Making the subsution $u = x^{2}$ and using a trig identity, we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{e^{-x^{2}} \sin^{2}(x^{2})}{x^{2}} \, \mathrm dx &= 2 \int_{0}^{\infty} \frac{e^{-x^{2}} \sin^{2}(x^{2})}{x^{2}} \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{e^{-u}\sin^{2}(u)}{u^{3/2}} \, \mathrm du \\ & = \frac{1}{2} \int_{0}^{\infty} \frac{e^{-u} \left(1-\cos(2u) \right)}{u^{3/2}} \, \mathrm du. \end{align}$$
For $\Re(s) >0$, we have $$ \begin{align} \int_{0}^{\infty} e^{-u} \left(1- \cos(2u) \right) u^{s-1} \, \mathrm du &= \int_{0}^{\infty}e^{-u} u^{s-1} \, \mathrm du - \int_{0}^{\infty}e^{-u} \cos(2u) u^{s-1} \, \mathrm du \\ &\overset{(1)}{=} \Gamma(s) - \frac{\Gamma(s) \cos \left(s \arctan 2 \right)}{5^{s/2}} \\ &= \Gamma(s) \left(1-\frac{\cos \left(s \arctan 2 \right)}{5^{s/2}} \right). \end{align}$$
The Mellin transform $$\int_{0}^{\infty} e^{-u} \left(1- \cos(2u) \right) u^{s-1} \, \mathrm du $$ defines an analytic function for $\Re(s) >-2$.
And the function $$\Gamma(s) \left(1-\frac{\cos \left(s \arctan 2 \right)}{5^{s/2}} \right) $$ can be analytically continued to $\Re(s) >-2$ since it has removable singularities at $s=0$ and $s=-1$.
Therefore, by the identity theorem, $$\int_{0}^{\infty} e^{-u} \left(1- \cos(2u) \right) u^{s-1} \, \mathrm du = \Gamma(s) \left(1-\frac{\cos \left(s \arctan 2 \right)}{5^{s/2}} \right) \, , \quad \Re(s) >-2.$$
So we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{e^{-x^{2}} \sin^{2}(x^{2})}{x^{2}} \, \mathrm dx &= \frac{1}{2} \, \Gamma \left(- \frac{1}{2} \right) \left(1- 5^{1/4} \cos \left(\frac{1}{2} \, \arctan 2 \right)\right) \\ &= -\sqrt{\pi} \left(1-5^{1/4} \sqrt{\frac{1+ \cos(\arctan 2)}{2}} \right) \\ &= -\sqrt{\pi} \left(1-5^{1/4} \sqrt{\frac{1+\frac{1}{\sqrt{5}}}{2}} \right) \\ &= - \sqrt{\pi} \left(1- \sqrt{\frac{\sqrt{5}+1}{2}}\right) \\ &= \sqrt{\pi} \left(\sqrt{\phi} - 1 \right). \end{align}$$
$(1)$ Prove that $\int_{0}^{+\infty} u^{s-1} \cos (a u) \:e^{-b u}\:du=\frac{\Gamma(s)\cos\left(s\arctan \left(\frac{a}{b}\right)\right)}{(a^2+b^2)^{s/2}}$