# Program for efficient computation of given functional:

I need to plot the following functional with accuracy:

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

And let us restrict $$s\in[0,1]$$

What is the most efficient way of computing this integral in Mathematica?

What is the nature of functional as $$x\rightarrow\infty$$ from the computation ?

I computed some relatively small values which suggest the function is oscillatory with damping. But I need big values greater than x=100 and at s=1

• Are you trying to use the Abel-Plana formula? Commented Feb 8, 2021 at 14:07
• @J.M. yes sir $\phantom{}$ Commented Feb 8, 2021 at 14:08
• People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this meta Q&A helpful Commented Feb 21, 2021 at 18:24
• @MichaelE2 I completely agree with you sir. But as you can see the code is simple but my computations doesn't agree with expected results. So I'm asking here without mentioning anything i.e. from scratch. Commented Feb 24, 2021 at 6:20
• It was just some advice, meant to be helpful, for getting more people to try out your integral. Commented Feb 24, 2021 at 14:31

Here's a start you may wish to work with:

(*
define f
*)
myF[z_, s_] := 1/(z^s Gamma[Sin[(Pi Gamma[z])/(2 z)]^2])
(*
define the integrand
*)
integrand[z_, s_] := (myF[z, s] - myF[Conjugate[z], s])/(
Exp[2 Pi Im@z] - 1)
(*
define integral expression in terms of z=x+Iy and real 0<s<1
note dz=Idy in the expression for integrating with respect
to y for z=x+I y
*)
myInt[x_?NumericQ, s_?NumericQ] :=
NIntegrate[I integrand[z, s] /. z -> x + I y, {y, 0, 4}]
(*
Plot Real (red) and Im (blue) component of integral function for
s=1/4 integrating from 1 to 2
*)
Plot[{Re@myInt[x, 1/4], Im@myInt[x,1/4]}, {x, 1, 2},
PlotStyle -> {Red, Blue}]


• thank you for the answer sir but the integral is from zero to infinity and answer should be only in real domain as the functions in functional deals with real valued functions. Commented Feb 12, 2021 at 8:26
• Why have you got an I in front of integrand[z, s] in myInt? Also, in the plot, I think you are computing myInt[x, 1/4] twice. You could divide the computation time by 2 using ReIm (see also ReImPlot). Commented Feb 28, 2021 at 8:09
• The i is in the integral because the integrand is f(z)dz and letting z=x+iy with x constant in integrating with respect to y we have dz=idy and the integral becomes i f(z(y))dy Commented Feb 28, 2021 at 10:08