A closed form for $\int_0^\infty e^{-a\,x} \operatorname{erfi}(\sqrt{x})^3\ dx$ Let $\operatorname{erfi}(x)$ be the imaginary error function
$$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$
Consider the following parameterized integral
$$I(a)=\int_0^\infty e^{-a\,x} \operatorname{erfi}(\sqrt{x})^3\ dx.$$
I found some conjectured special values of $I(a)$ that are correct up to at least several hundreds of digits of precision:
$$I(3)\stackrel{?}{=}\frac{1}{\sqrt{2}},\ \ I(4)\stackrel{?}{=}\frac{1}{4\sqrt{3}}.$$


*

*Are these conjectured values correct?

*Is there a general closed-form formula for $I(a)$? Or, at least, are there any other closed-form special values of $I(a)$ for simple (e.g. integer or rational) values of $a$?

 A: Let $I = [0,1]$ and notice
$$\text{erfi}(t) = \frac{2}{\sqrt{\pi}} \int_0^t e^{z^2} dz \quad\implies\quad \text{erfi}(\sqrt{t}) = \frac{2}{\sqrt{\pi}}\sqrt{t} \int_I e^{tz^2} dz$$
The integral $\mathscr{I}$ we want can be rewritten as:
$$\begin{align}
\mathscr{I} 
= & \int_0^\infty e^{-at} \sqrt{t}^3 \left(\int_I e^{tz^2}dz\right)^3 dt\\
= & \frac{8}{\sqrt{\pi}^3}
\int_{I^3} dx dy dz\left[\int_0^\infty \sqrt{t}^3e^{-(a-x^2-y^2-z^2)t}) dt\right]\\
= & \frac{8}{\sqrt{\pi}^3} \Gamma(\frac52)
\int_{I^3} \frac{dx dy dz}{\sqrt{a - x^2 - y^2 - z^2}^5}\\
= & \frac{8}{\pi} \frac{\partial^2}{\partial a^2} 
\int_{I^3} \frac{dx dy dz}{\sqrt{a - x^2 - y^2 - z^2}}
\end{align}$$
Since the maximum value of $x^2 + y^2 + z^2$ on $I^3$ is $3$, above rewrite is valid
when $a > 3$.
To compute the integral, we will split the cube $I^3$ into 6 simplices according to the
sorting order of $x, y, z$. On any one of these simplices, say the one for
$1 \ge x \ge y \ge z \ge 0$, introduce parameters $(\rho,\lambda,\mu) \in I^3$ such that $(x, y, z) = (\rho,\rho\lambda,\rho\lambda\mu)$, we have:
$$\begin{align}
\mathscr{I}
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^3} 
\frac{\rho^2 \lambda d\rho d\lambda d\mu}{\sqrt{a - \rho^2 - \lambda^2 \rho^2 ( 1 + \mu^2})}\\
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^2} 
\frac{\rho^2 d\rho d\mu}{\rho^2(1+\mu^2)} \left[ - \sqrt{a - \rho^2 - \lambda^2 \rho^2 ( 1 + \mu^2}) \right]_{\lambda=0}^1\\
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^2} 
\frac{\rho^2 d\rho d\mu}{\rho^2(1+\mu^2)} \left[ \sqrt{a - \rho^2 } - \sqrt{a - \rho^2 ( 2 + \mu^2}) \right]\\
= & \frac{24}{\pi} \frac{\partial}{\partial a} \int_{I^2} \frac{d\rho d\mu}{1+\mu^2} 
\left[ 
\frac{1}{\sqrt{a - \rho^2 }} - 
\frac{\frac{1}{\sqrt{2+\mu^2}}
}{\sqrt{\frac{a}{2+\mu^2} - \rho^2})} 
\right]\\
= & \frac{24}{\pi} \frac{\partial}{\partial a} \int_{I} \frac{d\mu}{1+\mu^2} 
\left[ 
\arcsin(\frac{1}{\sqrt{a}}) -
\frac{1}{\sqrt{2+\mu^2}} \arcsin(\sqrt{\frac{2+\mu^2}{a}})
\right]\\
= & \frac{24}{\pi} \int_{I} \frac{d\mu}{1+\mu^2} 
\left[ 
\frac{-\frac{1}{2\sqrt{a}^3}}{\sqrt{1 - \frac{1}{a}}} -
\frac{-\frac{1}{2\sqrt{a}^3}}{\sqrt{1 - \frac{2+\mu^2}{a}}}
\right]\\
= & \frac{12}{\pi a } \int_{I} \frac{d\mu}{1+\mu^2} 
\left[ \frac{1}{\sqrt{a-2-\mu^2}} - \frac{1}{\sqrt{a-1}} \right]\\
\stackrel{\color{blue}{[1]}}{=} & \frac{12}{\pi a}
\left[
  \frac{1}{\sqrt{a-1}}\arctan \frac{\sqrt{a-1} \mu}{\sqrt{a - 2 -\mu^2}}  
- \frac{1}{\sqrt{a-1}} \arctan \mu
\right]_{\mu=0}^1
\\
= & \frac{12}{\pi a\sqrt{a-1}} \left[ \arctan\left(\sqrt{\frac{a-1}{a-3}}\right) - \frac{\pi}{4}
\right]\\
= & \frac{12}{\pi a\sqrt{a-1}}
\arctan\left(
\frac{\sqrt{\frac{a-1}{a-3}}-1}{\sqrt{\frac{a-1}{a-3}}+1}
\right)\\
\stackrel{\color{blue}{[2]}}{=} & 
\frac{6}{\pi a\sqrt{a-1}} \arctan\frac{1}{\sqrt{(a-1)(a-3)}}
\end{align}$$
Notes


*

*$\color{blue}{[1]}$ I am lazy, I get the leftmost integral in RHS from Wolfram alpha instead of deriving it myself.

*$\color{blue}{[2]}$ Let $u = \sqrt{\frac{a-1}{a-3}}$, we have
$$\begin{align}
2\arctan\frac{u-1}{u+1} 
= & \arctan\frac{2\frac{u-1}{u+1}}{1-\left(\frac{u-1}{u+1}\right)^2}
= \arctan\frac{u^2 - 1}{2u}\\
= & \arctan\frac{\frac{a-1}{a-3}-1}{2\sqrt{\frac{a-1}{a-3}}}
= \arctan\frac{1}{\sqrt{(a-1)(a-3)}}
\end{align}$$
A: I will use, on one hand, the intermediate result of Robert Israel:
$$I(a)=\frac{6}{a\sqrt{\pi}}\int_0^{\infty}e^{(1-a)x^2}\operatorname{erfi}^2 x\;dx,$$
and on the other hand, an evaluation from Prudnikov-Brychkov-Marychev (Vol.2, formula 2.8.19.8):
\begin{align}
J(p,b,c)&=\int_0^{\infty}e^{-px^2}\operatorname{erf}bx\operatorname{erf}cx\;dx=\\&=\frac{1}{\sqrt{\pi p}}\arctan\frac{bc}{\sqrt{p(b^2+c^2+p)}}.
\end{align}
Assuming that there are no surprises in the analytic continuation (this in principle can be worked out), the comparison of two formulas gives 
\begin{align}
I(a)&=-\frac{6}{a\sqrt{\pi}}\,J({a-1},i,i)=\\&=
\frac{6}{\pi a \sqrt{a-1}}\arctan\frac{1}{\sqrt{(a-1)(a-3)}}.
\end{align}
This is valid for $a>3$. The case $a=3$ can be treated by replacing $\arctan$ by its limiting value $\frac{\pi}{2}$.
A: I don't know, but you might substitute $\sqrt{x} = t$ to get 
$$ \int_0^\infty 2 \exp(-a t^2)\; \text{erfi}(t)^3 t\ dt$$
and for $a \ge 3$, integrate by parts:
$$ \dfrac{6}{a \sqrt{\pi}} \int_0^\infty \exp((1-a)t^2) \; \text{erfi}(t)^2\ dt$$
