On the evaluation of $\int_0^\infty e^{-x^2} \sin \left (\frac{1}{4x^2} \right ) \, dx$ When attempting to find an alternative solution to this question here, which called for the evaluation of the integral
$$\int_0^\infty \frac{x \sin x}{1 + x^4} \, dx$$
using real methods, I ran up against the following improper integral
$$\int_0^\infty e^{-x^2} \sin \left (\frac{1}{4 x^2} \right ) \, dx. \tag1$$
A value for this improper integral can be found. It is
$$\frac{\sqrt{\pi}}{2} \exp \left (-\frac{1}{\sqrt{2}} \right ) \sin \left (\frac{1}{\sqrt{2}} \right ),$$
and is what I am having trouble in finding.
I have tried a number of the various tricks one typically employs when attempting to find such integrals such as Feynman trick of differentiating under the integral sign, series solution, and so on, all to no avail (perhaps I missed something here).
One method that looked promising was to use properties for the (inverse) Laplace transform, namely
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty (\mathcal{L} f)(s) \cdot (\mathcal{L}^{-1} g)(s) \, ds.$$
After enforcing a change of variable $x \mapsto \dfrac{1}{2 \sqrt{x}}$ we have
\begin{align*}
\int_0^\infty e^{-x^2} \sin \left (\frac{1}{4 x^2} \right ) \, dx &= \frac{1}{4} \int_0^\infty \frac{e^{-1/(4x)}}{x^{3/2}} \cdot \sin x \, dx\\
&= \int_0^\infty \mathcal{L} \{\sin x\} \cdot \mathcal{L}^{-1} \left \{\frac{e^{-1/(4x)}}{x^{3/2}} \right \} \, ds\\
&= \frac{1}{4} \int_0^\infty \frac{1}{s^2 + 1} \cdot \frac{2 \sin (\sqrt{s})}{\sqrt{\pi}} \, ds\\
&= \frac{1}{\sqrt{\pi}} \int_0^\infty \frac{s \sin s}{1 + s^4} \, ds, \tag2
\end{align*}
where in the last line a substitution of $s \mapsto s^2$ has been made.
While this is a perfectly valid approach the only problem is the integral one ends up with in (2) is exactly the integral one started out with.
So my question is

Is it possible to evaluate the integral given in (1) using real methods that does not depend on evaluating the integral given in (2)?

 A: Following tired's suggestion, the problem is equivalent to finding
$$\text{Im}\int_{0}^{+\infty}\exp\left(-x^2+\frac{i}{4x^2}\right)\,dx =\frac{1}{2}\text{Im}\int_{-\infty}^{+\infty}\exp\left(-\left(x-\frac{1-i}{2\sqrt{2}\,x}\right)^2-\frac{1-i}{\sqrt{2}}\right)\,dx$$
and through a rotation of the integration line this can be easily computed through the Cauchy-Schlömilch substitution, the best known instance of Glasser's master theorem:
$$\forall a\in\mathbb{R}^+,\qquad \int_{0}^{+\infty}\exp\left(-x^2-\frac{a}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2}\,e^{-2\sqrt{a}}.$$
A: Considering $$ I=\int e^{-x^2} \cos \left (\frac{1}{4 x^2} \right ) \, dx\qquad  \qquad J=\int e^{-x^2} \sin \left (\frac{1}{4 x^2} \right ) \, dx$$
$$K=I+iJ=\int e^{-x^2+\frac{i}{4 x^2}}\,dx\qquad  \qquad \qquad  \qquad L=I-iJ=e^{-x^2-\frac{i}{4 x^2}}\,dx$$
$$-x^2+\frac{i}{4 x^2}=-\left(x^2-\frac{i}{4 x^2} \right)=-\left(x+\frac{\sqrt{-i}}{2 x} \right)^2+\sqrt{-i}$$
$$-x^2-\frac{i}{4 x^2}=-\left(x^2+\frac{i}{4 x^2} \right)=-\left(x+\frac{\sqrt{i}}{2 x} \right)^2-\sqrt{i}$$
All of that makes
$$K=\frac{ \sqrt{\pi }}{4} \left(e^{-(-1)^{3/4}}
   \left(-\text{erf}\left(\frac{(-1)^{3/4}}{2 x}-x\right)-1\right)+e^{(-1)^{3/4}}
   \left(\text{erf}\left(x+\frac{(-1)^{3/4}}{2 x}\right)+1\right)\right)$$
$$L=\frac{\sqrt{\pi }}{4}  \left(e^{-\sqrt[4]{-1}}
   \left(1-\text{erf}\left(\frac{\sqrt[4]{-1}}{2 x}-x\right)\right)+e^{\sqrt[4]{-1}}
   \left(\text{erf}\left(x+\frac{\sqrt[4]{-1}}{2 x}\right)-1\right)\right)$$ and then $I$ and $J$ and finally your result since 
$J=\frac{1}{2} i (L-K)$
Edit
Using Wolfram Alpha
$$\int_0^\infty e^{-x^2+\frac{a}{x^2}}\,dx=\frac{\sqrt{\pi }}{2}  e^{-2 \sqrt{-a}}$$
A: $$
\begin{align}
\int_0^\infty e^{-x^2-\frac i{4x^2}}\,\mathrm{d}x
&=\frac{i^{1/4}}{\sqrt2}\int_0^\infty e^{-\frac{i^{1/2}}2\left(x^2+\frac1{x^2}\right)}\,\mathrm{d}x\tag1\\
&=\frac{i^{1/4}}{\sqrt2}e^{-i^{1/2}}\int_0^\infty e^{-\frac{i^{1/2}}2\left(x-\frac1x\right)^2}\,\mathrm{d}x\tag2\\
&=\frac{i^{1/4}}{\sqrt2}e^{-i^{1/2}}\int_0^\infty e^{-\frac{i^{1/2}}2\left(x-\frac1x\right)^2}\frac1{x^2}\,\mathrm{d}x\tag3\\
&=\frac{i^{1/4}}{2\sqrt2}e^{-i^{1/2}}\int_{-\infty}^\infty e^{-\frac{i^{1/2}}2u^2}\,\mathrm{d}u\tag4\\
&=\frac{\sqrt\pi}2e^{-i^{1/2}}\tag5\\[3pt]
&=\frac{\sqrt\pi}2e^{-\frac1{\sqrt2}}\left(\cos\left(\frac1{\sqrt2}\right)-i\sin\left(\frac1{\sqrt2}\right)\right)\tag6
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto\frac{i^{1/4}}{\sqrt2}x\quad$(this requires Cauchy's Theorem)
$(2)$: multiply by $e^{-i^{1/2}}e^{i^{1/2}}$
$(3)$: substitute $x\mapsto\frac1x$
$(4)$: average $(2)$ and $(3)$
$(5)$: evaluate integral$\quad$(this requires Cauchy's Theorem)
$(6)$: expand complex exponential
Both of the applications of Cauchy's Theorem above use the contour
$$
[0,R]\cup\left[Re^{i\left[0,\frac\pi8\right]}\right]\cup\left[Re^{i\frac\pi8},0\right]
$$
Looking at the imaginary parts of $(6)$, we get
$$
\int_0^\infty e^{-x^2}\sin\left(\frac1{4x^2}\right)\,\mathrm{d}x=\frac{\sqrt\pi}2e^{-\frac1{\sqrt2}}\sin\left(\frac1{\sqrt2}\right)\tag7
$$
