Deje $I = [0,1]$ y aviso
$$\text{erfi}(t) = \frac{2}{\sqrt{\pi}} \int_0^t e^{z^2} dz \quad\implies\quad \text{erfi}(\sqrt{t}) = \frac{2}{\sqrt{\pi}}\sqrt{t} \int_I e^{tz^2} dz$$
La integral de la $\mathscr{I}$ queremos puede ser reescrita como:
$$\begin{align}
\mathscr{I}
= & \int_0^\infty e^{-at} \sqrt{t}^3 \left(\int_I e^{tz^2}dz\right)^3 dt\\
= & \frac{8}{\sqrt{\pi}^3}
\int_{I^3} dx dy dz\left[\int_0^\infty \sqrt{t}^3e^{-(a-x^2-y^2-z^2)t}) dt\right]\\
= & \frac{8}{\sqrt{\pi}^3} \Gamma(\frac52)
\int_{I^3} \frac{dx dy dz}{\sqrt{a - x^2 - y^2 - z^2}^5}\\
= & \frac{8}{\pi} \frac{\partial^2}{\partial a^2}
\int_{I^3} \frac{dx dy dz}{\sqrt{a - x^2 - y^2 - z^2}}
\end{align}$$
Puesto que el valor máximo de $x^2 + y^2 + z^2$$I^3$$3$, por encima de reescritura es válido
al $a > 3$.
Para calcular la integral, vamos a dividir el cubo de $I^3$ a 6 simplices de acuerdo a la
orden de clasificación de $x, y, z$. En cualquiera de estos simplices, dicen el uno para
$1 \ge x \ge y \ge z \ge 0$, introducir los parámetros de $(\rho,\lambda,\mu) \in I^3$ tal que $(x, y, z) = (\rho,\rho\lambda,\rho\lambda\mu)$, tenemos:
$$\begin{align}
\mathscr{I}
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^3}
\frac{\rho^2 \lambda d\rho d\lambda d\mu}{\sqrt{a - \rho^2 - \lambda^2 \rho^2 ( 1 + \mu^2})}\\
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^2}
\frac{\rho^2 d\rho d\mu}{\rho^2(1+\mu^2)} \left[ - \sqrt{a - \rho^2 - \lambda^2 \rho^2 ( 1 + \mu^2}) \right]_{\lambda=0}^1\\
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^2}
\frac{\rho^2 d\rho d\mu}{\rho^2(1+\mu^2)} \left[ \sqrt{a - \rho^2 } - \sqrt{a - \rho^2 ( 2 + \mu^2}) \right]\\
= & \frac{24}{\pi} \frac{\partial}{\partial a} \int_{I^2} \frac{d\rho d\mu}{1+\mu^2}
\left[
\frac{1}{\sqrt{a - \rho^2 }} -
\frac{\frac{1}{\sqrt{2+\mu^2}}
}{\sqrt{\frac{a}{2+\mu^2} - \rho^2})}
\right]\\
= & \frac{24}{\pi} \frac{\partial}{\partial a} \int_{I} \frac{d\mu}{1+\mu^2}
\left[
\arcsin(\frac{1}{\sqrt{a}}) -
\frac{1}{\sqrt{2+\mu^2}} \arcsin(\sqrt{\frac{2+\mu^2}{a}})
\right]\\
= & \frac{24}{\pi} \int_{I} \frac{d\mu}{1+\mu^2}
\left[
\frac{-\frac{1}{2\sqrt{a}^3}}{\sqrt{1 - \frac{1}{a}}} -
\frac{-\frac{1}{2\sqrt{a}^3}}{\sqrt{1 - \frac{2+\mu^2}{a}}}
\right]\\
= & \frac{12}{\pi a } \int_{I} \frac{d\mu}{1+\mu^2}
\left[ \frac{1}{\sqrt{a-2-\mu^2}} - \frac{1}{\sqrt{a-1}} \right]\\
\stackrel{\color{blue}{[1]}}{=} & \frac{12}{\pi a}
\left[
\frac{1}{\sqrt{a-1}}\arctan \frac{\sqrt{a-1} \mu}{\sqrt{a - 2 -\mu^2}}
- \frac{1}{\sqrt{a-1}} \arctan \mu
\right]_{\mu=0}^1
\\
= & \frac{12}{\pi a\sqrt{a-1}} \left[ \arctan\left(\sqrt{\frac{a-1}{a-3}}\right) - \frac{\pi}{4}
\right]\\
= & \frac{12}{\pi a\sqrt{a-1}}
\arctan\left(
\frac{\sqrt{\frac{a-1}{a-3}}-1}{\sqrt{\frac{a-1}{a-3}}+1}
\right)\\
\stackrel{\color{blue}{[2]}}{=} &
\frac{6}{\pi a\sqrt{a-1}} \arctan\frac{1}{\sqrt{(a-1)(a-3)}}
\end{align}$$
Notas
- $\color{blue}{[1]}$ Soy perezoso, me sale más a la izquierda de la integral en el lado derecho de Wolfram alpha, en lugar de derivar a mí mismo.
-
$\color{blue}{[2]}$ Deje $u = \sqrt{\frac{a-1}{a-3}}$, tenemos
$$\begin{align}
2\arctan\frac{u-1}{u+1}
= & \arctan\frac{2\frac{u-1}{u+1}}{1-\left(\frac{u-1}{u+1}\right)^2}
= \arctan\frac{u^2 - 1}{2u}\\
= & \arctan\frac{\frac{a-1}{a-3}-1}{2\sqrt{\frac{a-1}{a-3}}}
= \arctan\frac{1}{\sqrt{(a-1)(a-3)}}
\end{align}$$
