It's been a while since I used residue theorem to evaluate anything. I remember that whenever we have a real valued function, we can use residue theorem to evaluate its integral with associated complex integral etc.
Anyway, I'd have to evaluate the following, already complex-valued intergal: $$\int \limits_{-\infty}^\infty \dfrac{\mathrm{e}^{-ixt}}{a^2 - x^2 - ibx}\mathrm{d}x$$
Here $a$, $b$, $t$ and $x$ are real. Edit: actually, also $t>0$ and $b>0$. Ok, I can find the poles of the function and evaluate the residues around them. I've been given a hint to use the residue theorem, but how do I apply it here? I think I could apply it if I was able to visualise the given integration path, "complete it" with another path (so that this path wouldn't contribute in the end).
But how to choose it? I have trouble visualising the given integration path, what does it look like?
Thanks a lot in advance!
Because $t \gt 0$, the contour needs to be a semicircle of radius $R$ in the lower half-plane - not in the upper half plane. That is, consider the contour integral
$$\oint_C dz \frac{e^{-i z t}}{a^2-z^2-i b z} $$
where $C$ is the above-described semicircle in the lower half-plane, positively oriented. Note that the poles of the integrand are both contained within $C$:
$$z_{\pm} = -i \frac{b}{2} \pm i \frac12 \sqrt{b^2-4 a^2}$$
The contour integral, positively oriented (i.e., traversed counterclockwise), is equal to
$$-\int_{-R}^R dx \frac{e^{-i x t}}{-x^2-i b x+a^2} + i R \int_{-\pi}^0 d\theta \, e^{i \theta} \frac{e^{-i t R e^{i \theta}}}{a^2-R^2 e^{i 2 \theta}-i b R e^{i \theta}} $$
We can show that the second integral vanishes as $R\to\infty$ by noting that the magnitude of the integral is bounded by
$$\frac1{R} \int_{-\pi}^0 d\theta \, e^{t R \sin{\theta}} \le \frac{2}{R} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{t R^2}$$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles within the contour $C$. As both poles are contained within $C$, we have that
$$\int_{-\infty}^{\infty} dx \frac{e^{-i x t}}{-x^2-i b x+a^2} = \begin{cases}\displaystyle \frac{2 \pi}{\sqrt{b^2-4 a^2}} e^{-b t/2} \sin{\left (\sqrt{b^2-4 a^2}\, t \right )} & b \gt 2 a\\ \displaystyle\frac{2 \pi}{\sqrt{4 a^2-b^2}} e^{-b t/2} \sinh{\left (\sqrt{4 a^2-b^2}\, t \right )} & b \lt 2 a \\ 2 \pi t \, e^{-b t/2} & b=2 a\end{cases} $$