How to give an approximation expression of a complicated expression using Mathematica?

Question

I came across a very complicated expression as the first following picture shows, it makes normal processing such as coordinate conversion difficult, so I thought it would be nice to use an approximation of it, but I also met some difficulties when using Series[] and NIntegrate[].

so So what methods do you usually use with Mathematica to deal with/simplify this complex expression?

The above is a sub-problem that I think can solve the problem, and my final problem is this(second picture shows)：i want plot streamline of {vr,vθ} but failed due to the complexity of the vθ.

code if you need

H=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), 
FractionBox[
SuperscriptBox[\(r\), \(2\)], \(2\)]]\(
\*SuperscriptBox[\(E\), \((\(-s\)\  + \ 3 \(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]
\*FractionBox[\(1 - 
\*SuperscriptBox[\(E\), \(-\[Tau]\)]\), \(\[Tau]\)] \[DifferentialD]\[Tau]\))\)] \[DifferentialD]s\)\)
Series[H,{r,0,3}]

code if you need

v\[Theta]=\[CapitalGamma]0/(2 \[Pi] r) \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Eta]\)]\(
\*SuperscriptBox[\(E\), \((\(-s\)\  + \ 3 \(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]
\*FractionBox[\(1 - 
\*SuperscriptBox[\(E\), \(-\[Tau]\)]\), \(\[Tau]\)] \[DifferentialD]\[Tau]\))\)] \[DifferentialD]s\)\)/H\[Infinity]; (*a function of r*)
vr=-a r + (6\[Nu])/r (1-E^(-((a r^2)/(2 \[Nu])))); (*a function of r*)


\[Eta]=(a r^2)/(2 \[Nu]);
H\[Infinity]=37.905;


\[CapitalGamma]0=1;a=1;\[Nu]=1;
{vx,vy}=TransformedField["Polar"->"Cartesian",{vr,v\[Theta]},{r,\[Theta]}->{x,y}]; (*won't work out*)
StreamPlot[{vx,vy},{x,-5,5},{y,-5,5}] 

---

@Mariusz Iwaniuk @Mauricio Fernández, Thanks for your answers, but still a little confusion: indeed before I asked here, I had tried to use Assume[] according to the hint Mathematica gives, below picture&code is my trial but failed, the only difference between your and mine is the condition Assume[] use, but why r∈Reals will fail?

the code of the picture:

Clear["Global`*"]

a=1;
\[Nu]=1;
H=Assuming[r\[Element]Reals,\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), 
FractionBox[\(a\ 
\*SuperscriptBox[\(r\), \(2\)]\), \(2\ \[Nu]\)]]\(
\*SuperscriptBox[\(E\), \((\(-s\)\  + \ 3 \(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]
\*FractionBox[\(1 - 
\*SuperscriptBox[\(E\), \(-\[Tau]\)]\), \(\[Tau]\)] \[DifferentialD]\[Tau]\))\)] \[DifferentialD]s\)\)];
Assuming[r\[Element]Reals,Series[H,{r,0,3}]]

Answer 1

Using comment of user: Mauricio Fernández:

H\[Infinity] = 37.905;
\[CapitalGamma]0 = 1;
a = 1;
\[Nu] = 1;
\[Eta] = (a r^2)/(2 \[Nu]);
f[r_] := Integrate[Exp[-s + 3*Integrate[(1 - Exp[-tau])/tau, {tau, 0, s}]], {s, 0, \[Eta]}] ;
integral = Normal@Assuming[r > 0, Series[f[r], {r, 0, 20}]]

v\[Theta] = \[CapitalGamma]0/(2 \[Pi] r) *(integral)/H\[Infinity]; (*a function of r*)
vr = -a r + (6 \[Nu])/
 r (1 - E^(-((a r^2)/(2 \[Nu])))); (*a function of r*)
{vx, vy} = TransformedField["Polar" -> "Cartesian", {vr, v\[Theta]}, {r, \[Theta]} -> {x, y}]; 
StreamPlot[{vx, vy}, {x, -5, 5}, {y, -5, 5}] 

Answer 2

Try the following: First, let us observe that one calculates the inner integral precisely:

Assuming[{s > 0}, Integrate[(1 - Exp[-t])/t, {t, 0, s}]]

(*  EulerGamma + Gamma[0, s] + Log[s]  *)

Thus, one only needs to calculate the integral of Exp[-s + EulerGamma + Gamma[0, s] + Log[s]. The latter cannot be calculated precisely. Then let us first do this numerically:

lst = ParallelTable[{r, 
    NIntegrate[
     Exp[-s + EulerGamma + Gamma[0, s] + Log[s]], {s, 0, r^2/2}]}, {r,
     0, 10, 0.1}];

Now we see two regimes of behavior with a crossover between them (look at the image below). At small r, one can make a polynomial approximation:

model = k1*r^(7/4) + k2*r^(37/20);
ff = FindFit[lst, model, {k1, k2}, r]
(*   {k1 -> 1.78231, k2 -> -1.39256}   *)

while at r>4, one has the horizontal asymptote

int=2.325...

Altogether, one finds this:

Show[{
  ListPlot[lst, AxesLabel -> {"r", "int"}],
  Plot[model /. ff, {r, 0, 1}, PlotStyle -> Red],
  Plot[2.325, {r, 8, 10}, PlotStyle -> Green]
  }]

yielding the following:

Have fun!