In this post How to speed up the plot of NIntegrate? it was suggested that we can speed up the plotting of NIntegrate by using NDSolve instead. Can a similar way be done to plot this type of function
f[x_] := NIntegrate[Exp[-x y z - y^(1/2) - z^(1/2)], {y, 0, ∞}, {z, 0, ∞}]; Plot[f[x], {x, 0, 10}] // Timing How can I apply NDSolve to this function?
How can I apply
NDSolveto this function?
Do you want to apply NDSolve or do you want to plot the integral faster?
The plot can be produced ≈ 20 faster using appropriate option settings. Here is an example:
Clear[f2]; f2[x_?NumericQ] := NIntegrate[Exp[-x y z - y^(1/2) - z^(1/2)], {y, 0, Infinity}, {z, 0, Infinity}, PrecisionGoal -> 2, AccuracyGoal -> 3, Method -> {Automatic, "SymbolicProcessing" -> 0}]; Plot[f2[x], {x, 0, 10}, PlotRange -> All]; // AbsoluteTiming (* 0.694531 *) 
NIntegrate "skip" certain singularities. If the integrals have more or less the same shape then "en bloc" numerical integration can be used; see mathematica.stackexchange.com/a/126041 . $\endgroup$ Note: This is an extended comment.
This integral can be expressed in terms of the MeijerG special function:
16/π*MeijerG[{{1},{}},{{1,1,3/2,3/2},{}},1/(16*x)] Plotting using this expression takes about 6 seconds, compared to 12 seconds using OPs code. To obtain this expression, I first made a change of variables, replacing y by u/z where u is a new variable, and then evaluated the resulting integral
Integrate[Exp[-x*u-(u/z)^(1/2)-z^(1/2)]/z,{u,0,∞},{z,0,∞}, Assumptions->{x>0}] which takes about 10 seconds and gives the MeijerG result.
Having a symbolic result has advantages besides plotting, for example if OP wanted to know about $x \to \infty$ asymptotics, they could use
Series[16/π*MeijerG[{{1},{}},{{1,1,3/2,3/2},{}},1/(16*x)],{x,∞,1}] (* (-EulerGamma+Log[16]+Log[x]+2 PolyGamma[0,1/2])/x+O[1/x]^(3/2) *) 