# Nature of function as $x\rightarrow\infty$

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 ( I'm just a student so, please excuse me if anything is trivial):

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 that We can't do anything beyond $$x=3$$ numerically.

• Why is this question of interest? Jan 1 at 18:03
• @GeraldEdgar it just pops out in my study of Abel plana summation for different test functions.
– TPC
Jan 1 at 18:25
• Tryng different methods for numerical integration, no problem for $0.218 \leq x \leq 3.381$ but outside this range, all of them just explode Jan 3 at 15:35
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• A short comment. The nature of the $I(x)$ depends on the nature of the integrand's numerator $Im[F(x+iy)-F(x-iy)]$. For x passing some point beyond the (first?) zero of I(x) the integrand behaves wildly. You can plot it. Working with Pari/GP the breakpoint is $x_0=3.208917736262569989928994080...$ for which $I(x_0)=0$. I think Abel-Plana is not a good tool for summation/integration of such kind of functions. You should try an adequate variable transformation first. What if $u=\Gamma(x)$?. Mar 1 at 15:08