# Plotting singular integral

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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• People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful – Michael E2 Nov 27 '15 at 17:58
• – user9660 Nov 27 '15 at 18:10
• @Lou Thank you I will have a look at them. – Wita Nov 27 '15 at 18:31
• 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

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.)