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I'm studying the limits and applicability of Abel Plana summation for different test functions ( class of functions ) . In doing so this just pops out and couldn't handle the said integral so asked here:

Consider the following function :

$$F(x)=\frac{\sin^2(\Gamma(x))\Gamma'(x)}{e^{\sin^2(\Gamma(x))}}$$

Now consider the following function :

$$I(x) =-i\int_0^\infty\mathrm dy \frac{F(x + \mathrm iy) − F(x −\mathrm iy)}{\mathrm e^{2πy}-1}$$

What is the nature of the $I(x)$ as $x\rightarrow\infty$?

( Is $I(x)\rightarrow 0$ as $x\rightarrow\infty$ true?)

( Is there an analytic way to show this to be true or false?)

Some values I computed :

$x=0.3, I= -0.4596$

$x=0.5, I= 0.3347$

$x=0.7, I= 0.1407$

$x=0.9, I= 0.0706$

$x=1 , I= 0.05211$

$x=1.5, I=0.02101$

$x=2 , I= 0.02518$

$x=3, I=0.06752$

It seems after $x=3$ we are unable to compute numerically

I also asked this question on MathOverflow with no response as follows:

https://mathoverflow.net/q/412913/145223

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  • $\begingroup$ $F(z)$ has to be bounded in $a\le\Re(z)\le$ if you want to apply Abel-Plana formula to the interval $[a,b]$. $\endgroup$
    – TravorLZH
    Jan 12, 2022 at 16:05
  • $\begingroup$ @TravorLZH that's not true, Abel plana holds for much weaker conditions see FWJ Olver "Asymptotics and special functions" $\endgroup$
    – TPC
    Jan 13, 2022 at 8:02

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