# Plot of a function defined by an integral

(First of all, this is my first Mathematica question. I'm not used to Mathematica that much. So, apologies in advance.)

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 $$F(z, s) = \dfrac{\sin^2[π\Gamma(z)/(2z)]}{z^s}$$.

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

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

The reason for the question is that the functional gives very massive values ( upto 10^100) after the value x=6 which I think are not correct. I don't know how to resolve this issue.So I'm posting this question for bigger accurate values

• you defined $F$ as taking one argument $z$ but you are calling in using 2 arguments? What is s? – Nasser Feb 24 '20 at 20:15
• @Nasser thanks for correcting see denominator $(z^s)$ – bambi Feb 24 '20 at 20:17
• If you looking for analytical integration, Mathematica can not do the integral at all f[z_, s_] := Sin[Gamma[z]/z]^2/z^s; Integrate[(f[x + I y, s] - f[x - I y, s])/(Exp[2 Pi y] - 1), y] For numerical integration, need numerical values for the parameters involved. – Nasser Feb 24 '20 at 20:23
• @Nasser I'm looking for sharp approximation of the integral – bambi Feb 25 '20 at 5:51
• sharp approximation of the integral I assume you mean by a numerical approximation. For this, numerical values are needed for all parameters other than the integration variable itself. If you provide example of such values, may be someone could help. – Nasser Feb 25 '20 at 10:07

## 1 Answer

You could do it numerically?

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


Then

ParallelTable[{x, II[x, s] // Im}, {s, 1, 2, 1/2}, {x, 1, 5, 0.05}] //
ListLinePlot[#, PlotRange -> All] & • thank you for the answer but as you can see the functional gives very massive values after the value x=6 which I think are not correct. So I'm posting this question for bigger accurate values – bambi Feb 26 at 15:28