I have to  plot this trigonometric function:

  

   

     

        v1[y1_, y2_] := 7.6 NIntegrate[-(1/Sqrt[(x1 - y1)^2 + (-y2)^2]) 2/(4 Pi^2)
                      (0.71 Cos[a + ArcTan[y2/(y1 - x1)]]^2 Sin[a + ArcTan[y2/(y1 - x1)]] - 
                       0.5 Sin[a + ArcTan[y2/(y1 - x1)]]^3)/(Sin[a + ArcTan[y2/( y1 - x1)]]^4 
                      (Cot[a + ArcTan[y2/(y1 - x1)]]^2 - 0.26 + 0.51 I) 
                      (Cot[a + ArcTan[y2/(y1 - x1)]]^2 - 0.26 -0.51 I)) 1/Cos[a],
                      {a, 0, Pi}, {x1, 1 , 2}, MaxRecursion -> 4]

if I do a `ContourPlot`of the real part of the function like 

    ContourPlot[Re[v1[y1,y2]],{y1,0,5},{y2,0,5}]

I obtain something like this with some white voids

[![enter image description here][1]][1]


So, after reading [How to plot the contour of f[x,y]==0 if always f[x,y]>=0][2] I increased the `MaxRecursion` and `PlotPoint` (here only two of of those made), but the problem remain.

     ContourPlot[Re[v1[y1,y2]],{y1,0,5},{y2,0,5},MaxRecursion->1,PlotPoint->100]

[![enter image description here][3]][3]
 
   
    ContourPlot[Re[v1[y1,y2]],{y1,0,5},{y2,0,5},MaxRecursion->2,PlotPoint->20,Mesh->All]

[![enter image description here][4]][4]

Here I can see is a problem of the mesh. I tried many combination of `MaxRecursion` and `PlotPoints`, but I can't find a solution.I only have to increase more? Because I tried a MR->2, PP->100 but after 2 days Mathematica didn't give me a result. So I ask you if there is a method to parallelize it? 



  [1]: https://i.sstatic.net/VQ42g.jpg
  [2]: http://mathematica.stackexchange.com/questions/32734/how-to-plot-the-contour-of-fx-y-0-if-always-fx-y-0?lq=1
  [3]: https://i.sstatic.net/hbqgY.jpg
  [4]: https://i.sstatic.net/ZWuVa.jpg