I want to solve any of the two integrals for the complex number $a$ \begin{aligned} I_1 & = \int\limits_{0}^{\infty} xe^{a/x^2 - x^2}\text{Erfi}(x/\sqrt{2}) dx\\ I_2 & = \int\limits_{0}^{\infty} \frac{1}{x}e^{a/x^2 - x^2}\text{Erfi}(x/\sqrt{2}) dx, \end{aligned} where \begin{aligned} a & \in \mathbb{C}\\ \text{Re}[a] & < 0. \end{aligned}
Solving one integral, I can get the other through the relation $$I_2 = \frac{d I_1}{da}.$$
My approach
I can not get an answer out of Mathematica. Below are some of the approaches I tried.
a) Integration by parts
I can simplify the integrand enormously by taking the derivative of Erfi, which can be done by integration of parts. I know that the definite integral $\int_{0}^{\infty}e^{a/z^2-z^2}z^n dz$ is related to the Bessel functions, but I need the indefinite integral when doing partial integration, which I can not find.
b) Using a known integral
I know that
\begin{aligned}
\int\limits_{0}^{\infty}e^{1/x^2 -b^2x^2}\text{Erfc}(\frac{1}{x})dx & = \frac{1}{b\sqrt{\pi}}\left[\sin(2b\cdot\text{Ci}(2b))-\cos(2b\cdot\text{si}(2b))\right] \\
\int\limits_{0}^{\infty}e^{1/x^2 -b^2x^2}\text{Erfc}(\frac{1}{x})xdx & = \frac{\pi}{2b}\left[H_1(2b) - Y_1(2b)-\frac{1}{b}\right], \quad |\text{arg}\,b| < \frac{\pi}{4} \\
\int\limits_{0}^{\infty}e^{1/x^2 -b^2x^2}\text{Erfc}(\frac{1}{x})\frac{dx}{x} & = \frac{\pi}{2}\left[H_0(2b) - Y_0(2b)\right], \quad |\text{arg}\,b| < \frac{\pi}{4} \\
\end{aligned}
where Ci and si are the Cosine and Sine integrals, and $H$ is the Struve function. What I try to do is to make the variable substitution $z\to\frac{1}{z}$, then to turn Erfi into Erf (and Erfc) with the same argument as in the above equation $z \to \frac{iz}{\sqrt{2}}$. I can not figure out the limits however when doing this last substitution.
c) Rewriting the integrand
I can rewrite the integrals as
\begin{aligned}
I_1 & = \int\limits_{0}^{\infty} e^{a(1/x - x)}\text{Erfi}(\sqrt{\sqrt{a}x/2}) dx\\
I_2 & = \int\limits_{0}^{\infty} \frac{1}{x}e^{a(1/x - x)}\text{Erfi}(\sqrt{\sqrt{a}x/2}) dx.
\end{aligned}