Numerical Landau Damping dispersion relation, Principal value numerical integral

For NIntegrate all parameters must have numerical values.

Gaussian = Exp[-(((v - V0)/VT)^2)/2]/(Sqrt[2*Pi]*VT);

integrand[v_, V0_, VT_, k_, w_] = D[Gaussian, v]/(w/k - v) //   Simplify

(*   (E^(-((v - V0)^2/(2 VT^2))) k (v - V0))/(Sqrt[2 \[Pi]] VT^3 (k v - w))   *)


nint[V0_, VT_, k_, w_] := 
  NIntegrate[integrand[v, V0, VT, k, w], {v, -Infinity, Infinity}, 
    Method -> "PrincipalValue", Exclusions -> (k v - w) == 0]

nint[1, 2, 3, 4]

(*   0.24312   *)

Have a look at Normaldistribution and central moment. Than take a look at statistical dispersion. Return to Mathematica documentation: descriptive statistics. They more often towards the reverse situation of given measurement values. This question is targeted towards a fundamental situation for example in plasmas with plane waves. There is not much interest in solving the equation. There is a historical discussion between Landau and Pavlov about the physicality of this dispersion. Pavlov did take a realistics position, Landau an antirealistic one with a preference for pedagogical reason.

There has been a general question group regarding this question. How to deal with complicated Gaussian integrals in Mathematica?.

Form there I took by chance a function for the calculation of this dispersion:

gaussMoment[fPre_, fExp_, vars_] := 
 Module[{coeff, dist, ai, \[Mu], norm}, 
  coeff = CoefficientArrays[fExp, vars, "Symmetric" -> True];
  ai = Inverse[2 coeff[[3]]];
  \[Mu] = -ai.coeff[[2]];
  dist = MultinormalDistribution[\[Mu], -ai];
  norm = 1/PDF[dist, vars] /. Thread[vars -> \[Mu]];
  Simplify[
   norm Exp[1/2 coeff[[2]].\[Mu] + coeff[[1]]] Distribute@
     Expectation[fPre, vars \[Distributed] dist]]]

The moments from there as example for consistency:

RepeatedTiming[gaussMoment[(x^2 + x^4 + x^6), -(x - 1)^2, {x}]]

{0.0028, (223 Sqrt[\[Pi]])/8}

For this question:

RepeatedTiming[gaussMoment[((x - q)^(-1)), -(x - 1)^2, {x}]]

{10.2, Sqrt[\[Pi]]
   Expectation[
   1/(-q + x), {x} \[Distributed] 
    MultinormalDistribution[{1}, {{1/2}}]]}

This is very fast and safe. I think they probably took their formulas from Matlab. In Matlab, more application is in concern for such integrals. As stated earlier in other question this integration can be done nicely numerical for a general parametric solution.

With Your parameters in my suggested integration method:

D[Exp[-(((v - V0)/VT)^2)/2]/(Sqrt[2*Pi]*VT), v]

-((E^(-((v - V0)^2/(2 VT^2))) (v - V0))/(Sqrt[2 \[Pi]] VT^3))

RepeatedTiming[
  gaussMoment[(((v - V0)/(w/k - v))), -(((v - V0)/VT)^2)/
    2, {x}]]/(Sqrt[2 \[Pi]] VT^3)

The result is

{0.00054/VT^3, (E^(
    1/2 (-((v - V0)^2/
        VT^2) + {-((v - V0)^2/(2 VT^2))}[[
         2]].(-Inverse[
            2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(
              2 VT^2))}[[2]])))
     Expectation[(k (-v + V0))/(
     k v - w), {x} \[Distributed] 
      MultinormalDistribution[-Inverse[
          2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(2 VT^2))}[[
          2]], -Inverse[2 {-((v - V0)^2/(2 VT^2))}[[3]]]]])/(Sqrt[
    2 \[Pi]] VT^3 PDF[
     MultinormalDistribution[-Inverse[
         2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(2 VT^2))}[[
         2]], -Inverse[
        2 {-((v - V0)^2/(2 VT^2))}[[3]]]], {-Inverse[
         2 {-((v - V0)^2/(2 VT^2))}[[3]]].{-((v - V0)^2/(2 VT^2))}[[
         2]]}])}