Estimating a quotient of improper integrals This was an extra credit question on my test on improper integrals. It was: 
Define $ \lfloor{x}  \rfloor$ to be the greatest integer less than or equal to $x$, where $x$ is a real number. Calculate:
\begin{equation}
 \left\lfloor{\ \frac{\int_0^{\infty} e^{-x^{2}}\,dx}{\int_0^{\infty} e^{-x^{2}}\cos 2x\, dx}}\ \right\rfloor
\end{equation}
I don't know how to start, since I don't think any of the integrals are elementary functions. Can anyone help?
Edit: He did give us the first integral as $\sqrt{\pi}/2$.
Note: This was the final $10$ point extra credit question. It was designed to be hard!
 A: If we define a new integral
$$I(t)=\int_0^\infty e^{-x^2} (\cos{(tx)}+i\sin{(tx)})\,dx$$
then
$$\Re{(I(t))}=\int_0^\infty e^{-x^2} \cos{(tx)}\,dx$$
So in order to calculate $I(t)$, we have that
$$\begin{align}I(t)&=\int_0^\infty e^{-x^2} (e^{itx})\, dx\\
&=\int_0^\infty e^{itx-x^2}\, dx\\
&=\int_0^\infty e^{-(x^2-itx)}\, dx\\
&=\int_0^\infty e^{-((x-\frac{it}2)^2-(\frac{it}2)^2)}\, dx\\
&=\int_0^\infty e^{-(x-\frac{it}2)^2-\frac{t^2}4}\, dx\\
&=e^{-\frac{t^2}4}\int_0^\infty e^{-(x-\frac{it}2)^2}\, dx \end{align}$$
Now using the substitution $u=x-\frac{it}2\to dx=du$ we have
$$I(t)=e^{-\frac{t^2}4}\int_{-\frac{it}2}^\infty e^{-u^2}\, du=e^{-\frac{t^2}4}\int_0^\infty e^{-u^2}\, du+e^{-\frac{t^2}4}\int_{-\frac{it}2}^0 e^{-u^2}\, du$$
Now the second integral can be evaluated by using the substitution $v=iu\to du=-idv$ which gives
$$e^{-\frac{t^2}4}\int_{-\frac{it}2}^0 e^{-u^2}\, du=e^{-\frac{t^2}4}\int_{\frac{t}2}^0 e^{-(-iv)^2}\, (-i)dv=ie^{-\frac{t^2}4}\int_0^{\frac{t}2} e^{v^2}\, dv$$
So we then have
$$I(t)=e^{-\frac{t^2}4}\int_0^\infty e^{-u^2}\, du+ie^{-\frac{t^2}4}\int_0^{\frac{t}2} e^{v^2}\, dv$$
Both of the integrands are real and strictly positive so we can say that
$$\Re{(I(t))}=e^{-\frac{t^2}4}\int_0^\infty e^{-u^2}\, du=e^{-\frac{t^2}4}\int_0^\infty e^{-x^2}\, dx=\int_0^\infty e^{-x^2} \cos{(tx)}\,dx$$
We then finally have an answer of
$$\left\lfloor{\ \frac{\int_0^{\infty} e^{-x^{2}}\,dx}{\int_0^{\infty} e^{-x^{2}}\cos {(2x)}\, dx}}\ \right\rfloor=\left\lfloor{\ \frac{\int_0^{\infty} e^{-x^{2}}\,dx}{e^{-\frac{(2)^2}4}\int_0^\infty e^{-x^2}\, dx}}\ \right\rfloor=\left\lfloor\frac1{e^{-1}}\right\rfloor=\left\lfloor e\right\rfloor=2$$
A: Without complex numbers - and maybe what your instructor was thinking of:
Write 
$$ f(t) = \int_0^\infty e^{-x^2}\cos (t x) \, dx.$$
Differentiate (with respect to $t$): $$f'(t) = -  \int_0^\infty e^{-x^2} x \sin (t x) \, dx.$$
[I'm hoping your instructor allows differentiation under the integral sign, 'as obvious.' If not, one can justify it with words such as 'dominated convergence' - but...]
Integrate (with respect to $x$) by parts,  with $du = e^{-x^2} (-2x)/2  \,dx$ and $v =\sin (t x)$:
 $$f'(t) =   \bigg( 1/2\, e^{-x^2} \sin(tx)\bigg|_0^\infty - 1/2\, \int_0^\infty e^{-x^2} t \cos (tx)\, dx   \bigg).$$
The first term comes to zero, and we can take the factor of $t$  outside of the integral,
so that one ends up with 
$$ f'(t)= {-t\over 2} f(t).$$
Divide by $f(t)$, integrate, and solve for the constant of integration, to obtain
$$ f(t) = f(0)\, e^{-t^2/4}.$$
[Qualms about the formal manipulation of dividing by $f(t)$ as $f(t)$ could be zero can be ignored, in this case, because of the so-called existence and uniqueness theorem of differential equations.]
In any event, as in Peter Foreman's answer, your ratio is 
$$ \left\lfloor  f(0) \over f(2) \right\rfloor = \left\lfloor  f(0) \over e^{-2^2/4}\, f(0) \right\rfloor =\left\lfloor  e  \right\rfloor = 2.$$ 
FWIW, differentiation under the integral sign is often referred to as the Feynman trick.
A: Here's my incomplete start.
$\begin{array}\\
\int_0^{\infty} e^{-x^{2}}\cos 2x dx
&=\int_0^{\infty} e^{-x^{2}}Re(e^{2ix}) dx\\
&=Re\int_0^{\infty} e^{-x^{2}+2ix} dx\\
&=Re\int_0^{\infty} e^{-x^{2}+2ix-i^2+i^2} dx\\
&=Re\int_0^{\infty} e^{-(x-i)^2-1} dx\\
&=Re\frac1{e}\int_0^{\infty} e^{-(x-i)^2} dx\\
\end{array}
$
This shows where the
$1/e$ comes from.
If we can show that
$Re\int_0^{\infty} e^{-(x-i)^2} dx
=\int_0^{\infty} e^{-x^{2}}dx
$
then we are done,
but I don't know how to do this.
