I am asked to evaluate, principal value of
$$\int_{-\infty}^\infty\frac{\cos(x)}{a^2-x^2} \, dx=\pi \frac{\sin (a)}{a},a>0$$
If we start from $$\oint\limits_{C}\frac{e^{iz}}{a^2-z^2}dz,a>0$$ the line $C$ is composed of the half circle $\Gamma$, pole circles at $-a,a, \gamma_1,\gamma_2$ whose circumferences are ($r,r_1,r_2$), and a portion of the $x$-axis. If we use the Cauchy remainder theorem, we get
$$
\begin{split}
\int_0^\pi \frac{e^{ir\cos \theta -r\sin \theta}}
{a^2-r^2e^{2-\theta}}
ire^{i\theta} \, d\theta
&+ \int_{-r}^{-a-r_2} f(x) \, dx
+ J_2 \\
&+ \int_{-a+r_2}^{a-r_1} f(x) \, dx
+ J_1
+ \int_{a+r_1}^r f(x) \, dx
= 0
\end{split}
$$
Since $\left|\int_0^\pi \frac{e^{ir\cos \theta -rsin \theta}}{a^2-r^2 e^{2-\theta}}ire^{i\theta} \, d\theta\right|\leq{\frac{\pi r}{r^2-a^2},(r>a)}$
We get $$\lim_{n \to \infty}\int_0^\pi \frac{e^{ir\cos \theta -r\sin \theta}}{a^2-r^2e^{2-\theta}}ire^{i\theta} \, d\theta=0$$
Evaluating residuum at $J_{1}$ and $J_{2}$ we get $$J_1=\operatorname{Res}f(a)=\lim_{x \to a}(a-x)\frac{e^{ix}}{(a-x)(a+x)} =\frac{e^{ia}}{2a}$$ and $$J_2= \operatorname{Res}f(-a)=\lim_{x \to -a}(a+x)\frac{e^{ix}}{(a-x)(a+x)}=\frac{e^{-ia}}{2a}$$ In my book the author got $J_{1}=\frac{\pi i}{2a}e^{ia}\land J_2=-\frac{\pi i}{2a} e^{-ia}$ Where does the $\pi i$ come from ? also, why – in the second one? Is it because the residuum is at $-a$? Then, adding those two gives us the result, but still, where does $\pi$ come from?
Best Answer
I suspect that the author meant to write $\pi i$ times the residue terms. And the residue at $z=a$ is given by
$$\lim_{z\to a}(z-a)\frac{e^{iz}}{a^2-z^2}=-\frac{e^{ia}}{2a}$$
So, in order to provide support of your analysis, let's start from scratch and evaluate the closed contour integral
$$\begin{align} 0&=\oint_C\frac{e^{iz}}{a^2-z^2}\,dz\\\\ &=\int_{-R}^{-a-r}\frac{e^{ix}}{a^2-x^2}\,dx+\int_\pi^0 \frac{e^{i(-a+re^{i\phi})}}{a^2-(-a+re^{i\phi})^2}\,ire^{i\phi}\,d\phi\\\\ &+\int_{-a+r}^{a-r}\frac{e^{ix}}{a^2-x^2}\,dx+\int_\pi^0 \frac{e^{i(a+re^{i\phi})}}{a^2-(a+re^{i\phi})^2}\,ire^{i\phi}\,d\phi\\\\ &+\int_{a+r}^R \frac{e^{ix}}{a^2-x^2}\,dx+\int_0^\pi \frac{e^{iRe^{i\phi}}}{a^2-(Re^{i\phi})^2}\,iRe^{i\phi}\,d\phi\tag1 \end{align}$$
The last integral on the right-hand side of $(1)$ vanishes as $R\to\infty$. And as $r\to 0^+$, the second and fourth integrals on the right-hand side of $(1)$ approach $-\frac{i\pi e^{-ia}}{2a}$ and $\frac{i\pi e^{ia}}{2a}$, respectively.
We find, therefore, that the Cauchy Principal Value of the integral of interest is
$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{\cos(x)}{a^2-x^2}\,dx\right)&=\lim_{r\to 0^+}\left(\int_{-\infty}^{-a-r}\frac{\sin(x)}{a^2-x^2}\,dx+\int_{-a+r}^{a-r}\frac{\sin(x)}{a^2-x^2}\,dx\\\\ +\int_{a+r}^\infty\frac{\sin(x)}{a^2-x^2}\,dx\right)\\\\ &=\frac{\pi\sin(a)}{a} \end{align}$$
as was to be shown.