Integral of cosine over a quadratic I need help with the following integral:
$$
\int_{-\pi}^{\pi}{\cos\left(\, ax\,\right) \over 1-bx^{2}}\,{\rm d}x
$$
The constants $a$ and $b$ are both real and positive.
Any help will be appreciated :)
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\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}

However,
  \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}
  where $\ds{\rm Ci}$ and $\ds{\rm Si}$ are the
  Cosine Integral and the Sine Integral , respectively.

Then,
\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}
