I have to show using Laplace Transform that:
$ \int_{0}^{\infty} \ \cos(x^2) \ dx = \frac{1}{2} \sqrt{\frac{\pi}{2}} $
I have no idea how to begin. The right hand side seems like it has something to do with the gamma function. Can anybody help me out?
In the improper-Riemann sense, $$ \int_{0}^{+\infty}\cos(x^2)\,dx =\int_{0}^{+\infty}\frac{\cos(x)}{2\sqrt{x}}\,dx=\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{1+s^2}\,ds = \frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{u^2}{1+u^4}\,du$$ where the middle equality is due to $\mathcal{L}(\cos x)(s)=\frac{s}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{2\sqrt{x}}\right)(s)=\frac{1}{2\sqrt{\pi s}}$.
The last integral can be computed through standard techniques (for instance, partial fraction decomposition) and it equals $\frac{\pi}{2\sqrt{2}}$, proving the claim.