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

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    $\begingroup$ One may write these Bessel functions as Whittaker functions (i.e. DLMF 10.36 - dlmf.nist.gov/10.39 especially 10.39.6), which leads to confluent hypergeometric equations, which are simply a fancy term to describe equations where the singularities merge. $\endgroup$
    – Asaf
    Nov 10, 2021 at 17:02
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    $\begingroup$ Did you try the trick of reducing the integral to a Gaussian one? Meaning writing $1/(x_1^2+x_2^2+a^2)^s$ in terms of $\int_0^{\infty}\exp(-t(x_1^2+x_2^2+a^2)) t^{s-1} dt$. $\endgroup$ Nov 10, 2021 at 17:45
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    $\begingroup$ Dear @Asaf, yes of course you are right. It is just that the summation I obtained of the K-Bessel functions is kind of messy and I would be happy to have a simpler formula to work with. In fact what one naturally finds is a linear combination of higher derivatives of K-Bessel functions but then using the usual cylinder identities (for cylinder functions) we can rewrite the higher derivatives in terms of $K_{\nu}$ where the order $\nu$ has been shifted. $\endgroup$ Nov 10, 2021 at 20:31
  • $\begingroup$ Dear @Abdelmalek, no I have not tried it. I can give it a try and see if some kind of change of order of integration really helps. $\endgroup$ Nov 10, 2021 at 20:33
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    $\begingroup$ It looks awfully similar to some intertwiner of some principle series, say in the line model (which it is probably is). The $\ell$ factor comes from some derivatives/rising operator. I would start with calculating that for $\ell=0$, which can be tackled with basic Fourier techniques (you literally get FT for some negative power of the norm, maybe convoluted over a ball/sphere, as you have some Bessel function there). The $\ell$ part will come from derivatives/connection formulas (if you don't want to relay on pure representation theory considerations). $\endgroup$
    – Asaf
    Nov 11, 2021 at 15:41

1 Answer 1


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 $Q$.

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    $\begingroup$ There must be a tiny error somewhere as you cannot have $t$ dependence in your final answer right? $\endgroup$
    – Asaf
    Nov 11, 2021 at 15:36
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    $\begingroup$ Thanks @Asaf, I just corrected the obvious typo, there was indeed no $t$ there. $\endgroup$ Nov 11, 2021 at 20:14
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    $\begingroup$ I'm glad my suggestion was useful. I have some doubt about the 2nd equation. Shouldn't it be derivatives with respect to $\zeta_1,\zeta_2$ rather than a multiplicative factor? Also, what would be the $SL_n$ analogue of this question? $\endgroup$ Nov 12, 2021 at 13:40
  • $\begingroup$ @AbdelmalekAbdesselam the $\mathrm{SL}_n$ analogue is about identities for Whittaker functions of $\mathrm{GL}_n(\mathbb{C})$, which are complicated in general, since these types of functions don't necessary have nice closed formulae. However there do exist recursive formulae for certain nice Whittaker functions; see e.g. my paper on archimedean newform theory for $\mathrm{GL}_n$. $\endgroup$ Nov 12, 2021 at 16:10
  • $\begingroup$ @PeterHumphries: Hi Peter! I see lots integrals in your article with $|{\rm det}\ g|^{s+\ldots}$ factors. Does that generalize to "flag-like" factors which include all minors coming down from the top left corner, a bit like in my MO answer mathoverflow.net/questions/393706/… (towards the end). $\endgroup$ Nov 12, 2021 at 17:37

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