Residue Theorem to solve $\int_{x=-\infty}^\infty \frac{sin(x \, l)}{x \, (x^2+a^2) \, (x^2+b^2)} \, e^{\, \mathrm{j} \,x\,c} \, \mathrm{d}x$ I would like to solve the following integral using residue theorem
$\int_{x=-\infty}^\infty \frac{sin(x \, l)}{x \, (x^2+a^2) \, (x^2+b^2)} \, e^{\, \mathrm{j} \,x\,c} \, \mathrm{d}x$
with $a^2 \in\mathbb{C}$, $\,b^2 \in\mathbb{C}$, $\,l \in \mathbb{R}, \, l>0$ and $\,c \in \mathbb{R}, \, c>0$.
I have troubles with the point $x=0$, since the residue becomes $0$ because $\mathrm{sin}(0)=0$. Can I simply ignore the removable discontinuity and use the Cauchy theorem to solve it? I will have two poles with positive imaginary part.
The original problem is a bit more complicated:
$\int_{x=-\infty}^\infty \int_{y=-\infty}^\infty \frac{sin(x \, l)}{x \, ((x^2+y^2)^2 - k^4)} \, e^{\, \mathrm{j} \,x\,c} \, e^{\, \mathrm{j} \,y\,d} \, \mathrm{d}x \, \mathrm{d}x$
with $k^4 \in\mathbb{C}$, $\,l \in \mathbb{R}, \, l>0$, $\,c \in \mathbb{R}, \, c>0$ and $\,d \in \mathbb{R}, \, d>0$.
I try to integrate first with respect to $x$ and then with respect to $y$ (or the other way around).
 A: When you have a singularity/branch cut at a point on your contour, you go around it and the limit $lim_{\epsilon\to 0+}\int^{\infty}_{\epsilon}f(x) dx+\int^{-\epsilon}_{-\infty}f(x) dx$ is called Cauchy's Principal Value. When you go around 0 you have to use a lemma from Complex Analysis
Lemma: Let $z_0$ be a pole of $f$ and let $\gamma:[\alpha,\beta]\to \mathbb{C}$ be the curve $\gamma(t)=z_0+re^{it}.$ Then,
$lim_{r\to 0}\int_{\gamma}{f(z)dz}=(\beta-\alpha)i Res[f,z=z_0]$
This gives us the usual Residue theorem when we take $\beta=2\pi$ and $\alpha=0$. In our case, it would be $\pi c_{-1}$ if you go above the pole or if you go below the pole it would be $-\pi c_{-1}$ but this time you get  contribution of $2\pi c_{-1}$ as the pole is included in the domain, thus yielding the same answer either way. 
Also are you sure about the conditions on $a,b$ or $k$? They may again be on the real axis which would then produce more detours.
When you do the actual contour integration your integrand should be $\frac{e^{izl}e^{izc}}{z(z^2+a^2)(z^2+b^2)}$ as that would allow you to use Jordan's Lemma by closing in the upper half plane. (Note the condition $l>0, c>0$ and that $\frac{1}{z(z^2+a^2)(z^2+b^2)}$ decays sufficiently fast. )
Now you then have to take the complex part of your answer which would yield the integral for $\frac{sin(xl)e^{ixc}}{x(x^2+a^2)(x^2+b^2)}$
