Fourier Transform of an odd function and the Contour Integration I have a question regarding the Fourier transform of an odd function. The function is:  
$ \begin{align} 
f(x) = \sin(\eta x)e^{-q|{x}|}, \quad \mathrm{where}\ q,\eta\in\mathbb{R}, q>0.
\end{align}$
So far, I have the following working:
$ \begin{align} 
F(k) &= \int^{\;\infty}_{-\infty} dx\;e^{-ikx}\;\sin({\eta}x)e^{-q|x|} \\
&= \int^{\;0}_{-\infty} dx\;e^{-ikx}\;\sin({\eta}x)e^{qx} + \int^{\infty}_{0} dx\;e^{-ikx}\;\sin({\eta}x)e^{-qx} \\
&= \int^{\;0}_{-\infty} dx\;e^{-ikx}\;\left[\frac{1}{2ix}\left(e^{i\eta x}-e^{-i\eta x}\right)\right]e^{qx} + \int^{\infty}_{0} dx\;e^{-ikx}\;\left[\frac{1}{2ix}\left(e^{i\eta x}-e^{-i\eta x}\right)\right]e^{-qx} \\
&= \frac{1}{2i}\int^{\;0}_{-\infty} dx\;e^{i(-k+\eta)x}\;\frac{e^{qx}}{x}-e^{i(-k-\eta)x}\;\frac{e^{qx}}{x} + 
\frac{1}{2i}\int^{\infty}_{0} dx\;e^{i(-k+\eta)x}\;\frac{e^{-qx}}{x}-e^{i(-k-\eta)x}\;\frac{e^{-qx}}{x} \\
\end{align}$
So my rationale is that because there is a modulus on the $x$, I have to split the domain up. And I get a mess like this. So my question is two-fold:


*

*Is there a simpler method to compute the Fourier transform in this case, keeping in mind that the function I am working with is odd? 

*If not, how would I go about computing the integral? I think I should be using contour integration, by considering the analytic continuation over the complex plane and closing the contour. But down the line, I am wondering if it is possible to use Jordan's Lemma? Since the integration limits are from $0$ to $\infty$ and $-\infty$ to $0$.


Thank you!
 A: The problem is a bit simpler than you illustrate.  Let me write the FT as
$$\begin{align}\int_{-\infty}^{\infty} dx \, \sin{\eta x} \, e^{-q |x|} \, e^{i k x} &= \int_{-\infty}^{\infty} dx \, \sin{\eta x} \, e^{-q |x|} \cos{k x} + i \int_{-\infty}^{\infty} dx \, \sin{\eta x} \, e^{-q |x|}  \sin{k x} \\ &= i \int_{-\infty}^{\infty} dx \, \sin{\eta x} \, \sin{k x} \,  e^{-q |x|} \\ &= \frac{i}{2} \int_{-\infty}^{\infty} dx \, \left [\cos{(\eta-k) x} - \cos{(\eta+k) x} \right ] \, e^{-q |x|} \\ &= i \int_{0}^{\infty} dx \, \left [\cos{(\eta-k) x} - \cos{(\eta+k) x} \right ] \, e^{-q x} \\ &= i \, \operatorname{Re} \int_{0}^{\infty} dx \, \left [e^{i(\eta-k) x} - e^{i(\eta+k) x} \right ] \, e^{-q x} \\ &= i \operatorname{Re} \left [\frac1{q-i(\eta-k)} - \frac1{q-i(\eta+k)}\right ] \end{align} $$
Therefore,

$$ \int_{-\infty}^{\infty} dx \, \sin{\eta x} \, e^{-q |x|} \, e^{i k x} = i 2 q \left [\frac1{q^2+(k-\eta)^2} - \frac1{q^2+(k+\eta)^2} \right ] $$

Note that contour integration would not have helped as there were no poles in the integrand.  Rather, I just exploited the symmetries of the integrand to extract the simple result provided by this integral.
