text: Q:
  
  Complex Integration poles real axis
  
  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
  $$
  -\frac{\pi}{b}\sin(ab).
  $$
  
  A:
  
  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...
  
  A:
  
  $\newcommand{\+}{^{\dagger}}%
   \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
   \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
   \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
   \newcommand{\dd}{{\rm d}}%
   \newcommand{\isdiv}{\,\left.\right\vert\,}%
   \newcommand{\ds}[1]{\displaystyle{#1}}%
   \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
   \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
   \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
   \newcommand{\ic}{{\rm i}}%
   \newcommand{\imp}{\Longrightarrow}%
   \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
   \newcommand{\pars}[1]{\left( #1 \right)}%
   \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
   \newcommand{\pp}{{\cal P}}%
   \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
   \newcommand{\sech}{\,{\rm sech}}%
   \newcommand{\sgn}{\,{\rm sgn}}%
   \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
   \newcommand{\ul}[1]{\underline{#1}}%
   \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
   \newcommand{\yy}{\Longleftrightarrow}$
  Assumming the integral meaning is "principal value $\pp$ of it":
  $$
  \pp\int_{-\infty}^{\infty}{\expo{\ic ax} \over x^{2} - b^{2}}\,\dd x
  =
  \int_{-\infty}^{\infty}\expo{\ic ax}\bracks{%
  {1 \over x^{2} - b^{2} + \ic 0^{+}} + \ic\pi\delta\pars{x^{2} - b^{2}}}\,\dd x
  \tag{1}
  $$
  
  \begin{align}
  &{1 \over x^{2} - b^{2} + \ic 0^{+}}
  = {1 \over \pars{x - \root{b^{2} - \ic 0^{+}}}\pars{x + \root{b^{2} - \ic 0^{+}}}}
  \\[3mm]&=
  {1 \over \bracks{x - \pars{b - \ic 0^{+}}}\bracks{x - \pars{-b + \ic 0^{+}}}}
  \quad
  \mbox{since}\ b > 0.\tag{2}
  \\[3mm]
  &\delta\pars{x^{2} - b^{2}} = {\delta\pars{x + b} + \delta\pars{x - b} \over 2b} 
  \quad
  \mbox{since}\ b > 0.\tag{3}
  \end{align}
  Since $a > 0$, we choose a "${\tt\mbox{half moon}}$" contour in the upper complex half plane:
  
  We replace $\pars{2}$ and $\pars{3}$ in expression $\pars{1}$:
  \begin{align}
  &\color{#0000ff}{\large%
  \pp\int_{-\infty}^{\infty}{\expo{\ic ax} \over x^{2} - b^{2}}\,\dd x}
  =
  2\pi\ic\,{\expo{\ic a\pars{-b + \ic 0^{+}}}
            \over
            \pars{-b + \ic 0^{+}} - \pars{b - \ic 0^{+}}}
  +
  {\ic \pi \over 2b}\,\bracks{\expo{\ic a\pars{-b}} + \expo{\ic a\pars{b}}}
  \\[3mm]&=
  -\,{\pi\ic\expo{-\ic ab} \over b} + {\ic \pi \over 2b}2\cos\pars{ab}
  =
  \bracks{-\,{\pi\ic \over b}\,\cos\pars{ab} - {\pi \over b}\,\sin\pars{ab}}
  +
  {\pi\ic \over b}\,\cos\pars{ab}
  \\[3mm]&=
  \color{#0000ff}{\large -\,{\pi \over b}\,\sin\pars{ab}}
  \end{align}
  
  A:
  
  A related problem. Here is how you advance
  i) Using partial fraction gives
  
  $$ \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = \frac{1}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iax} }{ \left( x-b \right) }}dx-\frac{1}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iax} }{ \left( x+b
   \right) }}dx.$$
  
  ii) Making change of variables $x-b=t$ and $x+b=t$ in the two integrals on the RHS yields
  
  $$ \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = \frac{e^{iab}}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iat} }{  t  }}dt - \frac{e^{-iab}}{2b}\,\int_{-\infty}^{\infty}{\frac {e^{iat} }{  t  }}dt=\\
  \frac{i\sin(ab)}{b}\int_{-\infty}^{\infty}{\frac {e^{iat} }{  t  }}dt. $$
  
  iii) the integral
  
  $$\int_{-\infty}^{\infty}{\frac {e^{iat} }{  t  }}dt =-i\pi$$
  
  is well known and exists in the Cauchy principal value sense. Check Fourier transform too. 
  iv) To evaluate the last integral using residue theorem, consider the complex integral
  $$ \int_C \frac{e^{iaz}}{z}dz, $$
  where $C$ is the contour (for $a>0$)
  
  