Complex Integral final steps How do we evaluate the Cauchy Principal value for:
$$ \int_{-\infty}^\infty\frac{\cos kx}{x-a}dx  $$ 
Given, a is real, k >${\ 0}$?
I thought of integrating from ${-\infty}$ to ${\ a}$ and then from ${\ a}$ to ${+\infty}$
Any help will be appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\mbox{P.V.}\int_{-\infty}^{\infty}{\cos\pars{3x} \over x^{2} - 25}\,\dd x =
\Re\int_{-\infty}^{\infty}
{\cos\pars{3x} \over x^{2} - \pars{25 + \ic 0^{+}}}\,\dd x
\\[5mm] = &\
\Re\int_{-\infty}^{\infty}{\cos\pars{3x} \over
\pars{x + \root{25 + \ic 0^{+}}}\pars{x - \root{25 + \ic 0^{+}}}}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\Re\int_{-\infty}^{\infty}{\expo{3\ic x} \over
\pars{x + 5 + \ic 0^{+}}\pars{x - 5 - \ic 0^{+}}}\,\dd x +
{1 \over 2}\,\Re\int_{-\infty}^{\infty}{\expo{-3\ic x} \over
\pars{x + 5 + \ic 0^{+}}\pars{x - 5 - \ic 0^{+}}}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\Re\pars{2\pi\ic\,{\expo{3\ic\bracks{5 + \ic 0^{+}}} \over 10 + \ic 0^{+}}} +
{1 \over 2}\,\Re\pars{-2\pi\ic\,{\expo{-3\ic\bracks{-5 - \ic 0^{+}}} \over -10 - \ic 0^{+}}} =
{1 \over 10}\,\pi\,\bracks{\Re\pars{\ic\expo{15\,\ic}} +
\Re\pars{\ic\expo{15\,\ic}}}
\\[5mm] = &\
\bbx{\ds{-\,{1 \over 5}\,\pi\sin\pars{15}}}
\end{align}
A: First note that the integral of interest $\int_{-\infty}^\infty \frac{\cos(3x)}{x^2-25}\,dx$ diverges.  Its Cauchy principal value ("CVP") does exist.  We will now present an approach to evaluate the CVP.

To apply Cauchy's Integral Theorem correctly we have 
$$\begin{align}
\oint_C\frac{e^{i3z}}{z^2-25}\,dz&=\int_{-R}^{-5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{-5+\epsilon}^{5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{5+\epsilon}^{R} \frac{e^{i3x}}{x^2-25}\,dx\\\\
&+\int_{\pi}^{0} \frac{e^{i3(-5+\epsilon e^{i\phi})}}{(-5+\epsilon e^{i\phi})^2-25}\,i\epsilon e^{i\phi}\,d\phi+\int_{\pi}^{0} \frac{e^{i3(5+\epsilon e^{i\phi})}}{(5+\epsilon e^{i\phi})^2-25}\,i\epsilon e^{i\phi}\,d\phi\\\\
&+\int_0^\pi \frac{e^{i3Re^{i\phi}}}{(Re^{i\phi})^2-25}\,iRe^{i\phi}\,d\phi\tag 1\\\\
&=0
\end{align}$$
As $R\to \infty$, it is easy to see that the last integral on the right-hand side of $(1)$ approaches $0$.  And as $\epsilon \to 0$, the fourth and fifth integrals approach $i(\pi/10) e^{-i15}$ and $-i(\pi/10)e^{i15}$, respectively.
Hence, letting $R\to \infty$ an d$\epsilon\to 0$, and taking the real part of $(1)$, we find that  

$$\text{PV}\left(\int_{-\infty}^\infty \frac{\cos(3x)}{x^2-25}\,dx\right)=-\frac{\pi \sin(15)}{5}$$

where
$$\text{PV}\left(\int_{-\infty}^\infty \frac{\cos(3x)}{x^2-25}\,dx\right)=\lim_{R\to \infty}\left(\int_{-R}^{-5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{-5+\epsilon}^{5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{5+\epsilon}^{R} \frac{e^{i3x}}{x^2-25}\,dx\right)$$
