and for which

$W_{0,0}^{\prime}(\zeta)=\zeta^{-\frac{1}{2}}\big{(}{\mathcal{E}}_{1,1}(\zeta)-\log(\zeta){\mathcal{E}}_{0,2}(\zeta)\big{)}.$

∎

###  A.2. Lower bounds for Whittaker functions

The next lemma shows that there are no zeroes of $M_{+}(x)$, for any
$x\in(0,\infty)$.

######  Lemma A.5.

Let $x>0$. We have the following.

  * •

For $\beta^{2}\leq 1/4$, then $M_{0,\mu}(x)$ is monotone increasing and

$M_{0,\mu}(x)>x^{\frac{1}{2}+\mu},\quad
M\left(\tfrac{1}{2}+\mu,1+2\mu,x\right)\geq e^{\frac{1}{2}x}.$

  * •

For $\beta^{2}>1/4$, then $|M_{0,i\nu}(x)|$ is monotone increasing and

$x|\Gamma(1+i\nu)|^{2}\frac{\sinh(\nu\pi)}{\nu\pi}\leq|M_{0,i\nu}(x)|^{2}\leq
x\cosh(x)|\Gamma(1+i\nu)|^{2}\frac{\sinh(\nu\pi)}{\nu\pi},$

with also

$\left|M\left(\tfrac{1}{2}+i\nu,1+2i\nu,x\right)\right|\geq{\rm
e}^{\frac{1}{2}x}|\Gamma(1+i\nu)|\sqrt{\frac{\sinh(\nu\pi)}{\nu\pi}}.$

###### Proof.

From [NIST], we have

$M_{0,\gamma}\left(2x\right)=2^{2\gamma+\frac{1}{2}}\Gamma\left(1+\gamma\right)\sqrt{x}I_{\gamma}\left(x\right).$

For $\beta^{2}\leq 1/4$, we have $\gamma=\mu$ and the conclusion is
straightforward, since we can use the power series representation of
$I_{\mu}(x)$ to obtain

$M_{0,\mu}(2x)>(2x)^{\frac{1}{2}+\mu}.$

On the other hand, for $\beta^{2}>1/4$, we have $\gamma=i\nu$ and

$M_{0,i\nu}\left(2x\right)=2^{2i\nu+\frac{1}{2}}\Gamma\left(1+i\nu\right)\sqrt{x}I_{i\nu}\left(x\right).$

Therefore,

$\displaystyle|M_{0,i\nu}(2x)|^{2}$
$\displaystyle=2x|\Gamma(1+i\nu)|^{2}I_{i\nu}(x)I_{-i\nu}(x)=2x|\Gamma(1+i\nu)|^{2}\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}I_{0}(2x\cos\theta)\cosh(2\nu\theta)\mathrm{d}\theta$

The upper and lower bound follow from the fact that $1\leq
I_{0}(x)\leq\cosh(x)$, for all $x\geq 0$. See [NIST] for the product formula
for $I_{i\nu}(x)I_{-i\nu}(x)$. ∎

###  A.3. Growth bounds and comparison estimates for $\beta^{2}>1/4$

In this subsection we treat the case $\beta^{2}>1/4$, so that $\mu=0$ and
$\nu=\sqrt{\beta^{2}-1/4}$.

######  Lemma A.6.

Denote $a:=\frac{1}{2}+i\nu$ and $b:=2a$. Then, there exists $C>0$ and
$N_{\nu,0}>0$ such that

${\rm e}^{-\frac{1}{8}\nu\pi}{\rm e}^{\frac{1}{2}{\rm
Re}\zeta}\leq\left|\frac{\Gamma(a)}{\Gamma(b)}M_{+}(\zeta)\right|\leq{\rm
e}^{\frac{1}{8}\nu\pi}{\rm e}^{\frac{1}{2}{\rm Re}\zeta},$

and

$\left|\frac{M_{\pm}^{\prime}(\zeta)}{M_{\pm}(\zeta)}-\frac{1}{2}\right|\leq\frac{1}{4},$

for all ${\rm Re}\zeta\geq N_{\nu,0}$.

###### Proof.

Let $\zeta\in{\mathbb{C}}$. We recall that

$\displaystyle M_{+}(\zeta)$ $\displaystyle={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{a}M(a,b,\zeta)={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{a}\frac{\Gamma(b)}{\Gamma(a)}\left({\rm e}^{-i\pi
a}U(a,b,\zeta)+{\rm e}^{i\pi a}{\rm e}^{\zeta}U(a,b,{\rm
e}^{i\pi}\zeta)\right).$

Moreover, we have that $U(a,b,\zeta)=\zeta^{-a}+\mathcal{E}_{1}(\zeta)$, where
further

$|\zeta^{a}{\mathcal{E}}_{1}(\zeta)|\leq\frac{2\beta^{2}}{|\zeta|}{\rm
e}^{\frac{2\beta^{2}}{|\zeta|}}.$

In the sequel, we write $x:=\frac{2\beta^{2}}{|\zeta|}$. Therefore, we can
write

$M_{+}(\zeta)=\frac{\Gamma(b)}{\Gamma(a)}{\rm
e}^{\frac{1}{2}\zeta}\left(\left[1+({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm e}^{i\pi}\zeta)\right]+{\rm
e}^{-\zeta}{\rm e}^{-i\pi
a}\left[1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right]\right)$ (A.2)

We shall focus on obtaining upper and lower bound estimates for

$\left|\left[1+({\rm e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm
e}^{i\pi}\zeta)\right]+{\rm e}^{-\zeta}{\rm e}^{-i\pi
a}\left[1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right]\right|$

when ${\rm Re}\zeta$ is large. To this end, we note that
$|\zeta^{a}{\mathcal{E}}_{1}(\zeta)|\leq 2x$, for $x\leq\frac{1}{2}$.
Moreover,

$|1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)|\leq{\rm e}^{\nu\frac{\pi}{16}},$

provided that $x\leq\frac{1}{2}\min\left\\{1,{\rm
e}^{\nu\frac{\pi}{16}}-1\right\\}$. Similarly, we also have that

$1+{\rm e}^{-\zeta}{\rm e}^{\nu\pi}\leq{\rm e}^{\nu\frac{\pi}{16}},$

for all ${\rm Re}\zeta>\nu\pi-\log({\rm e}^{\frac{\nu\pi}{16}}-1)$. Hence,

$\left|\left[1+({\rm e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm
e}^{i\pi}\zeta)\right]+{\rm e}^{-\zeta}{\rm e}^{-i\pi
a}\left[1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right]\right|\leq{\rm
e}^{\frac{\nu\pi}{8}}.$

On the other hand, for $x\leq\min\left\\{\frac{1}{4}\left(1-{\rm
e}^{-\frac{1}{8}\nu\pi}\right),\frac{1}{2}\right\\}$, we have that

$|\zeta^{a}{\mathcal{E}}_{1}(\zeta)|\leq\frac{1}{2}\left(1-{\rm
e}^{-\nu\frac{\pi}{8}}\right),$

and also

$\left|{\rm e}^{-\zeta}{\rm
e}^{\nu\pi}\left(1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right)\right|\leq\frac{1}{2}\left(1-{\rm
e}^{-\nu\frac{\pi}{8}}\right),$

provided that ${\rm Re}\zeta>\nu\pi+\log\left(\frac{4}{1-{\rm
e}^{-\frac{1}{8}\nu\pi}}\right)$. Therefore, we can lower bound

$\left|\left[1+({\rm e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm
e}^{i\pi}\zeta)\right]+{\rm e}^{-\zeta}{\rm e}^{-i\pi
a}\left[1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right]\right|\geq{\rm
e}^{-\frac{\nu\pi}{8}}.$

We choose $N_{\nu}>0$ so that all the above conditions are satisfied when
${\rm Re}\zeta\geq N_{\nu}$. For the second part of the lemma, we take a
$\frac{\mathrm{d}}{\mathrm{d}\zeta}$ derivative in (A.2) to obtain

$\displaystyle M_{+}^{\prime}(\zeta)$
$\displaystyle=\frac{1}{2}M_{+}(\zeta)+\frac{\Gamma(b)}{\Gamma(a)}{\rm
e}^{\frac{1}{2}\zeta}\left(\frac{a}{\zeta}({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm e}^{i\pi}\zeta)+{\rm e}^{i\pi}({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}^{\prime}({\rm e}^{i\pi}\zeta)\right)$
$\displaystyle\quad+\frac{\Gamma(b)}{\Gamma(a)}{\rm e}^{-\frac{1}{2}\zeta}{\rm
e}^{i\pi
a}\left(\frac{a}{\zeta}\zeta^{a}{\mathcal{E}}_{1}(\zeta)+\zeta^{a}{\mathcal{E}}_{1}^{\prime}\zeta)-1-\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right).$

Since $|\zeta^{a}{\mathcal{E}}_{1}(\zeta)|\leq\frac{2\beta^{2}}{|\zeta|}{\rm
e}^{\frac{2\beta^{2}}{|\zeta|}}$ and
$|\zeta^{a}{\mathcal{E}}_{1}^{\prime}(\zeta)|\leq\frac{4\beta^{2}}{|\zeta|}{\rm
e}^{\frac{2\beta^{2}}{|\zeta|}}$, confer [NIST], we find that

$\left|\frac{a}{\zeta}({\rm e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm
e}^{i\pi}\zeta)+{\rm e}^{i\pi}({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}^{\prime}({\rm
e}^{i\pi}\zeta)\right|\leq\left(\frac{|a|}{|\zeta|}+2\right)\frac{2\beta^{2}}{|\zeta|}{\rm
e}^{\frac{2\beta^{2}}{|\zeta|}}\leq 6x,$

and

$\left|{\rm e}^{i\pi
a}\left(\frac{a}{\zeta}\zeta^{a}{\mathcal{E}}_{1}(\zeta)+\zeta^{a}{\mathcal{E}}_{1}^{\prime}\zeta)-1-\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right)\right|\leq\left(1+\left(\frac{|a|}{|\zeta|}+3\right)\frac{2\beta^{2}}{|\zeta|}{\rm
e}^{\frac{2\beta^{2}}{|\zeta|}}\right)\leq 5,$

for $|\zeta|\geq|a|$ and $x\leq\frac{1}{2}$. Therefore,

$\displaystyle\left|\frac{M_{+}^{\prime}(\zeta)}{M_{+}(\zeta)}-\frac{1}{2}\right|$
$\displaystyle\leq\left|\frac{\frac{a}{\zeta}({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm e}^{i\pi}\zeta)+{\rm e}^{i\pi}({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}^{\prime}({\rm e}^{i\pi}\zeta)}{1+({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm e}^{i\pi}\zeta)+{\rm e}^{-\zeta}{\rm
e}^{-i\pi a}\left(1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right)}\right|$
$\displaystyle+{\rm e}^{-{\rm Re}(\zeta)}\left|\frac{{\rm e}^{i\pi
a}\left(\frac{a}{\zeta}\zeta^{a}{\mathcal{E}}_{1}(\zeta)+\zeta^{a}{\mathcal{E}}_{1}^{\prime}\zeta)-1-\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right)}{1+({\rm
e}^{i\pi}\zeta)^{a}{\mathcal{E}}_{1}({\rm e}^{i\pi}\zeta)+{\rm e}^{-\zeta}{\rm
e}^{-i\pi a}\left(1+\zeta^{a}{\mathcal{E}}_{1}(\zeta)\right)}\right|,$

which can be made arbitrarily small due to the previous bounds for ${\rm
Re}(\zeta)$ sufficiently large. ∎

######  Lemma A.7.

Let $y_{0}\in[0,1]$ such that $2my_{0}\leq N_{\nu,0}$. Then, there exists
$\varepsilon_{0}>0$ such that

$\left|\frac{M_{+}(y_{0}-i\varepsilon)}{M_{+}(y_{0}+i\varepsilon)}\right|\leq{\rm
e}^{\frac{5}{4}\nu\pi},$

for all $\varepsilon\leq\varepsilon_{0}$.

###### Proof.

Let $\theta=\arg(y_{0}-i\varepsilon)\in\left[-\frac{\pi}{2},0\right]$. Recall
that for $a=\frac{1}{2}+i\nu$ and $b=2a$, for $\zeta\in{\mathbb{C}}$,

$\displaystyle M_{+}(\zeta)$ $\displaystyle={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{a}M(a,b,\zeta).$

Therefore, we can estimate

$\left|\frac{M_{+}(y_{0}-i\varepsilon)}{M_{+}(y_{0}+i\varepsilon)}\right|=\left|{\rm
e}^{-i\varepsilon}{\rm e}^{i\theta}{\rm
e}^{-2\nu\theta}\frac{M(a,b,2m(y_{0}-i\varepsilon))}{M(a,b,2m(y_{0}+i\varepsilon))}\right|\leq{\rm
e}^{\nu\pi}\left|\frac{M(a,b,2m(y_{0}-i\varepsilon))}{M(a,b,2m(y_{0}+i\varepsilon))}\right|.$

Now, since
$\frac{\mathrm{d}}{\mathrm{d}\zeta}M(a,b,\zeta)=\frac{1}{2}M(a+1,b+1,\zeta)$,
which is entire in $\zeta\in{\mathbb{C}}$, we have that

$M(a,b,2m(y_{0}+i\varepsilon))=M(a,b,2my_{0})+\int_{0}^{\varepsilon}imM(a+1,b+1,2m(y_{0}+is))\mathrm{d}s.$

We can further bound the error term by noting that $|2m(y_{0}+is)|\leq
N_{\nu}+10\beta^{2}$, for all $|s|\leq|\varepsilon|$. As a result, there
exists $C_{\nu}$ such that $|M(a+1,b+1,2m(y_{0}+is))|\leq C_{\nu}$, for all
$|s|\leq|\varepsilon|$. Therefore,

$\displaystyle\left|\int_{0}^{\varepsilon}imM(a+1,b+1,2m(y_{0}+is))\mathrm{d}s\right|$
$\displaystyle\leq|M(a,b,2my_{0})|\frac{C_{\nu}m|\varepsilon|}{|M(a,b,2my_{0})|}$
$\displaystyle\leq C_{\nu}|M(a,b,2my_{0})|{\rm
e}^{-my_{0}}\sqrt{\frac{\nu\pi\cosh\nu\pi}{\sinh\nu\pi}}m|\varepsilon|$
$\displaystyle\leq(1-{\rm e}^{-\frac{1}{8}\nu\pi})|M(a,b,2my_{0})|,$

for all $0\leq|\varepsilon|\leq\varepsilon_{0}=\frac{1-{\rm
e}^{-\frac{1}{8}\nu\pi}}{m}\sqrt{\frac{\sinh\nu\pi}{\nu\pi\cosh\nu\pi}}C_{\nu}$.
Consequently, we have that

$\left|\frac{M(a,b,2m(y_{0}-i\varepsilon))}{M(a,b,2m(y_{0}+i\varepsilon))}\right|\leq{\rm
e}^{\frac{1}{4}\nu\pi}.$

∎

######  Lemma A.8.

Let $N_{\nu,0}$ be given as above and $N_{\nu,1}>0$. Let $\sigma\in\\{+,-\\}$.
If $N_{\nu,1}<N_{\nu,0}$, then, there exists $\varepsilon_{0}>0$ such that

$|M_{\sigma}(y_{0})|{\rm
e}^{-\frac{1}{8}\nu\pi}\leq|M_{\sigma}(y_{0}+i\varepsilon)|\leq|M_{\sigma}(y_{0})|{\rm
e}^{\frac{1}{8}\nu\pi},$

for all $y_{0}\in[0,1]$ such that $N_{\nu,1}\leq 2my_{0}\leq N_{\nu,0}$, and
all $0\leq|\varepsilon|\leq\varepsilon_{0}$.

###### Proof.

The result follows from the Fundamental Theorem of Calculus, the asymptotic
expansions of $M_{\sigma}$ and $M_{\sigma}^{\prime}$ for small arguments from
Lemma tal and the lower bounds on $|M_{\sigma}|$ from Lemma Qual. More
precisely, assume without loss of generality that $0\leq\varepsilon$ and note
that

$M_{\sigma}(y_{0}+i\varepsilon)=M_{\sigma}(y_{0})+\int_{0}^{\varepsilon}\frac{\mathrm{d}}{\mathrm{d}s}M_{\sigma}(y_{0}+is)\mathrm{d}s.$

Thanks to the asymptotic expansions for small arguments we next estimate

$\left|\int_{0}^{\varepsilon}\frac{\mathrm{d}}{\mathrm{d}s}M_{\sigma}(y_{0}+is)\mathrm{d}s\right|\leq\int_{0}^{\varepsilon}|M_{\sigma}^{\prime}(y_{0}+is)|\mathrm{d}s\leq
C_{\nu}\int_{0}^{\varepsilon}\frac{(2m)^{\frac{1}{2}}}{|y_{0}+is|^{\frac{1}{2}}}\mathrm{d}s\leq
C_{\nu}(2m\varepsilon)(2my_{0})^{-\frac{1}{2}}.$

Using the lower bound
$(2my_{0})^{\frac{1}{2}}\leq\sqrt{\frac{\nu\pi\cosh\nu\pi}{\sinh\nu\pi}}|M_{\sigma}(y_{0})|$
we have that

$\left|\int_{0}^{\varepsilon}\frac{\mathrm{d}}{\mathrm{d}s}M_{\sigma}(y_{0}+is)\mathrm{d}s\right|\leq
C_{\nu}|M_{\sigma}(y_{0})|\frac{\varepsilon}{y_{0}}\leq
C_{\nu}N_{\nu,2}^{-1}2m\varepsilon.$

We now choose $\varepsilon_{0}=\frac{N_{\nu,2}}{2mC_{\nu}}(1-{\rm
e}^{-\frac{1}{8}\nu\pi})$. The conclusion of the lemma follows swiftly for all
$\varepsilon\leq\varepsilon_{0}$. ∎

######  Lemma A.9.

Let $y_{0}\in[0,1]$ and $0\leq\varepsilon\leq\frac{\beta}{m}$. Then,

  * •

If $my_{0}\leq 3\beta$, there exists $\varepsilon_{0}>0$ such that

$\left(m|y_{0}+i\varepsilon|\right)^{\frac{1}{2}}\lesssim|M_{\pm}(y_{0}+i\varepsilon)|$

for all $0\leq\varepsilon\leq\varepsilon_{0}$.

  * •

If $my_{0}\geq 3\beta$,

$1\lesssim{|M_{\pm}(y_{0}+i\varepsilon)|}.$

###### Proof.

For the first part of the Lemma, recall
$M_{+}(\zeta)=e^{-\frac{1}{2}\zeta}\zeta^{a}M\left(a,b,\zeta\right)$, the
lemma follows once we obtain lower bounds on
$e^{-\frac{1}{2}\zeta}M\left(a,b,\zeta\right)$. For this, note that since
$\frac{\mathrm{d}}{\mathrm{d}\zeta}M(a,b,\zeta)=\frac{1}{2}M(a+1,b+1,\zeta)$,
which is entire in $\zeta\in{\mathbb{C}}$, we have that

$M(a,b,2m(y_{0}+i\varepsilon))=M(a,b,2my_{0})+\int_{0}^{\varepsilon}imM(a+1,b+1,2m(y_{0}+is))\mathrm{d}s.$

We further bound the error term by noting that $|2m(y_{0}+is)|\leq 10\beta$,
for all $|s|\leq|\varepsilon|$. As a result, there exists $C>0$ such that
$|M(a+1,b+1,2m(y_{0}+is))|\leq C$, for all $|s|\leq|\varepsilon|$. Therefore,
using the lower bounds on $|M(a,b,2my_{0})|$ from Lemma A.5,

$\displaystyle\left|\int_{0}^{\varepsilon}imM(a+1,b+1,2m(y_{0}+is))\mathrm{d}s\right|$
$\displaystyle\leq|M(a,b,2my_{0})|\frac{Cm|\varepsilon|}{|M(a,b,2my_{0})|}$
$\displaystyle\leq
C|M(a,b,2my_{0})||\Gamma(1+i\nu)|\sqrt{\frac{\nu\pi}{\sinh\nu\pi}}m|\varepsilon|.$

In particular, there exists $\varepsilon_{0}>0$ such that for all
$0\leq\varepsilon\leq\varepsilon_{0}$,

$\displaystyle e^{-my_{0}}|M(a,b,2m(y_{0}+i\varepsilon))|$ $\displaystyle\geq
e^{-my_{0}}|M(a,b,2my_{0})|\left(1-Cm\frac{\varepsilon}{|M(a,b,2my_{0})|}\right)$
$\displaystyle\geq\frac{1}{2}\frac{1}{|\Gamma(1+i\nu)|}\sqrt{\frac{\sinh(\nu\pi)}{\nu\pi}},$

and the first part of the lemma follows. As for the second statement, it is a
direct consequence of Lemma A.6 and the fact that $|M_{\pm}(\cdot)|$ is
bounded in compact domains (it is entire). ∎

###  A.4. Growth bounds and comparison estimates for $\beta^{2}=1/4$

######  Lemma A.10.

Let $\beta^{2}=1/4$ and let $\mu=\sqrt{1/4-\beta^{2}}$. Denote
$a:=\frac{1}{2}$ and $b:=2a=1$. Then, there exists $N_{0}>0$ such that

$\left|\frac{W_{0,0}(\zeta)}{M_{0,0}(\zeta)}\right|\leq 2\sqrt{\pi}{\rm
e}^{-{\rm
Re}\zeta},\quad\left|\frac{W_{0,0}^{\prime}(\zeta)}{W_{0,0}(\zeta)}+\frac{1}{2}\right|\leq\frac{1}{4},\quad\frac{1}{2}e^{-\frac{1}{2}{\rm
Re}\zeta}\leq|W_{0,0}(\zeta)|\leq\frac{3}{2}{\rm e}^{-\frac{1}{2}{\rm
Re}\zeta}$

for all ${\rm Re}\zeta\geq N_{0}$.

###### Proof.

Let $\zeta\in{\mathbb{C}}$ and $\delta>0$. We recall that

$\displaystyle M_{0,0}(\zeta)$ $\displaystyle={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{\frac{1}{2}}M(1/2,1,\zeta)$ $\displaystyle={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{1/2}\frac{\Gamma(1)}{\Gamma(1/2)}\left(-iU(1/2,1,\zeta)+i{\rm
e}^{\zeta}U(1/2,1,{\rm e}^{i\pi}\zeta)\right),$

while also

$W_{0,0}(\zeta)={\rm e}^{-\frac{1}{2}\zeta}\zeta^{\frac{1}{2}}U(1/2,1,\zeta).$

Thus, we have that

$\frac{W_{0,0}(\zeta)}{M_{0,0}(\zeta)}=-i\sqrt{\pi}\frac{U(1/2,1,\zeta)}{{\rm
e}^{\zeta}U(1/2,1,{\rm
e}^{i\pi}\zeta)-U(1/2,1,\zeta)}=-i\sqrt{\pi}\frac{1}{{\rm
e}^{\zeta}\frac{U(1/2,1,{\rm e}^{i\pi}\zeta)}{U(1/2,1,\zeta)}-1}.$

Now, we also recall that
$U(1/2,1,\zeta)=\zeta^{-\frac{1}{2}}\left(1+\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}(\zeta)\right)$,
with $|\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}(\zeta)|\leq\frac{1}{2|\zeta|}{\rm
e}^{\frac{1}{2|\zeta|}}$. Therefore, we have the lower bound

$\left|\frac{U(1/2,1,{\rm
e}^{i\pi}\zeta)}{U(1/2,1,\zeta)}\right|\geq\frac{3}{4},$

for $|\zeta|$ sufficiently large. Moreover, $\frac{3}{4}{\rm e}^{{\rm
Re}\zeta}-1\geq\frac{1}{2}{\rm e}^{{\rm Re}\zeta}$, for all ${\rm Re}\zeta\geq
2\ln 2$. The desired conclusion follows.

For the second part of the Lemma, since
$W_{0,0}(\zeta)=e^{-\frac{1}{2}\zeta}\zeta^{\frac{1}{2}}\left(\zeta^{-\frac{1}{2}}+{\mathcal{E}}_{1}(\zeta)\right)$,
we note that

$W_{0,0}^{\prime}(\zeta)=-\frac{1}{2}W_{0}(\zeta)+\frac{1}{2}\frac{W_{0}(\zeta)}{\zeta}+e^{-\frac{1}{2}\zeta}\left(-\frac{1}{2\zeta}+\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}^{\prime}(\zeta)\right),$

where we recall that
$\left|\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}^{\prime}(\zeta)\right|\leq\frac{1}{4|\zeta|}e^{\frac{1}{2|\zeta|}}\leq
x$, for $x=\frac{1}{2|\zeta|}\leq\frac{1}{2}$. Hence,

$\left|\frac{W_{0,0}^{\prime}(\zeta)}{W_{0,0}(\zeta)}+\frac{1}{2}\right|\leq
x+2x\left|\frac{1}{1+\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}(\zeta)}\right|\leq
2x\leq\frac{1}{4},$

for all $x\leq\frac{1}{8}$.

For the third statement of the Lemma, note that
$W_{0,0}(\zeta)=e^{-\frac{1}{2}\zeta}\left(1+\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}(\zeta)\right)$,
the conclusion follows for $|\zeta|$ large enough so that
$\left|\zeta^{\frac{1}{2}}{\mathcal{E}}_{1}(\zeta)\right|\leq\frac{1}{2}$. ∎

######  Lemma A.11.

Let $\beta^{2}=1/4$. Denote $a:=\frac{1}{2}\pm\mu$ and $b:=2a=1$. Then, for
all $\epsilon>0$ there exists $\delta_{0}>0$ such that

$\left|\frac{M_{0,0}(\zeta)}{W_{0,0}(\zeta)}\right|\leq\epsilon,$

for all $\zeta\in{\mathbb{C}}$ such that $|\zeta|\leq\delta$.

###### Proof.

We use the functional relation between the Whittaker functions and the
modified Bessel functions in order to extract the correct asymptotic behaviour
of the functions near the origin and estimate the quotient precisely. In this
direction, recall that

$M_{0,0}(2\zeta)=\sqrt{2\zeta}I_{0}(\zeta),\quad
W_{0,0}(2\zeta)=\sqrt{\frac{2\zeta}{\pi}}K_{0}(\zeta),$

where $I_{0}(\zeta)$ and $K_{0}(\zeta)$ denote the modified Bessel functions
of order 0. Moreover, we have that

$K_{0}(\zeta)=-\left(\ln\left(\frac{\zeta}{2}\right)+\varsigma\right)I_{0}(\zeta)+2\sum_{k=1}^{\infty}\frac{I_{2k}(\zeta)}{k},$

where

$I_{2k}(\zeta)=\left(\frac{\zeta}{2}\right)^{2k}\sum_{j\geq
0}\frac{\left(\frac{\zeta^{2}}{4}\right)^{j}}{j!(2k+j)!}.$

In particular, one observes that $|I_{2k}(\zeta)|\leq I_{2k}(|\zeta|)$.
Moreover, under the observation that $(2k+j)!\geq(2k)!j!$, we can bound

$\left|2\sum_{k=1}^{\infty}\frac{I_{2k}(z)}{k}\right|\leq
2\sum_{k=1}^{\infty}I_{0}(|\zeta|)\frac{\left(\frac{1}{2}|\zeta|\right)^{2k}}{k(2k)!}\leq
2I_{0}(|\zeta|)\left(\cosh\frac{1}{2}|\zeta|-1\right).$

With this, together with the fact that $I_{0}(\cdot)$ is analytic in
${\mathbb{C}}$ and $I_{0}(\zeta)\rightarrow 1$ when $\zeta\rightarrow 0$, we
have that

$|K_{0}(\zeta)|\geq-\frac{1}{2}\ln\left(\frac{1}{2}|\zeta|\right),$

for $|\zeta|$ sufficiently small. The conclusion follows, since for $|\zeta|$
sufficiently small we have

$\left|\frac{M_{0,0}(\zeta)}{W_{0,0}(\zeta)}\right|\leq-\frac{3}{\ln\left(\frac{1}{4}|\zeta|\right)}$

∎

######  Lemma A.12.

Let $\beta^{2}=1/4$ and let $y_{0}\in[0,1]$ such that $N_{2}\leq 2my_{0}\leq
N_{1}$. Then, for all $\epsilon>0$ there exists $\varepsilon_{0}>$ such that

$\left|\frac{W_{0}(y_{0}-i\varepsilon)}{M_{0}(y_{0}-i\varepsilon)}-\frac{W_{0}(y_{0})}{M_{0}(y_{0})}\right|\leq\epsilon,\quad\left|\frac{W_{0}(y_{0}-i\varepsilon)}{M_{0}(y_{0}-i\varepsilon)}\right|\leq
C$

for all $\varepsilon\leq\varepsilon_{0}$ and some $C>0$. In particular,

$\left|{\rm
Im}\left(\frac{W_{0}(y_{0}-i\varepsilon)}{M_{0}(y_{0}-i\varepsilon)}\right)\right|\leq\epsilon,$

###### Proof.

It follows from the continuity of the functions involved, plus the fact that
$M_{0,0}(x)$ does not vanish and $W_{0,0}(x)$ is bounded, for any $x>0$ such
that $0<N_{2}\leq x\leq N_{1}<\infty$. ∎

######  Lemma A.13.

There exists $\delta_{2}>0$ such that

$|\zeta|^{\frac{1}{2}}\left(1+\big{|}\log|\zeta|\big{|}\right)\lesssim|W_{0,0}(\zeta)|,$

for all $|\zeta|\leq\delta_{2}$.

###### Proof.

Recall that $W_{0,0}(\zeta)=\sqrt{\frac{\zeta}{\pi}}K_{0}(\zeta/2)$ and the
fact that $|K_{0}(\zeta)|\geq-\frac{1}{2}\log\left(\frac{|\zeta|}{2}\right)$
for $\zeta\rightarrow 0$. Then,

$|W_{0,0}(\zeta)|\geq\frac{1}{2\sqrt{\pi}}|\zeta|^{\frac{1}{2}}\big{|}\log|\zeta|-\log
4\big{|}\geq\frac{1}{20\sqrt{\pi}}|\zeta|^{\frac{1}{2}}\big{|}\log|\zeta|\big{|}\geq\frac{1}{40\sqrt{\pi}}|\zeta|^{\frac{1}{2}}\left(1+\big{|}\log|\zeta|\big{|}\right),$

for $|\zeta|$ sufficiently small. ∎

######  Lemma A.14.

Let $y_{0}\in[0,1]$ and $0\leq\varepsilon\leq\frac{\beta}{m}$. Then,

  * •

If $my_{0}\leq 3\beta$, there exists $\varepsilon_{0}>0$ such that

$\left(m|y_{0}+i\varepsilon|\right)^{\frac{1}{2}}\lesssim|M_{0}(y_{0}+i\varepsilon)|$

for all $0\leq\varepsilon\leq\varepsilon_{0}$.

  * •

If $my_{0}\geq 3\beta$,

$1\lesssim{|M_{\pm}(y_{0}+i\varepsilon)|}.$

###### Proof.

The proof uses the ideas from Lemma A.9 together with the bounds from Lemma
A.15. We omit the details. ∎

###  A.5. Growth bounds and comparison estimates for $\beta^{2}<1/4$

In this subsection we consider the case $\beta^{2}<1/4$, for which
$\mu=\sqrt{1/4-\beta^{2}}$ with $\mu\in\left(0,\frac{1}{2}\right)$ and
$\nu=0$.

######  Lemma A.15.

Denote $a_{\pm}:=\frac{1}{2}\pm\mu$ and $b_{\pm}:=2a_{\pm}$. Then,

$\lim_{{\rm Re}(\zeta)\rightarrow+\infty}\left|\frac{\Gamma(a)}{\Gamma(b)}{\rm
e}^{-\frac{1}{2}\zeta}M_{+}(\zeta)\right|=1.$

Moreover, let $C_{\mu}=2^{-4\mu}\frac{\Gamma(1-\mu)}{\Gamma(1+\mu)}$. There
exists $N_{\mu,0}>0$ such that

$\left|\frac{M_{-}(\zeta)}{M_{+}(\zeta)}-2^{-4\mu}\frac{\Gamma(1-\mu)}{\Gamma(1+\mu)}\right|\leq\min\left(5,\frac{\tan\mu\pi}{1+\tan\mu\pi}\right)\frac{C_{\mu}}{4}.$

for all ${\rm Re}\zeta\geq N_{\mu,0}$.

###### Proof.

Let $\zeta\in{\mathbb{C}}$ and $\delta>0$. We recall that

$\displaystyle M_{\pm}(\zeta)$ $\displaystyle={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{a_{\pm}}M(a_{\pm},b_{\pm},\zeta)$
$\displaystyle={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{a_{\pm}}\frac{\Gamma(b_{\pm})}{\Gamma(a_{\pm})}\left({\rm
e}^{-i\pi a_{\pm}}U(a_{\pm},b_{\pm},\zeta)+{\rm e}^{i\pi a_{\pm}}{\rm
e}^{\zeta}U(a_{\pm},b_{\pm},{\rm e}^{i\pi}\zeta)\right).$

Moreover, we have that
$U(a_{\pm},b_{\pm},\zeta)=\zeta^{-a_{\pm}}+\mathcal{E}_{\pm}(\zeta)$, where
further

$|\zeta^{a_{\pm}}{\mathcal{E}}_{\pm}(\zeta)|\leq\frac{2\beta^{2}}{|\zeta|}{\rm
e}^{\frac{2\beta^{2}}{|\zeta|}}.$

In the sequel, we write $x:=\frac{2\beta^{2}}{|\zeta|}$. Therefore, we can
write

$M_{\pm}(\zeta)=\frac{\Gamma(b_{\pm})}{\Gamma(a_{\pm})}{\rm
e}^{\frac{1}{2}\zeta}\left(\left[1+({\rm
e}^{i\pi}\zeta)^{a_{\pm}}{\mathcal{E}}_{\pm}({\rm e}^{i\pi}\zeta)\right]+{\rm
e}^{-\zeta}{\rm e}^{-i\pi
a_{\pm}}\left[1+\zeta^{a_{\pm}}{\mathcal{E}}_{\pm}(\zeta)\right]\right)$

Now, since $b_{\pm}=2a_{\pm}$, we have that

$\frac{\Gamma(b_{\pm})}{\Gamma(a_{\pm})}=\pi^{-\frac{1}{2}}2^{2a_{\pm}-1}\Gamma\left(a_{\pm}+\frac{1}{2}\right),$

confer, [NIST]. Therefore,

$\frac{\displaystyle{\Gamma(b_{-})}/{\Gamma(a_{-})}}{\displaystyle{\Gamma(b_{+})}/{\Gamma(a_{+})}}=2^{-4\mu}\frac{\Gamma(1-\mu)}{\Gamma(1+\mu)}$

and we note that

$\frac{M_{-}(\zeta)}{M_{+}(\zeta)}=2^{-4\mu}\frac{\Gamma(1-\mu)}{\Gamma(1+\mu)}\frac{1+({\rm
e}^{i\pi}\zeta)^{a_{-}}{\mathcal{E}}_{-}({\rm e}^{i\pi}\zeta)+{\rm
e}^{-\zeta}{\rm e}^{-i\pi
a_{-}}\left[1+\zeta^{a_{-}}{\mathcal{E}}_{-}(\zeta)\right]}{1+({\rm
e}^{i\pi}\zeta)^{a_{+}}{\mathcal{E}}_{+}({\rm e}^{i\pi}\zeta)+{\rm
e}^{-\zeta}{\rm e}^{-i\pi
a_{+}}\left[1+\zeta^{a_{+}}{\mathcal{E}}_{+}(\zeta)\right]}.$

Moreover, we observe that that
$|\zeta^{a}{\mathcal{E}}_{1}(\zeta)|\leq\frac{4\beta^{2}}{|\zeta|}$, for
$|\zeta|\geq 4\beta^{2}$. Hence, for any $\delta>0$,

$\left|({\rm e}^{i\pi}\zeta)^{a_{\pm}}{\mathcal{E}}_{\pm}({\rm
e}^{i\pi}\zeta)+{\rm e}^{-\zeta}{\rm e}^{-i\pi
a_{\pm}}\left(1+\zeta^{a_{\pm}}{\mathcal{E}}_{\pm}(\zeta)\right)\right|\leq\delta$

provided that ${\rm Re}\zeta>N_{\mu,0}$ for some $N_{\mu,0}>0$. ∎

######  Lemma A.16.

Denote $a_{\pm}:=\frac{1}{2}\pm\mu$ and $b_{\pm}:=2a_{\pm}$. Then,

$\lim_{\zeta\rightarrow 0}\frac{M_{+}(\zeta)}{M_{-}(\zeta)}=0.$

Therefore, there exists $\delta_{\mu,1}>0$ such that

$\left|\frac{M_{+}(\zeta)}{M_{-}(\zeta)}\right|\leq\min\left(\frac{1}{5M(a_{-},b_{-},2N_{\mu,0})},\frac{1}{3C_{\mu}}\right)$

for all $\zeta\in{\mathbb{C}}$ such that $|\zeta|\leq\delta_{\mu,1}$.

###### Proof.

We recall once again that

$M_{\pm}(\zeta)={\rm
e}^{-\frac{1}{2}\zeta}\zeta^{a_{\pm}}M(a_{\pm},b_{\pm},\zeta).$

Hence, we directly compute

$\frac{M_{+}(\zeta)}{M_{-}(\zeta)}=\zeta^{2\mu}\frac{M(a_{+},b_{+},\zeta)}{M(a_{-},b_{-},\zeta)}.$

Since $M(a_{\pm},b_{\pm},\zeta)\rightarrow 1$ for $\zeta\rightarrow 0$, and
$2\mu>0$, the conclusion follows for $|\zeta|$ small enough. ∎

######  Lemma A.17.

Denote $a_{\pm}:=\frac{1}{2}\pm\mu$ and $b_{\pm}:=2a_{\pm}$. Let
$y_{0}\in[0,1]$ such that $N_{\mu,1}\leq 2my_{0}\leq N_{\mu,0}$, for some
$N_{\mu,1}\in{\mathbb{R}}$. Then, there exists $\varepsilon_{0}>0$ such that

$\left|\frac{M_{\pm}(y_{0}+i\varepsilon)}{M_{\pm}(y_{0})}-1\right|\leq\frac{\sin\mu\pi}{5},$

for all $0<|\varepsilon|\leq\varepsilon_{0}$.

###### Proof.

Assume without loss of generality that $\varepsilon>0$. Then,

$M_{\pm}(y_{0}+i\varepsilon)=M_{\pm}(y_{0})+\int_{0}^{\varepsilon}\frac{\mathrm{d}}{\mathrm{d}s}M_{\pm}(y_{0}+is)\mathrm{d}s.$

Thanks to the asymptotic expansions for small arguments we next estimate

$\left|\int_{0}^{\varepsilon}\frac{\mathrm{d}}{\mathrm{d}s}M_{\pm}(y_{0}+is)\mathrm{d}s\right|\leq\int_{0}^{\varepsilon}|M_{\pm}^{\prime}(y_{0}+is)|\mathrm{d}s\leq
C_{\mu}\int_{0}^{\varepsilon}\frac{(2m)^{\frac{1}{2}\pm\mu}}{|y_{0}+is|^{\frac{1}{2}\mp\mu}}\mathrm{d}s\leq
C_{\mu}(2m\varepsilon)(2my_{0})^{-\frac{1}{2}}.$

Using the lower bound $(2my_{0})^{\frac{1}{2}\pm\mu}\leq M_{\pm}(y_{0})$ we
have that

$\left|\int_{0}^{\varepsilon}\frac{\mathrm{d}}{\mathrm{d}s}M_{\pm}(y_{0}+is)\mathrm{d}s\right|\leq
C_{\mu}M_{\pm}(y_{0})\frac{\varepsilon}{y_{0}}\leq
C_{\mu}M_{\pm}(y_{0})N_{2}^{-1}2m\varepsilon.$

Hence,

$M_{\pm}(y_{0})\left(1-C_{\mu}\frac{\varepsilon}{N_{2}}\right)\leq|M_{\pm}(y_{0}+i\varepsilon)|\leq
M_{\pm}(y_{0})\left(1+C_{\mu}\frac{\varepsilon}{N_{2}}\right)$

and now choose $\varepsilon_{0}>0$ sufficiently small, so that the conclusion
of the lemma follows swiftly for all $\varepsilon\leq\varepsilon_{0}$. ∎

######  Lemma A.18.

Let $y_{0}\in[0,1]$ and $0\leq\varepsilon\leq\frac{\beta}{m}$. Then,

  * •

If $my_{0}\leq 3\beta$, there exists $\varepsilon_{0}>0$ such that

$\left(m|y_{0}+i\varepsilon|\right)^{\frac{1}{2}\pm\mu}\lesssim|M_{\pm}(y_{0}+i\varepsilon)|$

for all $0\leq\varepsilon\leq\varepsilon_{0}$.

  * •

If $my_{0}\geq 3\beta$,

$1\lesssim{|M_{\pm}(y_{0}+i\varepsilon)|}.$

###### Proof.

The proof uses the ideas from Lemma A.9 together with the bounds from Lemma
A.15. We omit the details. ∎

## Acknowledgments

The research of MCZ was partially supported by the Royal Society URF\R1\191492
and EPSRC Horizon Europe Guarantee EP/X020886/1.

## References