Asymptotic of $I(x)=\int_{-\infty}^{x} \exp(-t^2 \operatorname {erfi}(t \sqrt{2\pi}) \operatorname {erf}(t \sqrt{2\pi})) dt$ for $x\to \infty$?

Question

Really I have tried to get any closed form of asymptotic series ( Taylor series ) of the below integrand function: $$I(x)=\int_{-\infty}^{x} \exp(-t^2 \operatorname {erfi}(t \sqrt{2\pi}) \operatorname {erf}(t \sqrt{2\pi})) dt$$ But I failed , I have used mathematica to evaluate its asymptotic but it dosn't Evaluate it , I have posted that function here to get help for plotting its asymptotic arround $x=0$ and for $x=\infty$ , But No result , Really the interesting thing about that function it present a probability density function because $I(+\infty)=0.994....$, Really I want to know if there is any approach to get its asymptotic series since Mathematica can't evaluate it ?

Note:The Motivation of this question is to know relationship between PDF of Studentdistribution and that function as shown in the below Plot , And also in the side of physics distribution of plasma physics

Answer 1

One can compute the Taylor series around $x=0$ the usual way. Compute the first few derivatives:

$$I^{(0)}(0) = \int_{-\infty}^{0} \exp \Bigr(-t^2 \operatorname {erfi}({t\sqrt{2\pi})}\operatorname {erf}({t\sqrt{2\pi}})\Bigr) dt \equiv I_0$$

$$I^{(1)}(0) = \exp \Bigr(-x^2 \operatorname {erfi}({x\sqrt{2\pi})}\operatorname {erf}({x\sqrt{2\pi}})\Bigr)\Bigr|_{x=0} = 1$$

$$I^{(2)}(0) = \exp \Bigr(-x^2 \operatorname {erfi}({x\sqrt{2\pi})}\operatorname {erf}({x\sqrt{2\pi}})\Bigr)\Bigr(\cdots\Bigr)\Bigr|_{x=0} = 0$$

etc. This gives the Taylor expansion

$$I(x) = I_0 + x + \cdots$$

where $I_0$ can be found numerically.

Answer 2

If we are concerned by the behavior around $x=0$, we could consider only the integral between $t=0$ and $t=x$.

Close to $t=0$, let us expand as a Taylor series $$-t^2 \text{erf}\left(\sqrt{2 \pi } t\right) \text{erfi}\left(\sqrt{2 \pi } t\right)=-8 t^4+O\left(t^{8}\right)$$ making $$\int_0^x\exp\Big(-t^2 \text{erf}\left(\sqrt{2 \pi } t\right) \text{erfi}\left(\sqrt{2 \pi } t\right)\Big)\,dt\sim\int_0^x\exp(-8t^4)\,dt=\frac{\Gamma \left(\frac{1}{4}\right)-\Gamma \left(\frac{1}{4},8 x^4\right)}{4\ 2^{3/4}}$$ which is almost $x-\frac{8 x^5}{5}+O\left(x^{9}\right)$.

For $x=0.1$, the approximate result is $0.09998400355$ while the exact value should be $0.09998400044$.

Answer 3

Let $f(t) = -t^2 \operatorname {erf}(t \sqrt {2 \pi}) \operatorname {erfi}(t \sqrt {2 \pi})$ and $x \to \infty$. The leading term for $I(x)$ is $\int_{\mathbb R} e^{f(t)} dt$. To approximate the remainder, make the change of variables $f(t) = f(x) + f'(x) u$. Then $t'(u) = 1 + O(|u|)$ for $u \to 0$. By Laplace's method, $$\int_x^\infty e^{f(t)} dt = e^{f(x)} \int_0^\infty t'(u) e^{f'(x) u} du \sim -\frac 1 {f'(x)} e^{f(x)}.$$ For the denominator, $f'(x) \sim -2 \sqrt 2 \, x^2 e^{2 \pi x^2}$.

Or you can apply integration by parts, same as here.