I need to evaluate the following multidimensional integral:
$$\int_{-\infty}^{\infty}dxdpdwdz\; e^{x^2+p^2-aw^2-az^2-ib(xz-pw)} \ln(2(p+z)^2+2(x+w)^2-1)$$
where $a$ and $b$ are known numerical parameters and can be plugged in. Since the integrand is even, I can rewrite:
$$2 \int_{0}^{\infty}dxdpdwdz\; e^{x^2+p^2-aw^2-az^2-ib(xz-pw)} \ln(2(p+z)^2+2(x+w)^2-1)$$
The integrant is highly oscillatory. I have never dealt with this kind of problems before. I was wondering if anyone can advise me on a direction. Maybe, there is a shortcut or some hammer-method to approach this problem. If it is hopeless, please, also let me know.
P.S.: I know that the result is supposed to be real. Also, when I use (probably, wrongly) a MonteCarlo-type integration, I get zero if the integration interval is sufficiently large ~[0, 60]. However, I know that this method is not supposed to work here...
I will assume that $\frac{b^2}{4} > a > 1$ and interpret the expression as the following conditonally converget iterated integral:
$$ I := \int_{\mathbb{R}^2} \mathrm{d}x_1\mathrm{d}p_1 \int_{\mathbb{R}^2} \mathrm{d}x_2\mathrm{d}p_2 \, e^{x_1^2+p_1^2-a(x_2^2+p_2^2) - ib(x_1p_2 - x_2p_1)} \log\left(2(x_1+x_2)^2+2(p_1+p_2)^2-1\right), $$
where $\log$ is the principal complex logarithm. Then substituting $(x_2,p_2) \mapsto (x_2-x_1, p_2-p_1)$,
$$ I = \int_{\mathbb{R}^2} \mathrm{d}x_1\mathrm{d}p_1 \int_{\mathbb{R}^2} \mathrm{d}x_2\mathrm{d}p_2 \, e^{-(a-1)(x_1^2+p_1^2)-a(x_2^2+p_2^2) + (2ax_1 + ibp_1)x_2 + (2ap_1 - ibx_1)p_2} \log\left(2(x_2^2+p_2^2)-1\right). $$
The inner integral converges if $a > 0$, which of course follows from the assumption. Now by applying the polar coordinate change $(x_2,p_2)=(r_2\cos\theta_2,r_2\sin\theta_2)$ to the inner integral, we get
\begin{align*} I &= \int_{\mathbb{R}^2} \mathrm{d}x_1\mathrm{d}p_1 \int_{0}^{\infty} \mathrm{d}r_2\int_{0}^{2\pi}\mathrm{d}\theta_2 \, r_2 e^{-(a-1)(x_1^2+p_1^2)-ar^2 + (2ax_1 + ibp_1)r_2\cos\theta_2 + (2ap_1 - ibx_1)r_2\sin\theta_2} \log\left(2r_2^2-1\right) \\ &= 2\pi \int_{\mathbb{R}^2} \mathrm{d}x_1\mathrm{d}p_1 \int_{0}^{\infty} \mathrm{d}r_2 \, r_2 e^{-(a-1)(x_1^2+p_1^2)-ar^2} I_0\left(r_2\sqrt{(4a^2-b^2)(x_1^2 + p_1^2)}\right) \log\left(2r_2^2-1\right), \end{align*}
where the integral with respect to $\theta_2$ is simplified using the modified Bessel function of the first kind and order 0. (See DLMF.10.31.1 as well.)
Now it becomes clear that the outer integral converges when $a > 1$ as assumed in the beginning. Then applying the polar coordinate change $(x_1,p_1)=(r_1\cos\theta_1,r_1\sin\theta_1)$ and switching the order of integration, the well-known integral formula (see DLMF.10.43.23, for instance) gives
\begin{align*} I &= (2\pi)^2 \int_{0}^{\infty} \mathrm{d}r_2 \int_{0}^{\infty} \mathrm{d}r_1 \, r_1 r_2 e^{-(a-1)r_1^2-ar^2} I_0\left(r_1r_2\sqrt{4a^2-b^2}\right) \log\left(2r_2^2-1\right) \\ &= \frac{2\pi^2}{a-1} \int_{0}^{\infty} \mathrm{d}r_2 \, r_2 e^{-\frac{b^2-4a}{4(a-1)}r_2^2} \log\left(2r_2^2-1\right). \end{align*}
The last step reveals another condition required for the integral $I$ to converge, namely $b^2 > 4a$, which is also implied by the assumption. Then substituting $u=2r_2^2-1$ and letting
$$ s := \frac{b^2-4a}{8(a-1)}, $$
the integral simplifies by
\begin{align*} I &= \frac{\pi^2}{2(a-1)} \int_{-1}^{\infty} \mathrm{d}u \, e^{-s(u+1)} \log u \\ &= \frac{\pi^2}{2(a-1)} e^{-s} \left( \int_{-1}^{\infty} \mathrm{d}u \, e^{-su} \log \left| u \right| + i\pi \int_{-1}^{0} \mathrm{d}u \, e^{-su} \right). \end{align*}
Finally, by using the integration formula
$$ \int_{-1}^{\infty} \mathrm{d}x \, e^{-sx}\log \left|x\right| = -\frac{\operatorname{Ei}(s)}{s}, \qquad s > 0, $$
where $\operatorname{Ei}(s)$ is the exponential integral, we obtain:
$$ I = \frac{4\pi^2}{b^2-4a} e^{-s} \left( - \operatorname{Ei}\left(s\right) + i\pi \left( e^s - 1 \right) \right). $$