Guía VII de Variable Compleja - Foro fmat.cl

Portal Matemático

¿Qué es FMAT?

Foro fmat.cl > Programa Enseñanza Universitaria > Comunidades Universitarias > Universidad de Santiago de Chile > Tópicos II

Reply to this topic

Start new topic

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: $$<br /> = \frac{\pi }<br />{{2\cos \frac{{a\pi }}<br />{2}}}<br />$$$

« Posterior · Tópicos II · Anterior »

Reply to this topic

Start new topic

1 usuario(s) está(n) leyendo esta discusión (1 invitado(s) y 0 usuario(s) anónimo(s))

0 miembro(s):

Modo de Ver: Estándar · Cambiar a: Lineal+ · Cambiar a: Outline

Suscribirse · Enviar por correo · Imprimir · Suscribirse a este foro

Versión Lo-Fi

Fecha y Hora actual: 4th April 2025 - 04:18 AM

Powered By IP.Board 2.2.1 © 2007 IPS, Inc.