Complex Analysis – Integration Poles on the Real Axis

Question

In class my professor said that
$$
\int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 – b^2}dx = -\frac{2\pi}{b}\sin(ab)
$$
where $a,b > 0$.
However, since the poles are on the real axis, isn't the integral equal to
$$
\pi i\sum_{\text{real axis}}\text{Res}(f(z); z_j)\mbox{?}
$$
If that is the case, the integral is
$$
-\frac{\pi}{b}\sin(ab).
$$

Answer 1

Define

$$f(z):=\frac{e^{iaz}}{z^2-b^2}\;,\;a,b\in\Bbb R^+\;,$$

$$C_R:=[-R,-b-\epsilon]\cup\gamma_{-b,\epsilon}\cup[-b+\epsilon,b-\epsilon]\cup\gamma_{b\epsilon}\cup[b+\epsilon,R]\cup\Gamma_R$$

with

$$\gamma_{r,s}:=\{r+se^{it}\;;\;0\le t\le \pi\}\;,\;r,s\in\Bbb R^+\;,\;\Gamma_R:=\{Re^{it}\;;\;0\le t\le \pi\}\;,\;\;R\in\Bbb R^+$$

Since $\;f(z)\;$ analytic on and within $\;C_R\;$ , we get

$$\oint\limits_{C_R}f(z)\,dz=0$$

By the corollary to the lemma in the second answer here we get

$$\begin{align*}\lim_{\epsilon\to 0}\int\limits_{\gamma_{-b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=-b}=i\pi\frac{e^{-iab}}{-2b}=-\frac{\pi i}{2b}e^{-iab}\\ \lim_{\epsilon\to 0}\int\limits_{\gamma_{b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=b}=i\pi\frac{e^{iab}}{2b}=\frac{\pi i}{2b}e^{iab}\end{align*}$$

And applying Jordan's Lemma the integral on $\;\Gamma_R\;$ goes to zero when $\;R\to \infty\;$ , so in the end

$$0=\lim_{R\to\infty\,,\,\epsilon\to 0}\oint\limits_{C_R}f(z)dz=\int\limits_{-\infty}^\infty f(x)dx-\frac{\pi i}{2b}\left(e^{iab}-e^{-iab}\right)\implies$$

$$\implies \int\limits_{-\infty}^\infty f(x)dx=-\frac\pi b\frac{e^{iab}-e^{-iab}}{2i}=-\frac\pi b\sin(ab)$$

and I get the same as you did...