$\displaystyle \int_{0}^{\infty} \cos \Big(x + \frac{1}{x} \Big) \frac{dx}{\sqrt{x}} = 2 \int^{\infty}_{0} \cos \Big(x^{2}+ \frac{1}{x^{2}} \Big) dx = 2 \ \Re \int^{\infty}_{0} e^{-i(x^{2}+\frac{1}{x^{2}})} dx $

and $\displaystyle \int_{0}^{\infty} \sin \Big(x + \frac{1}{x} \Big) \frac{dx}{\sqrt{x}} = 2 \int^{\infty}_{0} \sin \Big(x^{2}+ \frac{1}{x^{2}} \Big) dx = -2 \ \Im \int^{\infty}_{0} e^{-i(x^{2}+\frac{1}{x^{2}})} dx $

Could I use $\displaystyle \int^{\infty}_{0} e^{-a(x^{2}+\frac{1}{x^{2}})} dx = \frac{\sqrt{\frac{\pi}{a}}}{2e^{2a}} $ and substitute i for a?