Necesito ayuda con la siguiente integral:
$$ \int_{-\pi}^{\pi}{\cos\left(\, ax\,\right) \over 1-bx^{2}}\,{\rm d}x $$
Las constantes $a$ y $b$ son reales y positivos.
Cualquier ayuda será apreciada :)
Respuesta
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Felix Marin
Puntos
32763
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#66f}{\large\int_{-\pi}^{\pi}{\cos\pars{ax} \over 1 - bx^{2}}\,\dd x} =\half\bracks{\int_{-\pi}^{\pi}{\cos\pars{ax} \over 1 - b^{1/2}x}\,\dd x +\int_{-\pi}^{\pi}{\cos\pars{ax} \over 1 + b^{1/2}x}\,\dd x} \\[5mm]&=\half\sum_{\sigma\ =\ \pm}\ \overbrace{\int_{-\pi}^{\pi}{\cos\pars{ax} \over 1 + \sigma b^{1/2}x}\,\dd x} ^{\ds{\color{#c00000}{1 + \sigma b^{1/2}x\equiv t\ \imp\ x = \sigma b^{-1/2}\pars{t - 1}}}} \\[5mm]&=\half\sum_{\sigma\ =\ \pm} \int_{1 - \sigma b^{1/2}\pi}^{1 + \sigma b^{1/2}\pi} {\cos\pars{\sigma ab^{-1/2}t - \sigma ab^{-1/2}} \over t}\,\sigma b^{-1/2}\,\dd t \\[5mm]&={b^{-1/2}\cos\pars{ab^{-1/2}} \over 2}\sum_{\sigma\ =\ \pm}\sigma \int_{1 - \sigma b^{1/2}\pi}^{1 + \sigma b^{1/2}\pi} {\cos\pars{ab^{-1/2}t} \over t}\,\dd t \\&+{b^{-1/2}\sin\pars{ab^{-1/2}} \over 2}\sum_{\sigma\ =\ \pm}\sigma \int_{1 - \sigma b^{1/2}\pi}^{1 + \sigma b^{1/2}\pi} {\sin\pars{ab^{-1/2}t} \over t}\,\dd t \\[5mm]&={b^{-1/2}\cos\pars{ab^{-1/2}} \over 2}\sum_{\sigma\ =\ \pm}\sigma \int_{ab^{-1/2} - \sigma a\pi}^{ab^{-1/2} + \sigma a\pi} {\cos\pars{t} \over t}\,\dd t \\&+{b^{-1/2}\sin\pars{ab^{-1/2}} \over 2}\sum_{\sigma\ =\ \pm}\sigma \int_{ab^{-1/2} - \sigma a\pi}^{ab^{-1/2} + \sigma a\pi} {\sin\pars{t} \over t}\,\dd t \end{align}
Sin embargo, \begin{align} \int_{\mu}^{\nu}{\cos\pars{t} \over t}\,\dd t &=-\int_{\nu}^{\infty}{\cos\pars{t} \over t}\,\dd t -\bracks{-\int_{\mu}^{\infty}{\cos\pars{t} \over t}\,\dd t} ={\rm Ci}\pars{\nu} - {\rm Ci}\pars{\mu} \\[5mm] \int_{\mu}^{\nu}{\cos\pars{t} \over t}\,\dd t &=\int_{0}^{\nu}{\sin\pars{t} \over t}\,\dd t -\int_{0}^{\mu}{\sin\pars{t} \over t}\,\dd t ={\rm Si}\pars{\nu} - {\rm Si}\pars{\mu} \end{align} donde $\ds{\rm Ci}$ y $\ds{\rm Si}$ son los Integral del coseno y la integral del seno respectivamente.
Entonces, \begin{align} &\color{#66f}{\large\int_{-\pi}^{\pi}{\cos\pars{ax} \over 1 - bx^{2}}\,\dd x} \\[5mm]&=\color{#66f}{\large b^{-1/2}\cos\pars{ab^{-1/2}}\braces{% {\rm Ci}\pars{a\bracks{b^{-1/2} + \pi}} - {\rm Ci}\pars{a\bracks{b^{-1/2} - \pi}}}} \\&\color{#66f}{\large \mbox{}+b^{-1/2}\sin\pars{ab^{-1/2}}\braces{% {\rm Si}\pars{a\bracks{b^{-1/2} + \pi}} - {\rm Si}\pars{a\bracks{b^{-1/2} - \pi}}}} \end{align}
