A calculus problem by Rohan Shinde

$I=\int_0^{\infty} \sin (x^2) \ln^2xdx$

Substituting $x^2=t$

$I=\frac 18\int_0^{\infty} t^{-\frac 12} \sin t \ln^2tdt$

Define $J(a)=\frac 18 \int_0^{\infty} t^{a-1}\sin t dt=\frac 18\int_0^{\infty} t^a \left(\sum_{k=0}^{\infty} \frac {(-1)^k t^{2k}\Gamma(k+1)}{k!\Gamma(2+2k)}\right)dt$

Using Ramanujan's Master theorem this evaluates to $J(a)=\frac {1}{16}\Gamma(s)\phi(-s)$ where $s=\frac {a+1}{2}$ and $\phi(k)=\frac {\Gamma(k+1)}{\Gamma(2+2k)}$

What we need to find $I$ is just $J''(1/2)$ which from above formulation would yield the answer as $\frac 14\sqrt {\frac {\pi}{128}}\left(\pi+2\psi^{(0)}\left(\frac 12\right)\right)^2$