Polynomial integral calculation, maybe residue theorem When evaluating a loop correction for a certain field theory, I arrived at the following integral
$$\int\limits_{\mathbb{R}^{1,3}}\frac{d^4 k}{(2\pi)^4}\frac{1}{\left((k^0)^2-\dfrac{|k|^4}{M^2}-m^2+i\epsilon\right)^2}$$ where $d^4 k = dk^0\,d^3k$ and the $i\epsilon$ prescription is for the definition of the integration path for the propagator. I'm a bit rusty when it comes to evaluating integrals like this and my first thought was to use the residue theorem but I don't really know where to start since one pole is on the upper half plane and the other on the lower half plane!
If there's even some better way of doing it beside using the residue theorem, I'll gladly have a look at them!
Edit
I intended to use the residue theorem first on $k^0$ and then, if necessary, on $k$. I thought even about Wick rotating to simplify the problem but still I don't really know where to start.
 A: For $\varepsilon > 0$ define $\phi_\varepsilon \colon (0,\infty) \to \mathbb{C}$ by
$$ \phi_\varepsilon (\mu) = \int \limits_\mathbb{R} \frac{\mathrm{d} \kappa}{2 \pi} \, \frac{1}{(\kappa^2 - \mu^2 + \mathrm{i} \varepsilon)^2} = \int \limits_\mathbb{R} \frac{\mathrm{d} \kappa}{2 \pi} \, \frac{1}{(\kappa - \sqrt{\mu^2 - \mathrm{i} \varepsilon})^2 (\kappa + \sqrt{\mu^2 - \mathrm{i} \varepsilon})^2}$$
using the principal branch of the square root (and all other power functions). Indeed, the integrand has one second-order pole in the upper and one in the lower half-plane. In order to apply the residue theorem we need to close the integration contour in either half-plane using a semi-circle of infinite radius. In this case both choices work, since the integrand decays fast enough in any direction to make the additional integrals vanish. We choose the upper path and use the formula for the residue at second order poles to obtain
$$\phi_\varepsilon (\mu) = 2 \pi \mathrm{i} \lim_{\kappa \to - \sqrt{\mu^2 - \mathrm{i} \varepsilon}} \frac{\mathrm{d}}{\mathrm{d} \kappa} \frac{1}{2 \pi (\kappa - \sqrt{\mu^2 - \mathrm{i} \varepsilon})^2} = \frac{- 2 \mathrm{i}}{(-2\sqrt{\mu^2 - \mathrm{i} \varepsilon})^3} = \frac{\mathrm{i}}{4 (\mu^2 - \mathrm{i} \varepsilon)^{3/2}} \stackrel{\varepsilon \to 0^+}{\longrightarrow} \frac{\mathrm{i}}{4 \mu^3}$$
for $\mu > 0$. Note that the integral can also be computed using partial fractions.
Your integral is
$$ I(m,M) \equiv \lim_{\varepsilon \to 0^+} \int \limits_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \phi_\varepsilon \left(\sqrt{m^2 + \frac{\lvert k \rvert^4}{M^2}}\right) \, .$$
The dominated convergence theorem allows us to move the limit inside the integral and introducing spherical coordinates we find
\begin{align}
I(m, M) &= \frac{\mathrm{i}}{8\pi^2} \int \limits_0^\infty \mathrm{d} k \, \frac{k^2}{\left(m^2 + \frac{k^4}{M^2}\right)^{3/2}} \stackrel{k = \sqrt{m M} u}{=} \frac{\mathrm{i}}{8 \pi^2} \left(\frac{M}{m}\right)^{3/2} \int \limits_0^\infty \mathrm{d} u \, \frac{u^2}{(1+u^4)^{3/2}} \\
&\!\!\!\!\!\!\stackrel{1 + u^4 = \frac{1}{t}}{=} \frac{\mathrm{i}}{32 \pi^2} \left(\frac{M}{m}\right)^{3/2} \int \limits_0^1 \mathrm{d} t \, [t(1-t)]^{-1/4} = \frac{\mathrm{i}}{32 \pi^2} \left(\frac{M}{m}\right)^{3/2} \operatorname{B} \left(\frac{3}{4},\frac{3}{4}\right) \\
&= \frac{\mathrm{i} \operatorname{\Gamma}^2 \left(\frac{3}{4}\right)}{16 \pi^{5/2}} \left(\frac{M}{m}\right)^{3/2} = \frac{\mathrm{i}}{8 \sqrt{\pi} \operatorname{\Gamma}^2 \left(\frac{1}{4}\right)} \left(\frac{M}{m}\right)^{3/2} ,
\end{align}
where we have used an integral representation of the beta function, its relation to the gamma function and the gamma reflection formula.
