br /> = \frac{1}<br />{4}\left( {\frac{{\Gamma \left( {\frac{n}<br />{2} + \frac{1}<br />{2}} \right)\Gamma \left( { - \frac{n}<br />{2} + \frac{1}<br />{2}} \right)}}<br />{{\Gamma \left( {\frac{n}<br />{2} + \frac{1}<br />{2} + - \frac{n}<br />{2} + \frac{1}<br />{2}} \right)}} + \frac{{\Gamma \left( { - \frac{n}<br />{2} + \frac{1}<br />{2}} \right)\Gamma \left( {\frac{n}<br />{2} + \frac{1}<br />{2}} \right)}}<br />{{\Gamma \left( { - \frac{n}<br />{2} + \frac{1}<br />{2} + \frac{n}<br />{2} + \frac{1}<br />{2}} \right)}}} \right)<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\Gamma \left( {\frac{n}<br />{2} + \frac{1}<br />{2}} \right)\Gamma \left( {1 - \left( {\frac{n}<br />{2} + \frac{1}<br />{2}} \right)} \right)<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\frac{\pi }<br />{{\sin \pi \left( {\frac{1}<br />{2} + \frac{n}<br />{2}} \right)}}<br />$$$ $TEX: $$<br /> = \frac{\pi }<br />{{2\cos \left( {\frac{{n\pi }}<br />{2}} \right)}}<br />$$$

master_c	Jan 10 2013, 05:01 PM Publicado: #10
Invitado	para el 7 otra solucion como a pertenece a (-1,1) manipularemos a nuestro favor dicho dato $TEX: $$<br />\int\limits_0^{ + \infty } {\frac{{\cosh ax}}<br />{{\cosh x}}dx} = \frac{1}<br />{2}\int\limits_{ - \infty }^{ + \infty } {\frac{{e^{ax} + e^{ - ax} }}<br />{{e^x + e^{ - x} }}dx} = \frac{1}<br />{2}\int\limits_0^{ + \infty } {\frac{{y^a + y^{ - a} }}<br />{{y^2 + 1}}dy} = \frac{1}<br />{2}\int\limits_0^{ + \infty } {\frac{y}<br />{{y^2 + 1}}\left( {y^{ - \left( {1 - a} \right)} + y^{ - \left( {1 + a} \right)} } \right)dy} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {e^{ - u} \sin \left( {yu} \right)} \left( {y^{ - \left( {1 - a} \right)} + y^{ - \left( {1 + a} \right)} } \right)dudy} <br />$$$ (se uso $TEX: $$<br />\int {e^{bx} \sin axdx} = \frac{{e^{bx} }}<br />{{a^2 + b^2 }}\left( {b\sin ax - a\cos ax} \right) + K<br />$$$ integrando por partes es facil probar) $TEX: $$<br />\int\limits_0^{ + \infty } {e^{ - au} \sin bu} du = \frac{b}<br />{{b^2 + a^2 }}<br />$$$ entonces $TEX: $$<br /> = \frac{1}<br />{2}\int\limits_0^{ + \infty } {e^{ - u} \left( {\int\limits_0^{ + \infty } {\left( {y^{ - \left( {1 - a} \right)} + y^{ - \left( {1 + a} \right)} } \right)\sin \left( {yu} \right)dy} } \right)du} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\int\limits_0^{ + \infty } {e^{ - u} \left( {\int\limits_0^{ + \infty } {y^{ - \left( {1 - a} \right)} \sin \left( {yu} \right)dy} + \int\limits_0^{ + \infty } {y^{ - \left( {1 + a} \right)} \sin \left( {yu} \right)dy} } \right)du} <br />$$$ basandome en un viejo post http://www.fmat.cl/index.php?showtopic=66445 $TEX: $$<br />\int\limits_0^{ + \infty } {y^{a - 1} \sin \left( {yu} \right)dy} = - \int\limits_0^{ + \infty } {y^{a - 1} \sin \left( { - yu} \right)dy} = - \Im \int\limits_0^{ + \infty } {y^{a - 1} e^{ - iyu} dy} <br />$$$ $TEX: $$<br /> = \Im \left( {\frac{i}<br />{u}\int\limits_0^{ + \infty } {\left( {\frac{k}<br />{{ui}}} \right)^{a - 1} e^{ - k} dk} } \right) = \Im \left( {\frac{i}<br />{u}\left( {\frac{1}<br />{{ui}}} \right)^{a - 1} \int\limits_0^{ + \infty } {k^{a - 1} e^{ - k} dk} } \right) = \Im \left( {i^{2 - a} \frac{{\Gamma \left( a \right)}}<br />{{u^a }}} \right) = - \frac{1}<br />{{u^a }}\Im \left( {i^{ - a} \Gamma \left( a \right)} \right)<br />$$$ $TEX: $$<br /> = - \frac{1}<br />{{u^a }}\Im \left( {\left( {\cos \frac{{a\pi }}<br />{2} - i\sin \frac{{a\pi }}<br />{2}} \right)\Gamma \left( a \right)} \right)<br />$$$ $TEX: $$<br /> = \frac{1}<br />{{u^a }}\Gamma \left( a \right)\sin \frac{{a\pi }}<br />{2}<br />$$$ idem para la otra integral $TEX: $$<br />\int\limits_0^{ + \infty } {y^{ - \left( {1 + a} \right)} \sin \left( {yu} \right)dy} = - \int\limits_0^{ + \infty } {y^{ - \left( {1 + a} \right)} \sin \left( { - yu} \right)dy} = - \Im \int\limits_0^{ + \infty } {y^{ - \left( {a + 1} \right)} e^{ - iyu} dy} <br />$$$ $TEX: $$<br /> = \Im \left( {\frac{i}<br />{u}\int\limits_0^{ + \infty } {\left( {\frac{k}<br />{{ui}}} \right)^{ - \left( {a + 1} \right)} e^{ - k} dk} } \right) = \Im \left( {\frac{i}<br />{u}\left( {\frac{1}<br />{{ui}}} \right)^{ - \left( {a + 1} \right)} \int\limits_0^{ + \infty } {k^{ - \left( {a + 1} \right)} e^{ - k} dk} } \right) = - \Im \left( {i^a u^a \Gamma \left( { - a} \right)} \right)<br />$$$ $TEX: $$<br /> = - u^a \Im \left( {\left( {\cos \frac{{a\pi }}<br />{2} + i\sin \frac{{a\pi }}<br />{2}} \right)\Gamma \left( { - a} \right)} \right)<br />$$$ $TEX: $$<br /> = - u^a \sin \frac{{a\pi }}<br />{2}\Gamma \left( { - a} \right)<br />$$$ entonces, se tiene $TEX: $$<br /> = \frac{1}<br />{2}\int\limits_0^{ + \infty } {e^{ - u} \left( {\frac{1}<br />{{u^a }}\Gamma \left( a \right)\sin \frac{{a\pi }}<br />{2} - u^a \sin \frac{{a\pi }}<br />{2}\Gamma \left( { - a} \right)} \right)du} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\Gamma \left( a \right)\sin \frac{{a\pi }}<br />{2}\int\limits_0^{ + \infty } {u^{ - a} e^{ - u} du} - \frac{1}<br />{2}\Gamma \left( { - a} \right)\sin \frac{{a\pi }}<br />{2}\int\limits_0^{ + \infty } {e^{ - u} u^a du} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\Gamma \left( a \right)\sin \frac{{a\pi }}<br />{2}\int\limits_0^{ + \infty } {u^{1 - a - 1} e^{ - u} du} - \frac{1}<br />{2}\Gamma \left( { - a} \right)\sin \frac{{a\pi }}<br />{2}\int\limits_0^{ + \infty } {u^{1 + a - 1} e^{ - u} du} <br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\Gamma \left( {1 - a} \right)\Gamma \left( a \right)\sin \frac{{a\pi }}<br />{2} - \frac{1}<br />{2}\Gamma \left( {1 + a} \right)\Gamma \left( {1 - \left( {a + 1} \right)} \right)\sin \frac{{a\pi }}<br />{2}<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\left( {\frac{\pi }<br />{{\sin a\pi }} - \frac{\pi }<br />{{\sin \left( {a + 1} \right)\pi }}} \right)\sin \frac{{a\pi }}<br />{2}<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\left( {\frac{\pi }<br />{{\sin a\pi }} - \frac{\pi }<br />{{ - \sin a\pi }}} \right)\sin \frac{{a\pi }}<br />{2}<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\frac{{2\pi }}<br />{{\sin a\pi }}\sin \frac{{a\pi }}<br />{2}<br />$$$ $TEX: $$<br /> = \frac{1}<br />{2}\frac{\pi }<br />{{\sin \frac{{a\pi }}<br />{2}\cos \frac{{a\pi }}<br />{2}}}\sin \frac{{a\pi }}<br />{2}<br />$$$ $TEX: $$<br /> = \frac{\pi }<br />{{2\cos \frac{{a\pi }}<br />{2}}}<br />$$$

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