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Let I have the following function,

$f = Q(a\Re (x + y))Q(b\Im (x + y))\log \left\{ {Q(a\Re (x + y))Q(b\Im (x + y))} \right\}$

Where, $x,y \in C$, $a,b\in R$ and $- m \le \Re (x),\Re (y),\Im (x),\Im (y) \le m$, $m$ is a finite real number.

And $Q\left( z \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_z^\infty {{e^{ - \frac{{{u^2}}}{2}}}du} $.

I want to show that $f$ is analytic on the bounded complex plane.

** Note: I know that the Q function is analytic on the real line. Does that information help while proving the analyticity of $f$? Any suggestions will be helpful.

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The answer is: of course not. Indeed, take any real $a\ne0$, any real $b$, and any real $m>0$. If $f$ were analytic in $x,y$ such that $-m\le\Re x,\Re y,\Im x,\Im y\le m$ then it would be analytic in $x$ at $x=0$ for $y=0$. Then the real and imaginary parts of the function $$\mathbb R^2\ni(s,t)\mapsto g(s,t):=Q(as)Q(bt)\,\ln\big(Q(as)Q(bt)\big)\in\mathbb C$$ would satisfy the Cauchy–Riemann equations at $(s,t)=(0,0)$ -- which the do not, because the partial derivative of $\Re g(s,t)$ in $s$ at $(s,t)=(0,0)$ is $a\,\dfrac{\ln(4/e)}{2\sqrt{2\pi}}\ne0$, while the partial derivative of $\Im g(s,t)[=0]$ in $t$ is $0$ everywhere.

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  • $\begingroup$ @losif Pinellas, if $a$ and $b$ both are positive (non-zero) can I say f is analytic on the complex plane? $\endgroup$
    – Samantha
    Commented Nov 4, 2020 at 17:56
  • $\begingroup$ @losif Pinellas By looking at the equation, I think for non-zero $a$, $b$ both $\Im g(s,t)[=0]$=$\Re g(s,t)[=0]$ and will satisfy Cauchy Raman equations.. And Hence it will be analytic in the bounded complex plane. Am I right? $\endgroup$
    – Samantha
    Commented Nov 4, 2020 at 18:09
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    $\begingroup$ @Samantha : In your question, it was assumed that $a$ and $b$ are arbitrary real numbers. As such, your question has been fully answered (negatively). If you want to pursue this further and impose further conditions, then you should ask such additional questions in separate posts. Anyhow, as now shown in the edited answer, your function is not analytic for any real $a$ and $b$ unless $a=b=0$. $\endgroup$
    – Iosif Pinelis
    Commented Nov 4, 2020 at 18:34
  • $\begingroup$ @losif Pinellas, I posted a new question here link $\endgroup$
    – Samantha
    Commented Nov 4, 2020 at 20:08

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