# An error message with NIntegrate and inability to plot the integrand

I took a function of x which depended on parameters on m,n,a,y and then first summed up the n from -Infinity to Infinity and then I set the other parameters to some random values and then I asked it to integrate and I got a funny message back -

 Sum [    (((-m^2 + n^2/a^2 - x^2)/(m^2 + n^2/a^2 + x^2) + (
2 n^2 (m^2 - a m^2 + n^2 - n^2/a + x^2 - a x^2))/(
a^3 (m^2 + n^2/a^2 + x^2)^2)) Gamma[-I x + (1 + y)/2] Gamma[
I x + (1 + y)/2] )/((m^2 + n^2 + x^2) Gamma[-I x] Gamma[
I x]) , {n, -Infinity, Infinity}] // FullSimplify

1/Sqrt[-m^2   x^2] x (Cot[\[Pi] Sqrt[-m^2 - x^2]] -  Cot[a \[Pi] Sqrt[-m^2 - x^2]] +
a \[Pi] Sqrt[-m^2 - x^2] Csc[a \[Pi] Sqrt[-m^2 - x^2]]^2)
Gamma[ 1/2 (1 - 2 I x + y)] Gamma[1/2 (1 + 2 I x + y)] Sinh[\[Pi] x]

m = 46.5675786575; a = 5; y = 0;

Integrate [  1/Sqrt[-m^2 - x^2]  x (Cot[\[Pi] Sqrt[-m^2 - x^2]] - Cot[a \[Pi] Sqrt[-m^2 - x^2]] + a \[Pi] Sqrt[-m^2 - x^2] Csc[a \[Pi] Sqrt[-m^2 - x^2]]^2) Gamma[ 1/2 (1 - 2 I x + y)] Gamma[   1/2 (1 + 2 I x + y)] Sinh[\[Pi] x]  , {x, 0, Infinity}]


And then Mathematica tells me this -

NIntegrate::izero : "Integral and error estimates are 0 on all \integration subregions. Try increasing the value of the MinRecursion \ option. If value of integral may be 0, specify a finite value for the \
AccuracyGoal option. !(*ButtonBox[\"[RightSkeleton]\", \ ButtonStyle->\"Link\", ButtonFrame->None, \ ButtonData:>\"paclet:ref/NIntegrate\", ButtonNote -> \ \"NIntegrate::izero\"])"

0.+ 0. I

• What does this mean?

• Even if I try to plot the above integrand as a function of $x$ (for say the above fixed values of m, a and y) then why is the plot not coming?

-
It means it got zero everywhere it checked and couldn't estimate the % error (can't have a % error of zero). If you don't think it's zero, then try out the suggestions. –  Michael E2 Mar 16 at 3:41
@MichaelE2 Actually I would be very happy if this integral somehow is $0$ for any value $m \in \mathbb{R}$, $a \in \mathbb{N}$ and $y=0$. But how do I be sure? –  user6818 Mar 16 at 7:13