calculus - Evaluate $int_{0}^{infty} !frac{1-e^{-2x^2}}{xe^{x^4}}mathrm{d}x$

How one can evaluate the following improper integral
\begin{equation}
\int_{0}^{\infty} \!\frac{1-e^{-2x^2}}{xe^{x^4}}\mathrm{d}x
\end{equation}
It seems that one can consider the function
\begin{equation}
I(\alpha)=\int_{0}^{\infty} \!\frac{1-e^{-\alpha x^2}}{xe^{x^4}}\mathrm{d}x

\end{equation}
and try to evaluate $I'(\alpha)=\int_{0}^{\infty} xe^{-\alpha x^2-x^4}\mathrm{d}x$. But this approach did not work for me.

Answer

$$
\int_0^\infty \frac{1-e^{-2x^2}}{xe^{x^4}}dx=\int_0^\infty \frac{1-e^{-2u}}{2ue^{u^2}}du
=\int_0^\infty e^{-u^2} du \sum_{k=0}^\infty(-1)^k \frac{(2u)^k}{(k+1)!}\\
=\sum_{k=0}^\infty(-1)^k\frac{2^{k-1}\Gamma\left(\frac{k+1}{2}\right)}{(k+1)!}
=\frac{\sqrt\pi}2\sum_{n=0}^\infty\frac1{(2n+1)n!}-\sum_{n=0}^\infty\frac{2^n}{(n+1)(2n+1)!!}\\
=\frac{\sqrt\pi}2 {}_1F_1\left(\small\frac12;\small\frac32;1\right)-\frac12 {}_2F_2\left(1,1;\small\frac32,2;1\right).
$$

According to WA, the first summand can be represented as:
$$
\frac\pi4 \text{Erfi}(1).
$$

Further simplification seems to be not possible.