The following integral appears naturally within the computation of the Fourier series coefficients of a real analytic $\mathrm{SL}\_2(\mathbb{C})$-Eisenstein series:
\begin{align\*}
\int\_{-\infty}^{\infty}\left(\int\_{-\infty}^{\infty} \frac{(x\_1+ i x\_2)^{\ell}}{(x\_1^2+x\_2^2+a^2)^s} e^{ i x\_1\zeta\_1}dx\_1 \right) e^{ i x\_2\zeta\_2} 
dx\_2,
\end{align\*}
where $i=\sqrt{-1}$, $\zeta\_1,\zeta\_2,a\in\mathbb{R}$, $\ell\in\mathbb{Z}\_{\geq 0}$ and $s\in\mathbb{C}$ with $\Re(2s-\ell-1)>1$. Using tables of one dimensional integrals, it is possible to write down this double integral as a "linear combination" of $K$-Bessel functions $K\_{\nu}\left(a\sqrt{\zeta\_1^2+\zeta\_2^2}\right)$ for various order $\nu$ which is very similar to some of the formulas which appear in Friedberg's paper [1], see below.


Here is ma question:


**Q:** Is it possible to express this double integral in a very concise way using an appropriate hypergeometric function (e.g. a Meijer G-function with the appropriate parameters or something alike) ?



> 
> [1] S. Friedberg, *On Maass wave forms and the imaginary quadratic Doi-Naganuma lifting*. Math. Ann. 263 (1983), no. 4, 483–508
> 
> 
> 



On a certain double integral appearing in the Fourier series coefficients of $\mathrm{SL}_2(\mathbb{C})$-Eisenstein series
Following @Abdelmalek's great advice in the comments above it is enough to compute the following triple integral:
\begin{align\*}
I=\frac{\pi^s}{\Gamma(s)}\int\_{0}^{\infty}\left(\int\_{-\infty}^{\infty}\int\_{-\infty}^{\infty} 
(x\_1+x\_2 i)^{\ell}\cdot e^{-\pi t(x\_1^2+x\_2^2+a^2)} e^{2\pi i (x\_1\zeta\_1+x\_2\zeta\_2)} dx\_1 dx\_2 \right) 
t^s\frac{dt}{t}.
\end{align\*}
So we have
\begin{align\*}
I&=\frac{\pi^s}{\Gamma(s)}\int\_{0}^{\infty}t^{-1}\cdot\left(\frac{i}{t}(\zeta\_1+i\zeta\_2)\right)^{\ell}\cdot 
e^{-\frac{\pi}{t}(\zeta\_1^2+\zeta\_2^2)}\cdot e^{-\pi ta^2}t^{s}\frac{dt}{t}\\
&=\frac{(i)^{\ell}\pi^s}{\Gamma(s)}(\zeta\_1+i\zeta\_2)^{\ell}\int\_{0}^{\infty}
e^{-\frac{\pi}{t}(\zeta\_1^2+\zeta\_2^2)-\pi ta^2}\cdot t^{s-\ell-1} \frac{dt}{t} \\
&=\frac{(i)^{\ell}\pi^s}{\Gamma(s)}(\zeta\_1+i\zeta\_2)^{\ell}\cdot 2\left(\frac{|\zeta|}{a}\right)^{s-\ell-1}
K\_{s-\ell-1}\left(2\pi|\zeta|a\right)
\end{align\*}
where $\zeta=\zeta\_1+i\zeta\_2$.


So modulo some minor mistakes made above, out of excitement, it means that the initial messy formulas that I had obtained could be vastly simplified, as what I was hoping for. In any case this new approach is just better and simpler!


I can't wait for somebody to publish a book with tables of double integrals (or even multiple integrals)!


**added** To answer partly to @Abdelmalek's comment below, for the second equality, I'm using the fact that
\begin{align\*}
\widehat{P(x)e^{-\pi t\langle x,x\rangle}}(y)=t^{-n/2}\cdot P(\frac{i}{t}y)e^{-\frac{\pi}{t}\langle y,y\rangle}
\end{align\*}
where $x\in\mathbb{R}^n$ is a length $n$ real vector, $\langle,\rangle$ is the usual inner product on $\mathbb{R}^n$, $P(x)$ is a spherical polynomial with respect to the standard Laplacian and $\widehat{}$ corresponds to the Fourier transform. Of course, using a change a of variable we may obtain a more general formula which applies to any inner product $\langle,\rangle\_Q$ associated to any positive definite quadratic form $Q$.


