Countour Integration I had to evaluate this integral $ \int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 - b^2}dx  $, according to Wolfram this had the following result $$ \int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 - b^2}  dx= \pi e^{ib}$$
However, when I integrated it:
$$\oint_C f(z) dz = \oint_{\gamma_{1}} f(z) dz + \oint_{\gamma_{2}} f(z) dz + \oint_{\Gamma} f(z) dz $$
The last term goes to zero applying Jordan's Lemma and to get the result of the integral over $\gamma_{1}$ and $\gamma_{2}$, in which $\gamma_{1}$ is the contour over the first pole and $\gamma_{2}$ is the contour over the second pole.
So, I get  $$\oint_{\gamma_{1}} f(z) dz = i \pi  \lim_{z \rightarrow -b} \frac{z \sin(z)}{(z-b)(z+b)}(z+b) =  i \pi \frac{\sin(-b)}{-2} $$
and
$$\oint_{\gamma_{2}} f(z) dz = i \pi  \lim_{z \rightarrow b} \frac{z \sin(z)}{(z-b)(z+b)}(z-b) =  i \pi \frac{\sin(b)}{2} $$
The result that I'm getting is:
$$\oint_C f(z) dz = i \pi \left( \frac{\sin(b)}{2} - \frac{\sin(-b)}{-2} \right) = \frac{\pi}{2} \left( e^{ib} - e^{-ib} \right)$$
That according to Wolfram is not the correct result.
What am I doing wrong?
Edit:
In this case, I used $$f(z) = \frac{z \sin(z)}{z^2 - b^2}$$
Edit 2:

Edit 3:
Using @Ted Shifrin 's suggestion to change $f(z)$ to a more well-behaved function when $|z|$ is very large, I got:
$f(z) = \frac{z \exp(iz)}{z^2 - b^2}$
So, I get  $$\oint_{\gamma_{1}} f(z) dz = i \pi  \lim_{z \rightarrow -b} \frac{z \exp(iz)}{(z-b)(z+b)}(z+b) =  i \pi \frac{\exp(-ib)}{2} $$
and
$$\oint_{\gamma_{2}} f(z) dz = i \pi  \lim_{z \rightarrow b} \frac{z \exp(iz)}{(z-b)(z+b)}(z-b) =  i \pi \frac{\exp(ib)}{2} $$
The result that I'm getting is:
$$\oint_C f(z) dz = i \pi \left( \frac{\exp(ib)}{2} + \frac{\exp(-ib)}{2} \right) = i \pi \cos(b)$$
and in conclusion:
$$\int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 - b^2}dx = \Im(i \pi cos(b)) = \pi \cos(b) $$
I still don't understand, according to some of you this integral doesn't converge how do I prove that?
This result is equal to @FelixMarin 's result and he used another method of integration, but is different of Wolfram's result and assuming that $b \in \Re $ this integral is supposed to diverge. What am I missing?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\left.\mrm{P.V.}\int_{-\infty}^{\infty}
{x\sin\pars{x} \over x^{2} - b^{2}}\,\dd x
\,\right\vert_{\,b\ \in\ \mathbb{R}}}
\\[5mm] = &\
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\sin\pars{x} \over x + b}\,\dd x +
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\sin\pars{x} \over x - b}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\sin\pars{x - b} + \sin\pars{x + b} \over x}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{2\sin\pars{x}\cos\pars{b} \over x}\,\dd x =
\bbx{\pi\cos\pars{b}} \\ &
\end{align}
