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Consider the following functional :

$$ I(x,s) =\int_0^\infty\mathrm dy \frac{F(x + \mathrm iy, s) − F(x −\mathrm iy, s)}{\mathrm e^{2πy}-1}, $$ where $ F(z, s) = \dfrac{\sin^2[π\Gamma(z)/(2z)]}{z^s} $.

Let us restrict $s\in[0,1]$

In Mathematica

f[z_, s_] = Sin[π Gamma[z]/(2z)]^2/z^s; 
II[x_, s_] := Integrate[(f[x + I y, s] - f[x - I y, s])/(Exp[2 Pi y] - 1), 
             {y, 0,Infinity}]

Can we get sharp numerical asymptotic of $I(x)$ as as $x\rightarrow \infty$?

Also, can we get quantitative upper and lower bound estimations on the functional ?

Update : Also try for $$F(z)= \log(z){\sin^2[π\Gamma(z)/(2z)]}$$

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  • $\begingroup$ Have you tried Series[]? $\endgroup$
    – Michael E2
    Commented Feb 25, 2020 at 19:20
  • $\begingroup$ @Michael E2 Thank you for comment, sir . I've not used it. Actually this is my first day and second question on this site . I'm not used to Mathematica . $\endgroup$
    – bambi
    Commented Feb 25, 2020 at 19:25
  • $\begingroup$ I’m away from my computer, I can’t try it. It may or may not work on Integrate[] $\endgroup$
    – Michael E2
    Commented Feb 25, 2020 at 20:30
  • $\begingroup$ @MichaelE2 doesn't seem to work trivially at least. The function f is rather pathological near the origin. !Mathematica graphics $\endgroup$
    – chris
    Commented Feb 25, 2020 at 20:33
  • $\begingroup$ The most important cases for my purpose is $s=1$ or very near to 1 and $s=0$ or very near to zero $\endgroup$
    – bambi
    Commented Feb 26, 2020 at 8:05

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