2025/07/03
Let $M_{\kappa,\mu}(z)$ and $W_{\kappa,\mu}(z)$ be the $M$- and $W$-Whittaker functions, respectively, as defined at DLMF$\S$13.14. In the theory of Maass-Poincaré series, by constructing \[\varphi_{s,k}(mz):=|4\pi my|^{-\frac k2} M_{\frac k2 \sgn(m),s-\frac 12}(4\pi |m|y)e(mx)\] and summing \[P_m(z):=\sum_{\gamma\in \Gamma_{\infty}\setminus \Gamma} \overline{\nu}(\gamma) (\varphi_{s,k}|_k \gamma)(z), \] we will facing the following integral formula when computing the Fourier expansion of $P_m$: \[ \frac 1{\Gamma(2s)}\int_{-\infty}^\infty e\lp\frac{cu}{1+u^2}+\alpha u \rp M_{\ell,s-\frac 12}\lp \frac{4\pi c}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell du=\] \[\left\{\begin{array}{ll} \frac{2\pi}{\Gamma(s+\ell)} W_{\ell,s-\frac12}(4\pi \alpha) \left|\frac c\alpha\right|^{\frac12} J_{2s-1}(4\pi\sqrt{\alpha c}),& \alpha>0;\\ \frac{2\pi^{1+s}c^s}{(s-\frac12)\Gamma(s-\ell)\Gamma(s+\ell)}, & \alpha=0;\\ \frac{2\pi}{\Gamma(s-\ell)} W_{-\ell,s-\frac12}(4\pi \alpha) \left|\frac c\alpha\right|^{\frac12} I_{2s-1}(4\pi\sqrt{\alpha c}),& \alpha<0. \end{array} \right. \] The equation above is recorded in Proposition 3.6 of [Jeon, Kang, Kim, Weak Maass-Poincaré series and weight $3/2$ mock modular forms, 2013], which actually cite from Lemma 5.5 in [Hejhal, The Selberg Trace Formula for $\mathrm{PSL}(2,\R)$, volume 2, 1983].
Today we revisit Hejhal's proof on this formula with the help of modern table of special functions in DLMF. A kind reminder is that \[\Phi(a,b,z)=M(a,b,z),\quad \Psi(a,b,z)=U(a,b,z), \] \[G(z)=z^s e^{-\frac z2} \Phi(s-\ell,2s,z)\Gamma(2s)^{-1}=M_{\ell,s-\frac 12}(z)\Gamma(2s)^{-1},\] and \[(4\pi |\alpha|)^s e^{-2\pi |\alpha|} \Psi(s\pm \ell,2s,4\pi |\alpha|)= W_{\mp \ell,s-\frac 12}(4\pi |\alpha|), \] where the left-hand-sides are functions in Hejhal's book and the right-hand-sides $M$ and $U$ functions are Kummer functions in DLMF.
Following Hejhal's proof, we first set \[F(c):=\int_{-\infty}^\infty e\lp\frac{cu}{1+u^2}\rp M_{\ell,s-\frac 12}\lp \frac{4\pi c}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell e(\alpha u) du.\] The goal is to find a differential equation of $F$, e.g. $F''(c)+\cdots F(c)=0$. So we need to take the second derivative of $F$. Note that there are two functions including $c$ and we need $(uv)''=u''+v''+2u'v'$.
For simplicity, we denote $M'(u)$ for the derivative $\frac{d}{du} M_{\ell,s-\frac12}(u)$. Since $M_{\ell,s-\frac12}(u)$ satisfies the differential equation (which is its definition), we have \[\frac{d^2}{du^2} M_{\ell,s-\frac12}(u)=\lp\frac 14-\frac \ell u -\frac{s(1-s)}{u^2}\rp M_{\ell,s-\frac12}(u). \]
Now we can compute $F''(c)$. \[F''(c)=\int_{-\infty}^\infty \lp\frac{-4\pi^2 u^2}{(1+u^2)^2}+\frac{4\pi^2}{(1+u^2)^2}-\frac{4\pi \ell}{c(1+u^2)}-\frac{s(1-s)}{c^2}\rp e\lp\frac{cu}{1+u^2}\rp M_{\ell,s-\frac 12}\lp \frac{4\pi c}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell e(\alpha u)du + \int_{-\infty}^\infty \lp\frac{16\pi^2 i u}{(1+u^2)^2}\rp e\lp\frac{cu}{1+u^2}\rp M_{\ell,s-\frac 12}'\lp \frac{4\pi c}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell e(\alpha u)du. \] Such a long equation!
Note that $\frac{1-u^2}{(1+u^2)^2}=\frac{d}{du} \frac{u}{1+u^2}$ and $\frac{d}{du} \lp\frac{1-iu}{1+iu}\rp^\ell=\frac{-2i\ell}{1+u^2} \lp\frac{1-iu}{1+iu}\rp^\ell$ (also true for $\ell\neq 0$). These results give the first integral above: \[F''(c)+\frac{s(1-s)}{c^2} F(c)=\frac{-2\pi i}{c}\int_{-\infty}^\infty \frac{d}{du} \lp e\lp\frac{cu}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell \rp M_{\ell,s-\frac 12}\lp \frac{4\pi c}{1+u^2} \rp e(\alpha u)du + \int_{-\infty}^\infty \lp\frac{16\pi^2 i u}{(1+u^2)^2}\rp e\lp\frac{cu}{1+u^2}\rp M_{\ell,s-\frac 12}'\lp \frac{4\pi c}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell e(\alpha u)du. \]
For the last integral, we have the derivative $\frac{d}{du} M_{\ell,s-\frac 12}\lp \frac{4\pi c}{1+u^2}\rp =\frac{-8\pi cu}{(1+u^2)^2} M_{\ell,s-\frac 12}'\lp \frac{4\pi c}{1+u^2}\rp$. By multiplying $\frac{-2\pi i}c$, it gives exactly the last integral. Therefore, we ended up with \[F''(c)+\frac{s(1-s)}{c^2} F(c)=\frac{-2\pi i}{c}\int_{-\infty}^\infty \frac{d}{du} \lp e\lp\frac{cu}{1+u^2} \rp \lp\frac{1-iu}{1+iu}\rp^\ell M_{\ell,s-\frac 12}\lp \frac{4\pi c}{1+u^2}\rp \rp e(\alpha u)du. \]
Applying the partial integral on the right hand side above (and the integrand has to be $0$ at infinities), we get the key step in Hejhal's book: \[F''(c)+\lp \frac{4\pi^2 \alpha}c+\frac{s(1-s)}{c^2}\rp F(c)=0. \]
The next step in Hejhal's book is by citing G. N. Watson's book A Treatise on the Theory of Bessel Functions in 1944. We shall find this in some modern resource. For example, we cite DLMF(10.13.2) that the solution $w(z)$ for \[w''+\lp \frac{\lambda^2}{4z}-\frac{\nu^2-1}{4z^2}\rp w=0\] is $w(z)=z^{\frac 12} \mathcal{C}_\nu(\lambda z^{\frac 12})$, where $\mathcal{C}_\nu$ stands for any linear combination of $J_\nu$ and $Y_\nu$ (Bessel functions).
We apply the above equation for $\lambda=4\pi\sqrt{\alpha}$ (if $\alpha>0$), $\lambda= i 4\pi\sqrt{|\alpha|} $ (if $\alpha < 0 $), and $\nu=2s-1$ to solve \[F(c)=\left\{ \begin{array}{ll} Ac^{\frac 12} J_{2s-1}(4\pi\sqrt{\alpha c})+Bc^{\frac 12} Y_{2s-1}(4\pi\sqrt{\alpha c}), & \alpha>0;\\ A' c^{\frac 12} I_{2s-1}(4\pi\sqrt{|\alpha| c})+B' c^{\frac 12} K_{2s-1}(4\pi\sqrt{|\alpha| c}), & \alpha < 0; \end{array} \right. \] There are some basic facts about $J,Y,I,K$ Bessel functions: $Y_\nu$ is a linear combination of $J_{\pm\nu}$, $K_\nu$ is a linear combination of $I_{\pm\nu}$, and $I_\nu(x)=e^{-\nu\pi i/2}J_{\nu}(ix)$. So we only need to determine the above constants. Moreover, we will sove the $\alpha=0$ case later.
In order to determine $A,B,A',B'$ above, Hejhal let $c\rightarrow 0^+$. We can study from this post and find that the indicial equation of $F$ is $r(r-1)+s(1-s)=0$, hence the solution is $r_1=s$, $r_2=1-s$.
In the original equation defining $F(c)$, when $c\rightarrow 0^+$, we have by DLMF(13.14.14): \[M_{\ell,s-\frac 12}\lp\frac{4\pi c}{1+u^2}\rp=(1+O(c))\lp\frac{4\pi c}{1+u^2}\rp^s. \] For $\exp(\frac{cu}{1+u^2})$ we just use $1+O(c)$. Then we get \[ \begin{align} F(c)&=(4\pi c)^s \int_{-\infty}^\infty (1+O(c)) (1+u^2)^{-s} \lp\frac{1-iu}{1+iu}\rp^\ell e(\alpha u) du\\ &=(4\pi c)^s \int_{-\infty}^\infty (1-iu)^{-s+\ell} (1+iu)^{-s-\ell} e(\alpha u) du+O(c^{s+1}) \end{align} \] and the integral now depends on $\alpha>0$ or $\alpha< 0$.
For $x\geq 0$ and $\re(\beta)+\re(\gamma)>1$, computing the integral \[\int_{-\infty}^\infty (1-iu)^{-\beta} (1+iu)^{-\gamma} e^{iux} du\] goes back to page 353 of Hejhal's book.
The contour deformation is a little bit hard for me, so we turn to another reference: [Bateman, Harry, & Bateman Manuscript Project. 1954. Tables of Integral Transforms, volume I. New York: McGraw-Hill Book Company]. In page 119, equation 3.2(12) gives that for $\re(\mu+\nu)>\frac 12$, \[ \int_{-\infty}^\infty (1-ix)^{-2\nu} (1+ix)^{-2\mu} e^{-ixy} dx=\left\{ \begin{array}{ll} 2^{1-\nu-\mu}\pi \Gamma(2\nu)^{-1} y^{\nu+\mu-1} W_{\nu-\mu,\frac 12-\nu-\mu}(2y),& y>0;\\ 2^{1-\nu-\mu}\pi \Gamma(2\mu)^{-1} |y|^{\nu+\mu-1} W_{\mu-\nu,\frac12-\nu-\mu}(2|y|),& y< 0. \end{array} \right. \] One may notice that the reference has a $-$ sign when $y>0$. It should be wrong, by testing e.g. $\mu=\nu=\frac12$ and $y=1$ - the result should be $\frac{\pi}{e}$ but not minus.
So we set $\nu=\frac{s-\ell}2$, $\mu=\frac{s+\ell}2$, $\re(\mu+\nu)=\re(s)>\frac 12$, and $y=-2\pi \alpha$ to get that for $c\rightarrow 0^+$, \[F(c)=(4\pi c)^s\left\{ \begin{array}{ll} \frac{\pi^{s} |\alpha|^{s-1} }{\Gamma(s-\ell)} W_{-\ell,s-\frac 12}(4\pi \alpha),& \alpha< 0;\\ \frac{\pi^s \alpha^{s-1} }{\Gamma(s+\ell)} W_{\ell,s-\frac 12}(4\pi \alpha),& \alpha>0 ; \end{array} \right.\quad +O(c^{s+1}). \] We have used the property that for the $W$-Whittaker function, $W_{\alpha,\beta}=W_{\alpha,-\beta}$.
Recall the solution in $J,Y,I,K$ Bessel functions before. By DLMF(10.7) and DLMF(10.30), when $c\rightarrow 0^+$ it gives $c^{\pm(2s-1)}$. Since we have assumed $\re(s)>\re(1-s)$, i.e. $\re(2s-1)>0$, we shall have $B=B'=0$ (otherwise the exponent of $c$ is $1-s$).
Since $J_{2s-1}$ and $I_{2s-1}$ both give $(2\pi)^{2s-1}|c\alpha|^{s-\frac 12}\Gamma(2s)^{-1}$, we have the following solution: \[F(c)=\Gamma(2s)\left\{ \begin{array}{ll} \frac{2\pi}{\Gamma(s-\ell)} \left|\frac c\alpha\right|^{\frac 12} W_{-\ell,s-\frac 12}(4\pi \alpha) I_{2s-1}(4\pi\sqrt{c|\alpha|}),& \alpha< 0;\\ \frac{2\pi }{\Gamma(s+\ell)} \left|\frac c\alpha\right|^{\frac 12} W_{\ell,s-\frac 12}(4\pi \alpha) J_{2s-1}(4\pi\sqrt{c\alpha}),& \alpha>0 . \end{array} \right.\]
The only thing left is for $\alpha =0$. We need to solve \[c^2 F''(c)+s(1-s)F(c)=0. \] This is a second-order Euler-Cauchy equation. One can see this Wolfram Mathworld page to solve \[F(c)=Ax^s+Bx^{1-s}. \] Again, the limit argument shows that for $c\rightarrow 0^+$, \[F(c)=(4\pi c)^s \int_{-\infty}^\infty (1-iu)^{-s+\ell} (1+iu)^{-s-\ell} du+O(c^{s+1}). \] By DLMF(5.12.8), we get \[F(c)=(4\pi c)^s \frac{2^{2-2s}\pi \Gamma(2s)}{(2s-1)\Gamma(s-\ell)\Gamma(s+\ell)}+O(c^{s+1}). \] This eliminates the possibility of $Bx^{1-s}$ and the main term is exactly $F(c)$. After simplification we have \[F(c)=\frac{4\pi^{1+s}\Gamma(2s)}{(2s-1)\Gamma(s-\ell)\Gamma(s+\ell)} c^s,\quad \text{if }\alpha=0. \]
We have finished the calculation of $F(c)$.