# How I can plot the first terms of Taylor series arround $x=0$ of the below given function in the form of integrand?

I have tried to plot the first term of taylor expansion of the below function but I didn't come up to the plot . Any help , Where is the problem in my code ?
The Function is : $$I(x)=\int_{-\infty}^{x} \exp(-t^2 \operatorname {erfi}({(\sqrt{2\pi})t))}\operatorname {erf}({(\sqrt{2\pi})t)}) dt$$

Clear[\[Lambda], Ze, Z, ZTaylor];
\[Lambda] = NIntegrate[Exp[-t^2* Erf[(Sqrt[2*Pi])*t]* Erfi[(Sqrt[2*Pi])*t]],{t,-Infinity,Infinity}];
Ze[a_] := NIntegrate[Exp[-t^2* Erf[(Sqrt[2*Pi])*t]* Erfi[(Sqrt[2*Pi])*t]],{t,-Infinity,a}];
Z[x_] := Integrate[Exp[-t^2* Erf[(Sqrt[2*Pi])*t]* Erfi[(Sqrt[2*Pi])*t]], {t, 0, x}];
ZTaylor[n_][x_] := Series[Z[x], {x, 0, n}];
TaylorSeries == ZTaylor[11][x] //TraditionalForm
Plot[Evaluate[{Normal[ZTaylor[11][x]], Ze[x]}],
With[{x = 1},
Plot[Z[x] , {x, (x-(8x^5/5)), (
-32/405 (-45+4Pi^2)x^9)}, {x, -1, 2},
PlotRange -> {0, 1.5}, ImageSize -> 400,

]]]

• The integral $$I(x)=\int_{-\infty}^{x} \exp(-t^2 \operatorname {erfi}({(\sqrt{2\pi})t))}\operatorname {erf}({(\sqrt{2\pi})t)}) dt$$ does not evaluate by Mathematica 12. You might want to work on this part first before doing everything else. – Nasser Dec 8 '19 at 23:38
• But am sure that integral converge and its value for x go to infinty is 0.994.... – zeraoulia rafik Dec 8 '19 at 23:42
• I am sure you are right. I mean Mathematica does not know how to integrate it. Since you are calling Integrate on it as part of your logic, this step does not complete. – Nasser Dec 8 '19 at 23:44
• What I should do ? what command I should use for evaluation ? – zeraoulia rafik Dec 8 '19 at 23:45

You can inactivate your integral and then use Series:

s = Series[
Inactive[Integrate][
Exp[-t^2 Erfi[Sqrt[2 π]t] Erf[Sqrt[2 π]t]],
{t, -Infinity, x}
],
{x, 0, 10}
];
s // TeXForm


$$\int _{-\infty }^0e^{-t^2 \text{erf}\left(\sqrt{2 \pi } t\right) \text{erfi}\left(\sqrt{2 \pi } t\right)}dt+x-\frac{8 x^5}{5}+\frac{\left(1290240-114688 \pi ^2\right) x^9}{362880}+O\left(x^{11}\right)$$

You can numerically evaluate the inactive integral:

s /. i:_Inactive[__] :> N @ Activate[i] //TeXForm


$$0.497318+x-\frac{8 x^5}{5}+\frac{\left(1290240-114688 \pi ^2\right) x^9}{362880}+O\left(x^{11}\right)$$

• Ok, Thanks , what about plot ? – zeraoulia rafik Dec 9 '19 at 0:11
• sorry your code dosn't work at me in wolfram cloud , I don't know why ? – zeraoulia rafik Dec 9 '19 at 0:35
• @CarlWoll Very interesting approach, but the series expansion for x->Infinity doesn't look promising... – Ulrich Neumann Dec 9 '19 at 7:45
• @CarlWoll I tried AsymptoticIntegrate[ Exp[-t^2 Erfi[Sqrt[2 \[Pi]] t] Erf[Sqrt[2 \[Pi]] t]], {t, x, Infinity}, {x, Infinity, 1}]  but Mathematica doesn't evaluate. – Ulrich Neumann Dec 9 '19 at 8:09
• @Carlwoll yes , That present a probability density function , I want to get its asymptotic series – zeraoulia rafik Dec 9 '19 at 9:15