# 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 $a$?

## 3 Answers

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}

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 $\frac{\pi}{2}$.

• The formula for $J$ feels like it should be within your capability to derive, no? – Ron Gordon Sep 22 '13 at 23:14
• @RonGordon Yes I think it is doable, especially knowing the answer. For example differentiate the left and right side with respect to $b$ and $c$. On the left we will get a gaussian integral, on the right an elementary function, so the comparison should be easy. Then it remains to fix the "integration constants", but I was too lazy for this. – Start wearing purple Sep 22 '13 at 23:23

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: $$\dfrac{6}{a \sqrt{\pi}} \int_0^\infty \exp((1-a)t^2) \; \text{erfi}(t)^2\ dt$$

• Did you get the final result ?? – what'sup Sep 22 '13 at 19:34
• I think this is more of a comment than an answer, unless you have worked out a valid result with this hint. – Ron Gordon Sep 22 '13 at 19:54