Showing an improper integral is positive without explicitly evaulating it I would like to know how one can show that
$$\int^\infty_0 e^{-x^2} \cos (x^2) \, dx > 0$$
without explicitly evaluating the improper integral (its evaluating can be found here).
For $0 < x < \sqrt{\dfrac{\pi}{2}}$, $e^{-x^2} \cos (x^2) > 0$ and we clearly have
$$\int^{\sqrt{\frac{\pi}{2}}}_0 e^{-x^2} \cos (x^2) \, dx > 0,$$ 
so the positive condition would be equivalent to showing that
$$\int^\infty_{\sqrt{\frac{\pi}{2}}} e^{-x^2} \cos (x^2) \, dx > 0.$$
Of course graphically one can see the dominate contribution to the integral (which is positive) comes from those values of $x$ close to the origin, but of course graphical evidence, while it may point one in the right direction, does not constitute a formal proof of the claim.
 A: Here it comes a trick I love. The given integral equals
$$ \mathcal{I}=\int_{0}^{+\infty}e^{-x}\cos(x)\frac{dx}{2\sqrt{x}} \stackrel{\mathcal{L}}{=}\int_{0}^{+\infty}\frac{s+1}{1+(s+1)^2}\cdot\frac{ds}{2\sqrt{\pi s}} $$
by the properties of the (inverse) Laplace transform and the RHS is pretty obviously positive.

A sharper lower bound can be deduced from the Cauchy-Schwarz inequality:
$$ \mathcal{I}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{u^2+1}{1+(u^2+1)^2}\,du=\frac{\sqrt{\pi}}{2}-\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{(u^2+1)\left[1+(u^2+1)^2\right]} $$
and
$$ \int_{0}^{+\infty}\frac{du}{(u^2+1)\left[1+(u^2+1)^2\right]}\leq \sqrt{\int_{0}^{+\infty}\frac{du}{(u^2+1)^2}\int_{0}^{+\infty}\frac{du}{\left[1+(u^2+1)^2\right]^2}}$$
leads to:
$$ \mathcal{I}\geq \frac{2}{3}. $$

The explicit computation of the $u$-integral through the residue theorem gives
$$\boxed{\mathcal{I}=\frac{\pi}{4}\sqrt{1+\sqrt{2}}\approx 0.6885.}$$

In general, for any $a\in\mathbb{R}$ we have
$$ \int_{0}^{+\infty}e^{-x^2}\cos(ax^2)\,dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}\sqrt{1+\frac{1}{\sqrt{1+a^2}}}\geq \frac{1}{2}\sqrt{\frac{\pi}{2}}.$$
