Prove $\int_{-\infty}^\infty \cos t^2 dt = \sqrt{\frac{\pi}{2}}$ So, I need to prove the identity 
$$\int_{-\infty}^\infty \cos t^2 dt = \int_{-\infty}^\infty \sin t^2 dt = \sqrt{\frac{\pi}{2}}$$
and as a hint I have the Gaussian integral
$$\int_{-\infty}^\infty e^{-xt^2} dt = \sqrt{\frac{\pi}{x}} \;\;\;\forall x>0.$$
I suspect I have to take the real/imaginary part of $e^{-t^2}$ at some point, but I can't quite figure how. I.e., $\int e^z dz = e^z$ gives me nothing. So, how do I do it?
 A: For a nice real analytic derivation, we can notice that:
$$I^2=\left(\int_{0}^{+\infty}\cos(t^2)\,dt\right)^2 = \frac{1}{2}\iint_{[0,+\infty)^2}\cos(s^2-t^2)\,ds\,dt = \frac{1}{4}\int_{0}^{+\infty}\int_{-v}^{v}\cos(uv)\,du\,dv$$
so:
$$ I^2 = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin v^2}{v}\,dv =\frac{1}{4}\int_{0}^{+\infty}\frac{\sin z}{z}\,dz=\frac{\pi}{8}.$$
Some care is needed in proving that $\iint_{[0,+\infty)^2}\cos(s^2+t^2)\,ds\,dt$ vanishes and the manipulations are legit, since $\frac{\sin z}{z}$ is a Riemann integrable function but it does not belong to $L^1(\mathbb{R})$. Beside that, this proof just mimics the proof that $\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}$ through the Fubini-Tonelli's theorem.
A: HINT:
$\Im e^{iz} = \sin(z) \implies \sin(z^2) = \Im e^{iz^2}$
A: Substitute $x=i$ in your hint and simplify the resulting complex number and then you need to consider real and imaginary parts of the integral. See here for a justification which it depends on your background. 
