\n\\begin{equation}
\n\\label{eqn:r1}
\nc_k = \\exp(i(\\pi\/6 + 2k\\pi \/ 6) \\hspace{10px} k=0, 1, \\ldots, 5.
\n\\end{equation}
\nDe hecho para emplear la Proposici\u00f3n \\ref{propo:semiplanoSuperio}, solo nos importan los puntos de la ecuaci\u00f3n
\n(\\ref{eqn:r1}) que est\u00e1n en el semiplano superior.
\n
![\"Regi\u00f3n](\"https:\/\/www.candaana.com\/news\/wp-content\/uploads\/2018\/02\/region2-300x154.png\")
\nSea $\\Gamma_R = [-R, R]+C_R$ definida de la siguiente manera:
\n\\begin{equation}
\n\\label{eqn:gammaCurva}
\n\\Gamma_R = \\left\\{
\n\\begin{array}{ccc}
\nt &\\mbox{si}& -R\\leq t < R\\\\ R\\exp(i\\theta) &\\mbox{si}& 0\\leq \\theta \\leq \\pi \\end{array} \\right. \\end{equation} Notemos que para $z\\in C_R$ se tiene $$ |f(z)| = \\left| \\dfrac{z^2}{z^6 +1 }\\right| = \\dfrac{|z|^2}{|x^6+1|}\\leq \\dfrac{|z|^2}{|z|^6-1} = \\dfrac{R^2}{R^6-1}. $$ As\u00ed $$ \\left| \\int_{C_R}f(z)dz \\right| \\leq \\int_{C_R}|f(z)| |dz|\\leq \\dfrac{R^2}{R^6-1}\\pi R \\longrightarrow 0 \\hspace{10px} \\mbox{cuando } R\\rightarrow+\\infty. $$ Hemos probado que $$ \\lim_{R\\rightarrow +\\infty} \\int_{C_R}f(z)dz = 0. $$ Ahora por la Proposici\u00f3n \\ref{propo:semiplanoSuperio} $$ \\int_{-\\infty}^{+\\infty}f(x) dx= 2\\pi i \\sum_{n = 0}^{2} \\res{z=z_n} f(z)). $$ Ahora calculemos los residuos. Notemos que $p(z_k)=z_k^2\\neq 0$, $q(z_k)=z_k^6+1=0$ y $p'(z_k) = 6z_k^5\\neq 0$ con $k=0,\\; 1,\\; 2$. Lo anterior significa que los $z_k$’s son polos simples de $f(z)=p(z)\/q(z)$. As\u00ed \\begin{eqnarray*} \\res{z=z_0} f(z) &=& \\res{z=\\exp(i\\pi\/6)}f(z) = \\dfrac{p(\\exp(i\\pi\/6))}{q'(\\exp(i\\pi\/6))} \\\\ &=& \\dfrac{\\exp(i\\pi\/6)^2}{6\\exp(i\\pi\/6)^5} = \\dfrac{\\exp(i2\\pi\/6)}{6\\exp(i5\\pi\/6)}\\\\ &=& \\dfrac{\\exp(i\\pi\/3)}{6\\exp(i5\\pi\/6)} = \\frac{1}{6}\\dfrac{\\frac{1}{2}+i\\frac{\\sqrt{3}}{2}}{-\\frac{\\sqrt{3}}{2}+i\\frac{1}{2}}\\\\ &=& \\frac{1}{6} \\dfrac{1+i\\sqrt{3}}{-\\sqrt{3}+i} = \\frac{1}{6} \\dfrac{1+i\\sqrt{3}}{-\\sqrt{3}+i}\\cdot \\dfrac{-\\sqrt{3}-i}{-\\sqrt{3}-i}\\\\ &=& \\dfrac{-4i}{6\\cdot 4} = \\dfrac{-i}{6} \\\\ &=& \\dfrac{i}{6i} \\end{eqnarray*} \\begin{eqnarray*} \\res{z=i} f(z) = \\dfrac{-1}{6i} =-\\dfrac{1}{6i} \\end{eqnarray*} \\begin{eqnarray*} \\res{z=\\exp(i5\\pi\/6)} f(z) &=& \\dfrac{\\exp(i10\\pi\/6)}{6\\exp(i25\\pi\/6)} = \\dfrac{\\exp(i5\\pi\/3)}{6\\exp(i\\pi\/6)}\\\\ &=&\\frac{1}{6} \\cdot \\dfrac{\\exp(-i\\pi\/3)}{\\exp(i\\pi\/6)} = \\frac{1}{6} \\cdot\\dfrac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\\\ &=& \\frac{1}{6i} \\end{eqnarray*}
\nPor lo que $$ \\int_{-\\infty}^{+\\infty}f(x) dx = 2\\pi i (\\frac{1}{6i}-\\frac{1}{6i}+\\frac{1}{6i}) =\\frac{\\pi}{6}. $$ Como $f$ es par $$ \\int_{0}^{+\\infty}f(x) dx= \\frac{1}{2}\\int_{-\\infty}^{+\\infty}f(x) dx = \\frac{\\pi}{12}. $$<\/p>\n
\n
Proposici\u00f3n<\/h4>\n
Sea $f(z)=p(z)\/q(z)$, donde $p$ y $q$ son polinomios. Suponga que los ceros de $q$ no est\u00e1n sobre el eje real y algunos de ellos est\u00e1n sobre el semiplano superior. Sea $\\Gamma _R = [-R, R]+C_R$ donde $C_R$ es la curva dada por $z(t) = Re^{it}$ con $0\\leq t \\leq \\pi$.
\nSi \\begin{equation} \\lim_{R\\rightarrow +\\infty}\\int_{C_R}f(z)dz=0.
\n\\label{eqn:intPropo} \\end{equation} Entonces \\begin{enumerate} \\item Si $z_1.\\ldots, z_n$ son ceros de $q$ en el semiplano superior \\begin{equation} VP\\int_{-\\infty}^{\\infty}f(x)dx = 2\\pi i \\sum_{k=1}^{n}\\res{z=z_k}f(z) \\label{eqn:vpPropo1} \\end{equation} \\item Si $f(x)$ es par \\begin{equation} \\int_{-\\infty}^{\\infty}f(x)dx = 2\\pi i \\sum_{k=1}^{n}\\res{z=z_k}f(z) \\label{eqn:vpPropo2} \\end{equation} y \\begin{equation} \\int_{0}^{\\infty}f(x)dx = 2\\pi i \\sum_{k=1}^{n}\\res{z=z_k}f(z) \\label{eqn:vpPropo3} \\end{equation} \\end{enumerate} Establecer una Proposici\u00f3n an\u00e1loga a la anterior. Sea $f(z)=p(z)\/q(z)$, donde $p$ y $q$ son polinomios. Suponga que los ceros de $q$ no est\u00e1n sobre el eje real y algunos de ellos est\u00e1n sobre el semiplano inferior. Sea $\\Gamma _R = [-R, R]+C_R$ donde $C_R$ es la curva dada por $z(t) = Re^{-it}$ con $0\\leq t \\leq \\pi$. Si se cumple el an\u00e1logo (\\ref{eqn:intPropo}). Entonces \\begin{enumerate} \\item Si $z_1.\\ldots, z_n$ son ceros de $q(z)$ en el semiplano inferior, entonces (\\ref{eqn:vpPropo1}) \\item Si $f(x)$ es par, se tiene (\\ref{eqn:vpPropo2}) y (\\ref{eqn:vpPropo3}). \\end{enumerate} \\begin{proof} Existe $R_0>0$ tal que $z_1$, $z_2$, $\\ldots$, $z_n$ est\u00e1n en el interior de $\\Gamma_{R_0}$. Por el Teorema de Cauchy del residuo para $R>R_0$
\n$$
\n\\int_{\\Gamma_{R}}f(z)dz = 2\\pi i \\sum_{k=1}^{n}\\res{z=z_k}f(z).
\n$$
\nA su vez
\n$$
\n\\int_{\\Gamma_{R}}f(z)dz = \\int_{-R}^{R}f(x)dx+\\int_{C_R}f(z)dz.
\n$$
\nLuego
\n$$
\n\\int_{-R}^{R}f(x)dx = 2\\pi i \\sum_{k=1}^{n}\\res{z=z_k}f(z) – \\int_{C_R}f(z)dz.
\n$$
\nTomando el l\u00edmite cuando $R\\rightarrow +\\infty$ tenemos
\n$$
\nVP \\int_{-R}^{R}f(x)dx = \\lim_{R\\rightarrow \\infty} \\int_{-R}^{R}f(x)dx = 2\\pi i \\sum_{k=1}^{n}\\res{z=z_k}f(z) – \\lim_{R\\rightarrow \\infty} \\int_{C_R}f(z)dz.
\n$$
\nPor (\\ref{eqn:intPropo}) tenemos que se satisface (\\ref{eqn:vpPropo1}).
\n\\end{proof}<\/p>\n
\n
Calcular $\\displaystyle \\int_{-\\infty}^{\\infty}\\dfrac{\\cos 3x}{(x^2+1)^2}dx$. Calcular la siguiente integra: $$ \\int_{0}^{+\\infty}\\dfrac{x^2}{x^6+1}. $$ Soluci\u00f3n. Nos interesan los ceros del denominador, es decir, los $z\\in \\mathbb{C}$ tal que $z^6+1=0$. Estos n\u00fameros son<\/p>\n","protected":false},"author":2,"featured_media":91,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[12],"class_list":["post-88","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematicas","tag-matematicas"],"_links":{"self":[{"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/posts\/88","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/comments?post=88"}],"version-history":[{"count":1,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/posts\/88\/revisions"}],"predecessor-version":[{"id":198,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/posts\/88\/revisions\/198"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/media\/91"}],"wp:attachment":[{"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/media?parent=88"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/categories?post=88"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.candaana.com\/blog\/wp-json\/wp\/v2\/tags?post=88"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nConsideremos $\\displaystyle h(z)=f(x)e^{iaz}$ con $f(z) = \\dfrac{1}{(z^2+1)^2}$ y $a=3$. Notemos que $h(z)$ es anal\u00edtica, salvo en $z=\\pm i$.\\\\
\nSi $R>1$ y $\\Gamma_R$ definida como en (\\ref{eqn:gammaCurva}). De este modo
\n$$
\n\\int_{\\Gamma_R}h(z)dz = 2\\pi i \\res{z=i}h(z),
\n$$
\nes decir,
\n$$
\n\\int_{-R}^R h(z)dz+\\int_{C_R} h(z)dz = 2\\pi i \\res{z=i}h(z),
\n$$
\nesto implica que
\n\\begin{equation}
\n\\label{eqn:hzRes}
\n\\int_{-R}^R h(z)dz = 2\\pi i \\res{z=i}h(z)-\\int_{C_R} h(z)dz
\n\\end{equation}
\nNotemos que para $z\\in C_R$
\n$$
\n|f(z)| = \\left|\\dfrac{1}{(z^2+1)^2}\\right|\\leq \\dfrac{1}{(|z|^2-1)^2} = \\dfrac{1}{(R^2-1)^2} \\longrightarrow 0 \\mbox{ cuando } R\\rightarrow \\infty.
\n$$
\nPor el Lemma de Jordan
\n$$
\n\\lim_{R\\rightarrow \\infty}\\int_{C_R}f(z)dz=0.
\n$$
\nDe (\\ref{eqn:hzRes}) tenemos que
\n$$
\n\\lim_{R\\rightarrow \\infty}\\int_{-R}^R\\dfrac{e^{3ix}}{(x^2+1)^2}dx=2\\pi i \\res{z=i} \\dfrac{e^{3iz}}{(z^2+1)^2}
\n$$
\nPor otro lado, consideremos la funci\u00f3n
\n$$
\n\\phi (z) = \\dfrac{e^{i3z}}{(z+i)^2}
\n$$
\nanal\u00edtica en $i$ y $\\phi(i)\\neq 0$, entonces
\n$$
\nf(z) = \\dfrac{\\phi(z)}{(z-i)^2}
\n$$
\ntiene un polo de orden 2. Consideremos los siguiente
\n$$
\n\\phi'(z)= \\dfrac{3i(z+i)e^{3iz}-2e^{3iz}}{(z+i)^3}.
\n$$
\nEvaluando tenemos en $z= i$, tenemos
\n$$
\n\\phi'(i) = \\dfrac{3i(i+i)e^{3i^2}-2e^{3i^2}}{(i+i)^3} =
\n\\dfrac{-6e^{-3}-2e^{-3}}{-8i} = \\dfrac{-8e^{-3}}{-8i} = \\dfrac{1}{ie^3}
\n$$
\nPor lo que
\n$$
\n2\\pi i \\res{z=i} f(z) = \\dfrac{2\\pi i}{ie^3} =\\dfrac{2\\pi }{e^3}.
\n$$
\nAs\u00ed
\n$$
\n\\lim_{R\\rightarrow \\infty}\\int_{-R}^R\\dfrac{\\cos 3x}{(x^2+1)^2}dx+i\\int_{-R}^R\\dfrac{\\sin 3x}{(x^2+1)^2}dx = \\dfrac{2\\pi }{e^3}.
\n$$
\nPor lo tanto
\n$$
\n\\int_{-\\infty}^{\\infty}\\dfrac{\\cos 3x}{(x^2+1)^2}dx = \\dfrac{2\\pi }{e}
\n\\hspace{10px}\\mbox{ y }\\hspace{10px}
\n\\int_{-\\infty}^{\\infty}\\dfrac{\\sin 3x}{(x^2+1)^2}dx = 0.
\n$$
\nDemostrar que
\n$$
\n\\int\\limits_{0}^{\\infty} \\dfrac{\\ln{x}}{(x^2+4)^2}dx = \\frac{\\pi}{32}(\\ln{2}-1).
\n$$
\n![\"Regi\u00f3n](\"https:\/\/www.candaana.com\/news\/wp-content\/uploads\/2018\/02\/reg1-300x158.png\")
\n$$
\nf(z) = \\dfrac{\\log{z}}{(z^2+4)^2}.
\n$$
\nPor el Teorema de Cauchy del Residuo tenemos
\n\\begin{eqnarray*}
\n\\int\\limits_{\\Gamma_{R ,\\rho}}f(z)dz
\n&=&
\n\\int\\limits_{L_1}f(z)dz + \\int\\limits_{C_\\rho}f(z)dz +\\int\\limits_{L_2}f(z)dz +\\int\\limits_{C_R}f(z)dz\\\\
\n&=&
\n2\\pi i\\res{z=2i}f(z).
\n\\end{eqnarray*}
\nLo anterior implica que
\n\\begin{eqnarray}
\n\\int\\limits_{L_1}f(z)dz + \\int\\limits_{L_2}f(z)dz =2\\pi i\\res{z=2i}f(z) -\\int\\limits_{C_\\rho}f(z)dz -\\int\\limits_{C_R}f(z)dz.
\n\\label{eqn:l1l2}
\n\\end{eqnarray}
\nSi $z\\in C_\\rho$, entonces
\n\\begin{eqnarray}
\n|\\log z| = |\\ln|z|+i\\theta| &\\leq & |\\ln|z||+|i\\theta| \\nonumber \\\\
\n&=& |\\ln \\rho|+|\\theta| \\nonumber\\\\
\n|\\log z| & \\leq & -\\ln\\rho+\\theta ,
\n\\label{eqn:logz}
\n\\end{eqnarray}
\npor otro lado, tenemos
\n$$
\n|(z^2+4)^2| = |z^2+4|^2\\geq ||z|-4|^2 = (\\rho-4)^2 ,
\n$$
\nes decir,
\n\\begin{equation}
\n\\dfrac{1}{|(z^2+4)^2|}\\leq \\dfrac{1}{(\\rho-4)^2}.
\n\\label{eqn:z24}
\n\\end{equation}
\nPor (\\ref{eqn:logz}) y (\\ref{eqn:z24}) tenemos que
\n$$
\n|f(z)| \\leq \\dfrac{-\\ln\\rho+\\theta}{(\\rho-4)^2}
\n$$
\n\\begin{eqnarray*}
\n\\left| \\int\\limits_{C_\\rho}f(z)dz \\right| &\\leq& \\int\\limits_{C_\\rho}|f(z)||dz|
\n\\leq
\n\\int\\limits_{0}^{\\pi} \\dfrac{-\\ln\\rho+\\theta}{(\\rho-4)^2}\\left( \\dfrac{\\pi \\rho}{2}\\right) d\\theta \\leq \\int\\limits_{0}^{\\pi} \\dfrac{-\\ln\\rho+\\theta}{(\\rho-4)^2}\\left( \\pi \\rho\\right) d\\theta \\\\
\n&=&
\n\\dfrac{\\pi \\rho}{(\\rho-4)^2}\\int\\limits_{0}^{\\pi} (-\\ln\\rho+\\theta) d\\theta = \\dfrac{\\pi \\rho}{(\\rho-4)^2} = \\dfrac{\\pi \\rho}{(\\rho-4)^2} (-\\theta\\ln\\rho +\\theta^2\/2) \\left|_{0}^{\\pi}\\right.\\\\
\n&=&
\n\\dfrac{\\pi \\rho}{(\\rho-4)^2} (-\\pi\\ln\\rho +\\pi^2\/2)
\n=\\dfrac{-\\pi^2\\rho\\ln\\rho}{(\\rho-4)^2}+\\dfrac{\\pi^3\\rho}{2(\\rho-4)^2}.
\n\\end{eqnarray*}
\nObservemos que
\n$$
\n\\lim_{\\rho \\to 0}\\dfrac{\\pi^3\\rho}{2(\\rho-4)^2} = 0.
\n$$
\nPor variable real (regla de L’Hopital) se tiene,
\n$$
\n\\lim_{\\rho \\to 0}(\\rho \\ln\\rho) = \\lim_{\\rho \\to 0}\\dfrac{\\ln \\rho}{\\rho^{-1}} = 0.
\n$$
\nSi $z\\in C_R$, entonces (por simetr\u00eda)
\n$$
\n\\left| \\int\\limits_{C_R}f(z)dz \\right| \\leq \\dfrac{-\\pi^2R\\ln R}{(R-4)^2}+\\dfrac{\\pi^3R}{2(R-4)^2}.
\n$$
\nPor variable real se puede ver que
\n$$
\n\\lim_{R \\to \\infty}\\dfrac{\\pi^3R}{2(R-4)^2} = \\lim_{R \\to \\infty}\\dfrac{\\pi^3\/R}{2(1-4\/R)^2}=0,
\n$$
\nadem\u00e1s
\n$$
\n\\lim_{R \\to \\infty}\\dfrac{-\\pi^2R\\ln R}{(R-4)^2} = \\lim_{R \\to \\infty}\\dfrac{-\\pi^2(\\ln R)\/R}{(1-4\/R)^2}=0.
\n$$
\nCalculemos el residuo
\n$$
\n\\res{z = 2i} f(z)= \\dfrac{\\log(z)\/(z+2i)^2}{(z-2i)^2}.
\n$$
\nNotemos que $\\phi(z) = \\log(z)\/(z+2i)^2$ es anal\u00edtica en $z=2i$ y $\\phi(2i)\\neq0$, entonces $z= 2i$ es un polo de orden 2 de $f(z)$. As\u00ed
\n$$
\n\\res{z=2i}f(z) = \\phi ‘(2i).
\n$$
\ncomo
\n$$
\n\\phi ‘(z) = \\dfrac{\\frac{1}{z}(z+2i)-2\\log(z)}{(z+2i)^3}.
\n$$
\n\\begin{eqnarray*}
\n\\phi ‘(2i) &=& \\dfrac{\\frac{1}{2i}(4i)-2\\log(2i)}{(4i)^3}=
\n\\dfrac{2-2\\ln(2)+i\\pi}{(4i)^3} \\\\
\n&=& \\dfrac{2-2\\ln(2)+i\\pi}{-64i}\\\\
\n&=& \\dfrac{1-\\ln(2)}{-32i}-\\dfrac{\\pi}{64}
\n\\end{eqnarray*}
\nAhora evaluemos la siguiente integral
\n\\begin{eqnarray*}
\n\\int\\limits_{L_1}f(z)dz+\\int\\limits_{L_2}f(z)dz &=&
\n\\int_{-R}^{-\\rho}\\dfrac{\\ln |t|+i\\pi}{(t^2+4)^2}dt+ \\int_{\\rho}^{R}\\dfrac{\\ln t}{(t^2+4)^2}dt\\\\
\n&=& \\int_{\\rho}^{R}\\dfrac{2\\ln |t|+i\\pi}{(t^2+4)^2}dt\\\\
\n&=& \\int_{\\rho}^{R}\\dfrac{2\\ln |t|}{(t^2+4)^2}+\\int_{\\rho}^{R}\\dfrac{i\\pi}{(t^2+4)^2}dt.
\n\\end{eqnarray*}
\nTomano el l\u00edmite cuando $\\rho\\to 0$ y $R\\to \\infty$
\n\\begin{eqnarray*}
\n\\int_{0}^{\\infty}\\dfrac{2\\ln |t|}{(t^2+4)^2}+\\int_{0}^{\\infty}\\dfrac{i\\pi}{(t^2+4)^2}dt &=& 2\\pi
\ni \\left[
\n\\dfrac{1-\\ln(2)}{-32i}-\\dfrac{\\pi}{64} \\right]\\\\
\n&=&
\n2\\pi
\n\\left[
\n\\dfrac{\\ln(2)-1}{32i}-\\dfrac{\\pi i}{64}
\n\\right].
\n\\end{eqnarray*}
\nPor lo tanto
\n$$
\n\\int_{0}^{\\infty}\\dfrac{2\\ln |t|}{(t^2+4)^2} = 2\\pi\\frac{\\ln(2)-1}{32i}
\n\\Longrightarrow
\n\\int_{0}^{\\infty}\\dfrac{\\ln |t|}{(t^2+4)^2} = \\pi\\frac{\\ln(2)-1}{32i}
\n$$
\nCalcular
\n$$
\n\\int_{0}^{2\\pi}\\dfrac{d\\theta}{1+a\\sin \\theta},
\n$$
\ncon $-1<a<1$. Tomando \\begin{eqnarray*} \\dfrac{1}{1+a\\sin \\theta} = \\dfrac{1}{1+a[\\frac{1}{2i}(z-\\frac{1}{z})]} = \\dfrac{1}{1+\\frac{a(z^2+1)}{2zi}} = \\dfrac{2iz}{az^2+2iz-a}. \\end{eqnarray*} As\u00ed \\begin{eqnarray*} \\int_{0}^{2\\pi}\\dfrac{d\\theta}{1+a\\sin \\theta} &=& \\int\\limits_{|z|=1}\\dfrac{2iz}{az^2+2iz-a}\\cdot \\dfrac{1}{iz}dz = \\int\\limits_{|z|=1}\\dfrac{2}{az^2+2iz-a}dz \\end{eqnarray*} Notemos que \\begin{eqnarray*} az^2+2iz-a = 0 \\Longrightarrow z &=& \\dfrac{-2i \\pm \\sqrt{(2i)^2+4a^2}}{2a} = \\dfrac{-2i \\pm \\sqrt{(2i)^2+4a^2}}{2a}\\\\ & =& \\dfrac{-2i \\pm \\sqrt{-4+4a^2}}{2a} = \\frac{1}{a}(-i\\pm \\sqrt{-1+a^2})\\\\ &=& \\frac{i}{a}(-1\\pm \\sqrt{a^2-1}). \\end{eqnarray*} Pongamos $\\displaystyle z_1 = \\frac{i}{a}(-1+ \\sqrt{1-a^2})$ y $\\displaystyle z_2 = \\frac{i}{a}(-1- \\sqrt{1-a^2})$. Notemos que $$ |z_2| >1.
\n$$
\nPor otro lado, como $|z_1z_2|=1$, entonces
\n$$|z_1|= 1\/|z_2|<1.$$
\nNuestro integrando es
\n$$
\nf(z) = \\dfrac{2\/a}{(z-z_1)(z-z_2)} = \\dfrac{\\dfrac{2\/a}{(z-z_2)}}{z-z_1}
\n= \\dfrac{\\phi(z)}{z-z_1}.
\n$$
\nCon $\\phi(z) = \\dfrac{2\/a}{(z-z_2)}$ anal\u00edtica en $z_1$ y $\\phi(z_1)\\neq 0$. Entonces $z_1$ es un polo de orden 1 de $f(z)$. As\u00ed
\n\\begin{eqnarray*}
\n\\res{z=z_1}f(z) &=& \\phi(z) = \\dfrac{2a}{\\frac{i}{a}(-1+ \\sqrt{1-a^2}) – (\\frac{i}{a}(-1- \\sqrt{1-a^2}))} \\\\
\n&=& \\dfrac{1}{\\sqrt{1-a}}
\n\\end{eqnarray*}
\nAplicando el Teorema de Cauchy del Residuo
\n$$
\n\\int_{0}^{2\\pi}\\dfrac{d\\theta}{1+a\\sin \\theta} = 2\\pi i \\left( \\dfrac{1}{i\\sqrt{1-a^2}}\\right) = \\dfrac{1}{\\sqrt{1-a^2}}.
\n$$<\/p>\n","protected":false},"excerpt":{"rendered":"