$$I=\int_0^1\sin(\pi \sqrt{-\ln x})\mathrm~ dx$$
My efforts:
I substitute $t=-\ln x$
And I got $$I=\int_0^\infty e^{-t}\sin(\pi \sqrt{t})\mathrm ~dt$$
But, how to proceed with : $\int_0^\infty e^{-t}\sin(\pi\sqrt{t})\mathrm dt$?
(P.S.: The result is $\dfrac{\pi^{3/2}}{2e^{\pi^2/4}}$)
Let $t=\sqrt{-\ln x}$, hence $x=e^{-t^2}$. Integration by part, and we get
$$I=\int_\infty^0\sin(\pi t)~d(e^{-t^2})=\sin(\pi t)~d(e^{-t^2})\bigg|_\infty^0-\pi\int_\infty^0\cos(\pi t)\cdot e^{-t^2}dt=\pi\int^\infty_0\cos(\pi t)\cdot e^{-t^2}dt$$
This is a Gaussian-like integral and can be solved by many ways
$$\int_0^\infty \cos(ax)\cdot e^{-x^2}~dx=\frac{\sqrt\pi}{2}\,e^{-a^2/4}$$
therefore,
$$\boxed{\int_0^1\sin(\pi \sqrt{-\ln x})\mathrm~ dx=\frac{\pi\sqrt\pi}{2}\,e^{-\pi^2/4}}$$