Plotting singular integral

Question

I need to plot the following singular double integration $u(x,t)$ for $t=0, t=0.2, t=0.4, t=1$ and $t=2$ in one figure as a line. That is, I need $u$-coordinate vertically and $x$-coordinate horizontally.

where $f(y)= \exp(-(y-4.68)^2/0.4)$. Due to singularities, I cannot plot $u(x,t)$ for $x=0..15$. I checked the NIntegrate Integration Strategies website. But, I could not manage.

t = 0.2;
m = 0.2;
f[y_] := Exp[-(y - 4.68)^2/0.4]
u[x_, t_, m_] := 
 NIntegrate[
  f[y] (2. Sin[(x^2 + y^2)/(8. (t - r)) + \[Pi]/4] Cos[( x y)/(
      4. (t - r))]) (Exp[-(x + y)^2/(8. r)] (x + y - 4. m r)/(
      4. \[Pi] r Sqrt[r (t - r)]) + (m^2 Sqrt[2.])/Sqrt[\[Pi] (t - r)]
       Exp[m (x + y) + 2. m^2 r] Erfc[(x + y + 4. m r)/(
       2. Sqrt[2 r])]), {y, 0., Infinity}, {r, 0, t}]

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What do you mean by "plot the following singular double integration"? Do you want to plot the sampling points of the integration process? Do the sampling points plots in the accepted answer of this question show what you want? — Anton Antonov, Nov 27 '15 at 21:58
@AntonAntonov sorry for late answer. Actually, I want to plot u(x,t) for t=0, t=0.2, t=0.4, t=1 and t=2 in one figure as a line. That is, I need u-coordinate vertically and x-coordinate horizontally(x from 0 to 15). — Wita, Nov 28 '15 at 13:45
@2che Ok, I see. Please update the question with the clarification in your comment. — Anton Antonov, Nov 29 '15 at 01:41
@2che I understand the motivation for the question better now. I run this calculation code Table[{t, x, u[x, t, 0.2]}, {t, {0, 0.2, 0.4, 1, 2}}, {x, 0, 15, 0.1}] overnight and after 10 hours it was still not finished. (Using PrecisionGoal->3.) — Anton Antonov, Nov 29 '15 at 15:46
@AntonAntonov Thank you for trying again. I am surprise to hear that it is still running ...I cannot imagine to run the codes for all different times together! — Wita, Nov 29 '15 at 16:20

score 3 · Answer 1 · answered Nov 27 '15 at 22:32

Here is a way to plot the sampling points of the integration process. (If this is what it is asked.)

First using the functions Sow and Reap and the option EvaluationMonitor we gather the sampling points:

Block[{x = 0.2, t = 0.2, m = 0.2}, 
 res = Reap[
    NIntegrate[
     f[y] (2. Sin[(x^2 + y^2)/(8. (t - r)) + \[Pi]/
           4] Cos[(x y)/(4. (t - r))]) (Exp[-(x + y)^2/(8. r)] (x + y - 
            4. m r)/(4. \[Pi] r Sqrt[r (t - r)]) + (m^2 Sqrt[2.])/
          Sqrt[\[Pi] (t - r)] Exp[
          m (x + y) + 2. m^2 r] Erfc[(x + y + 4. m r)/(2. Sqrt[2 r])]), {y, 
      0., Infinity}, {r, 0, t}, PrecisionGoal -> 3, 
     EvaluationMonitor :> Sow[{y, r}]]];
 ]

Note that I have decreased the precision goal.

Here we separate the integral estimate from the gathered sampling points:

integralEstimate = res[[1]];
samplingPoints = res[[2, 1]];

Because of the coordinates magnitudes the following plot takes the logarithms of the sampling points coordinates.

Graphics[{Point[
   samplingPoints /. {x_?NumericQ, y_?NumericQ} :> {Log[x + 0.0001], 
      Log[y]}]}, AspectRatio -> 1, Frame -> True]

If we add a z-axis we can see the order in which the points are sampled.

k = 0;
Graphics3D[
 Point[{samplingPoints /. {x_?NumericQ, 
      y_?NumericQ} :> {Log[x + 0.0001], Log[y], k++}}], 
 BoxRatios -> {1, 1, 1}, Axes -> True]

(It will help rotating the 3D plot in Mathematica to get a better impression of the points distribution.)

Plotting singular integral

1 Answers1