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Let $\langle;\rangle$ be the usual scalar product on $L^2(\Bbb R^2)$. How to prove that the function $\lambda \mapsto \langle T_{\lambda}f; g\rangle$ is holomorphic on $\Bbb C^+_*=\{z\in\Bbb C:\text{Re}(z)>0\}$ where $f,g\in L^2(\Bbb R^2)$ and

$T_{\lambda}f(x,y)= \int_{\Bbb R^2}f(x-u,y-v) \psi\left( \frac{\lambda-\mu}{2\lambda},1,\frac{\lambda}{2}(u^2+v^2)\right) e^{-\frac{\lambda}{4}\ (u^2+v^2\ ) - i\frac{\lambda}{2} (yu-xv)} \ \ \ \ dudv,$

and $\psi\left( a,c,z\right)\sim z^{-a}$ as $z\to\infty$. we suppose that $\lambda\to \psi\left( \frac{\lambda-\mu}{2\lambda},1,\frac{\lambda}{2}(u^2+v^2)\right)$ is holomorphic on $\Bbb C^+_*=\{z\in\Bbb C:\text{Re}(z)>0\} $ and $\mu\in \Bbb C$

Here $\psi$ is the Kummer funtion noted sometimes $U(a,b,c)$ (https://en.wikipedia.org/wiki/Confluent_hypergeometric_function) Thank you a lot in advance

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    $\begingroup$ The answer should just boil down to "differentiate under the integral sign". The details should be fairly routine, I would think? What have you tried? This might be better suited to math.stackexchange. $\endgroup$ Commented Sep 28, 2022 at 10:01
  • $\begingroup$ I have found a big problem to do this $\endgroup$ Commented Sep 28, 2022 at 11:05
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    $\begingroup$ I would try: Morera's theorem. No need to differentiate, but you do need to interchange the order of integration. $\endgroup$ Commented Sep 28, 2022 at 17:32
  • $\begingroup$ Yeah. But how to do that $\endgroup$ Commented Sep 28, 2022 at 17:43
  • $\begingroup$ @GeraldEdgar. How to justify this? $\endgroup$ Commented Sep 29, 2022 at 11:36

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