A Near Mirror

Consider this function : $f(a) = \int\limits_0^{\infty} e^{-x^2} \cos a x \ \mathrm{d}x.$ It is well defined for any $a\in \mathbb{R}$ , differentiate it with respect to $a$ and integrate by parts to get that : $f'(a) = \int\limits_0^{\infty}- x e^{-x^2} \sin a x \ \mathrm{d}x = \bigg[\frac{e^{-x^2} \sin a x}{2}\bigg]_0^{\infty} - \frac{a}{2} \int\limits_{0}^{\infty} e^{-x^2} \cos a x \ \mathrm{d}x= \frac{-a f(a)}{2}.$ So : $f'(a)= \frac{-af(a)}{2}$ , by separation of variables we get that : $\ln(f(a))=\int \frac{\mathrm{d}f}{f(a)} = -\int \frac{a}{2} \ \mathrm{d}a= \frac{a^2}{4} +C \Longrightarrow f(a)= f(0)e^{-\frac{a^2}{4}} \ \ \ \ \forall a\in \mathbb{R}.$ Therefore : $\left\lfloor \frac{\int_0^{\infty} e^{-x^2} \ \mathrm{d}x }{\int_0^{\infty}e^{-x^2} \cos 2 x\ \mathrm{d}x}\right\rfloor= \left\lfloor \frac{f(0)}{f(1)} \right\rfloor =\lfloor e\rfloor = \boxed{2}.$

There are different ways of solving such mundane problem. Without thinking too much one can directly apply gamma function in its standard format. As for numerator the result is a 1/2(sqrt(pi)) and on the other hand , the denominator can also be converted into the closed form of a gamma function by applying a suitable " perfect square" trick. Then one can simply approximate cosine function as the real part of a plane wave exp(2ix). Finally, the denominator apart from 1/2(sqrt(pi)) will also contain "1/(e=2.71828)". And in the end, we end up with 2.71828. Since we have to calculate floor value so one just takes out 2.

This problem can also be evaluated by Laplace integral formalism and contour integral too.

I do not know, how to write here in proper math's symbols.

The integrand in the numerator describes half of a Gaussian with standard deviation $\sigma = \sqrt 2$ .

The integrand in the denominator describes that same Gaussian multiplied by $\cos 2x$ , which remains close to 1 for small values of $x$ , and reaches zero for $x = \pi/2 \approx 1.1\sigma$ .

It is well-known that about $64\%$ of the mass of a Gaussian lies within $\sigma$ from its center; in this case, between $x = 0$ and $x \approx 1.5$ . The multiplication by $\cos 2x$ decreases this percentage somewhat, and there is a negative contribution to the integral between $\pi/2 < x < \pi$ , but that is relatively small.

Therefore it is a reasonable guess that $0.33 < \frac{\text{denominator}}{\text{numerator}} < 0.50,$ which is sufficient to conclude that the answer should be $\boxed{2}$ .

The upper integral is a gaussian integral which can be evaluated using the gamma function evaluated at 3/2; then using the recursive formula for gamma function. It can also be derived using cartesian or polar corrdinates.

The lower integral can be evaluated using the integral of $f(z) = e^{-z^{2}}$ along the contour which is a rectangle with corners a+0i, a+bi, -a+bi, -a+0i. Since f(z) is entire, by Cauchy-Goursat Theorem the integral of f(z) around the contour is zero. Get the sum of the horizontal and vertical legs of the contour.