2
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ Are you trying to use the Abel-Plana formula? $\endgroup$ Commented Feb 8, 2021 at 14:07
  • $\begingroup$ @J.M. yes sir $\phantom{}$ $\endgroup$
    – bambi
    Commented Feb 8, 2021 at 14:08
  • 1
    $\begingroup$ 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 $\endgroup$
    – Michael E2
    Commented Feb 21, 2021 at 18:24
  • $\begingroup$ @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. $\endgroup$
    – bambi
    Commented Feb 24, 2021 at 6:20
  • 1
    $\begingroup$ It was just some advice, meant to be helpful, for getting more people to try out your integral. $\endgroup$
    – Michael E2
    Commented Feb 24, 2021 at 14:31

1 Answer 1

2
$\begingroup$

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}]

enter image description here

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – bambi
    Commented Feb 12, 2021 at 8:26
  • $\begingroup$ 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). $\endgroup$
    – anderstood
    Commented Feb 28, 2021 at 8:09
  • 1
    $\begingroup$ 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 $\endgroup$
    – Dominic
    Commented Feb 28, 2021 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.