Estimating a quotient of improper integrals

Question

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!

The numerator is a well-known integral, and equals $\sqrt{\pi}/2$. The denominator can be calculated by integrating by parts. — corey979, Apr 07 '19 at 20:01
If you are familiar with integrating complex functions, you could make use of the fact $$ cos(x) = \frac{e^{-ix} + e^{ix}}{2}.$$ — ViktorStein, Apr 07 '19 at 20:01
According to Wolfy, the numerator is $\sqrt{\pi}/2$ and the denominator is $\sqrt{\pi}/(2e)$ so the ratio is $e$ so the integer part is $2$. — marty cohen, Apr 07 '19 at 20:07
This is a much more straightforward derivation for the numerator than the link in my previous comment. — corey979, Apr 07 '19 at 20:08
When you know what the teacher had in mind, let us know what it was! Thanks. — peter a g, Apr 09 '19 at 21:18

Peter Foreman · Answer 1 · 2019-04-07T20:42:02.810

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 $$\left\lfloor{\ \frac{\int_0^{\infty} e^{-x^{2}}\,dx}{\int_0^{\infty} e^{-x^{2}}\cos {(2x)}\, dx}}\ \right\rfloor=\left\lfloor{\ \frac{\int_0^{\infty} e^{-x^{2}}\,dx}{e^{-\frac{(2)^2}4}\int_0^\infty e^{-x^2}\, dx}}\ \right\rfloor=\left\lfloor\frac1{e^{-1}}\right\rfloor=\left\lfloor e\right\rfloor=2$$

Is it of any importance that "the integrands are strictly positive"? — user, Apr 07 '19 at 20:37
Not really, but I thought stating this was useful as it didn't change the sign of any values within the working. — Peter Foreman, Apr 07 '19 at 20:38
It seems that the only relevant issue is that the real part of the integral is non-zero. — user, Apr 07 '19 at 20:41

peter a g · Answer 2 · 2019-04-15T13:59:41.417

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.

score 0 · Answer 3 · answered Apr 07 '19 at 20:16

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.

Estimating a quotient of improper integrals

3 Answers3