Prove that $$\int_0^{\infty} {e^{-t^{2}}}\cos(t^2) \, dt = \frac{1}{4} \sqrt{\pi} \sqrt{1+\sqrt{2}}$$
by integrating the function ${e^{-z^{2}}}$ in the counterclockwise direction around the boundary of the region {z : |z| $\leq$ R, 0 $\leq$ Arg z $\leq$ $\frac{\pi}{8}$ } and letting R $\to$ ${\infty}$
I believe I am supposed to parametrize this equation; perhaps ${e^{-t^{2}}}$ = ${e^{-(t+ib)^{2}}}$ and ${\cos(2bt-i\sin(2bt))dt}$ ?
Disclaimer: This proof DOES NOT use the Cauchy Integration Formula, neither Complex Analysis.
Let $$ I(a)=\int_{-\infty}^\infty \mathrm{e}^{-(1+a i)x^2}dx, \quad a\ge 0. $$ We need to show that $\,\frac{1}{2}\mathrm{Re}\, I(1)=\frac{\pi}{4}\sqrt{1+\sqrt{2}}$.
Squaring $I(a)$, and using the polar coordinates change of variables $dx\,dy=r\,dr d\vartheta$, we obtain $$ \big(I(a)\big)^2=\int_{-\infty}^\infty\int_{-\infty}^\infty \mathrm{e}^{-(1+a i)(x^2+y^2)}dx\,dy=2\pi\int_{0}^\infty \mathrm{e}^{-(1+a i)r^2}r\,dr=\pi\int_{0}^\infty \mathrm{e}^{-(1+a i)t}dt=\frac{\pi}{1+ai}, $$ and hence,
$$ I(a)=\pm\sqrt{\frac{\pi}{1+ai}}=\pm\sqrt{\frac{\pi(1-ai)}{1+a^2}}=\pm\frac{\sqrt{\pi}}{\sqrt{1+a^2}}\left(\sqrt{\frac{\sqrt{1+a^2}+1}{2}}-i\sqrt{\frac{\sqrt{1+a^2}-1}{2}}\right), $$ and therefore $$ f(a)=\int_0^\infty\mathrm{e}^{-x^2}\cos (a x^2)\,dx = \frac{1}{2}\mathrm{Re}\,I(a)=\pm \frac{\sqrt{\pi}}{2\sqrt{1+a^2}}\sqrt{\frac{\sqrt{1+a^2}+1}{2}}. $$ Observe now that, since $\mathrm{Re}\,I(a)$ is continuous in $a$ and does not vanish (since the real part of $I(a)$ does not vanish), for every $a$ real, then either we choose the plus sign, for all $a$, or the minus sign for all $a$. We choose the plus sign, since $I(0)>0$. Hence, $$ f(1)=\int_0^\infty\mathrm{e}^{-x^2}\cos (x^2)\,dx = \frac{1}{2}\mathrm{Re}\,I(1)= \frac{\sqrt{\pi}}{2\sqrt{1+1}}\sqrt{\frac{\sqrt{1+1}+1}{2}}=\frac{\sqrt{\pi(1+\sqrt{2}})}{4}. $$ In general $$ \int_0^\infty\mathrm{e}^{-x^2}\cos (ax^2)\,dx =\sqrt{\frac{\pi\big(\sqrt{1+a^2}+1\big)}{8(1+a^2)}}, \quad \int_0^\infty\mathrm{e}^{-x^2}\sin (ax^2)\,dx =\sqrt{\frac{\pi\big(\sqrt{1+a^2}-1\big)}{8(1+a^2)}}. $$