Principal value of $\int_{-\infty}^{+\infty}\dfrac{e^{ix}dx}{x(a^2-x^2)}$ Please can you give me a hint on the following exercise?
Is the contour $[-R,R]\cup C_R$, where $C_R$ is the semicircle centered at the origin of radio $R$ passing through $(0,-R)$ with a clockwise orientation, appropriate to calculate the PV of
$$I=\displaystyle\int_{-\infty}^{+\infty}\dfrac{e^{ix}dx}{x(a^2-x^2)},\qquad 0<a$$?
If so, find that integral using that contour. Otherwise, find another contour and calculate the integral justifying correctly.  
Writing $$\dfrac{e^{ix}}{x(a^2-x^2)}=\dfrac{\cos x}{x(a^2-x^2)}+i \dfrac{\sin x}{x(a^2-x^2)},$$
my idea is to try to find these integrals. 
$$\displaystyle\int_{-\infty}^{+\infty}\dfrac{\cos x dx}{x(a^2-x^2)}, \quad,\displaystyle\int_{-\infty}^{+\infty}\dfrac{\sin x dx}{x(a^2-x^2)}$$
Is there another way to do it?
Thank you.
 A: 
I worked on this problem using a 'simple' approach. This was too large for a comment.

Well, we have the following integral:
$$\mathcal{I}_\text{n}\left(\text{k},\beta\right):=\int_{-\text{n}}^\text{n}\frac{\exp\left(\text{k}x\right)}{x\left(x^2-\beta^2\right)}\space\text{d}x\tag1$$
Where $\exp\left(\cdot\right)$ is the Exponential function.
Using Partial fractions, we have:
$$\frac{\exp\left(\text{k}x\right)}{x\left(x^2-\beta^2\right)}=\frac{1}{2\beta^2}\cdot\frac{\exp\left(\text{k}x\right)}{x+\beta}+\frac{1}{2\beta^2}\cdot\frac{\exp\left(\text{k}x\right)}{x-\beta}-\frac{1}{\beta^2}\cdot\frac{\exp\left(\text{k}x\right)}{x}\tag2$$
Using the linearity of the integral, we can write:
$$\mathcal{I}_\text{n}\left(\text{k},\beta\right)=$$
$$\frac{1}{2\beta^2}\underbrace{\int_{-\text{n}}^\text{n}\frac{\exp\left(\text{k}x\right)}{x+\beta}\space\text{d}x}_{=\space\text{I}_1}+\frac{1}{2\beta^2}\underbrace{\int_{-\text{n}}^\text{n}\frac{\exp\left(\text{k}x\right)}{x-\beta}\space\text{d}x}_{=\space\text{I}_2}-\frac{1}{\beta^2}\underbrace{\int_{-\text{n}}^\text{n}\frac{\exp\left(\text{k}x\right)}{x}\space\text{d}x}_{=\space\text{I}_3}\tag3$$
Let $\text{u}=\text{k}x$, so we get:
$$\text{I}_3=\int_{-\text{kn}}^{\text{kn}}\frac{\exp\left(\text{u}\right)}{\text{u}}\space\text{du}=\left[\text{Ei}\left(\text{u}\right)\right]_{-\text{kn}}^{\text{kn}}=\text{Ei}\left(\text{kn}\right)-\text{Ei}\left(-\text{kn}\right)\tag4$$
Where $\text{Ei}\left(\cdot\right)$ is the Exponential integral.
Let $\text{s}=\text{k}\left(x-\beta\right)$, so we get:
$$\text{I}_2=\exp\left(\beta\text{k}\right)\int_{\text{k}\left(\beta-\text{n}\right)}^{\text{k}\left(\text{n}-\beta\right)}\frac{\exp\left(\text{s}\right)}{\text{s}}\space\text{ds}=\exp\left(\beta\text{k}\right)\left[\text{Ei}\left(\text{s}\right)\right]_{\text{k}\left(\beta-\text{n}\right)}^{\text{k}\left(\text{n}-\beta\right)}=$$
$$\exp\left(\beta\text{k}\right)\left(\text{Ei}\left(\text{k}\left(\text{n}-\beta\right)\right)-\text{Ei}\left(\text{k}\left(\beta-\text{n}\right)\right)\right)\tag5$$
Let $\text{p}=\text{k}\left(x+\beta\right)$, so we get:
$$\text{I}_1=\exp\left(-\beta\text{k}\right)\int_{-\text{k}\left(x+\beta\right)}^{\text{k}\left(\text{n}+\beta\right)}\frac{\exp\left(\text{s}\right)}{\text{s}}\space\text{ds}=\exp\left(-\beta\text{k}\right)\left[\text{Ei}\left(\text{s}\right)\right]_{-\text{k}\left(x+\beta\right)}^{\text{k}\left(\text{n}+\beta\right)}=$$
$$\exp\left(-\beta\text{k}\right)\left(\text{Ei}\left(\text{k}\left(\text{n}+\beta\right)\right)-\text{Ei}\left(-\text{k}\left(x+\beta\right)\right)\right)\tag6$$

Now, it is not hard to show that when $\text{k}=i$ (where $i^2=-1$), we have:
$$\lim_{\text{n}\to\infty}\text{I}_3=\lim_{\text{n}\to\infty}\left(\text{Ei}\left(\text{n}i\right)-\text{Ei}\left(-\text{n}i\right)\right)=2\pi i\tag7$$

