1
$\begingroup$

How to show that \begin{align} \left| \frac{\int_0^\infty \cos(ax)e^{-x^4} dx}{\int_0^\infty \cos(bx)e^{-x^4} dx} \right| \le 1 \end{align} if $a\ge b \ge 0$.

This is what I did.

One has to show then that \begin{align} \frac{ \left|\int_0^\infty \cos(ax)e^{-x^4} dx \right| }{ \left|\int_0^\infty \cos(bx)e^{-x^4} dx \right|} \le 1 \end{align}

Or that \begin{align} \left|\int_0^\infty \cos(ax)e^{-x^4} dx \right| \le \left|\int_0^\infty \cos(bx)e^{-x^4} dx \right| \end{align} But how to show the last inequality?

$\endgroup$
4
$\begingroup$

You cannot, because your inequality does not hold: just take $a=4$ and $b=3.5$.

If we take the function $$ \psi: \xi \mapsto \int_{0}^{+\infty}\cos(\xi x)e^{-x^4}\,dx $$ this is its graph over $[0,10]$:

$\hspace{2cm}$enter image description here

and it is not monotonic in absolute value, even if fast-decaying.

$\endgroup$
  • $\begingroup$ Thanks. You see the reason I thought that this inequality was correct is because I know it is true when the exponential is of degree 1 or degree 2. That since $\int_0^\infty \cos(ax) e^{-x} dx=\frac{1}{1+a^2}$ and $\int_0^\infty \cos(ax) e^{-x^2} dx= 0.5*\sqrt{\pi} e^{a^2/4}$. With this we get \begin{align} \frac{\int_0^\infty \cos(ax) e^{-x} }{\int_0^\infty \cos(bx) e^{-x} } =\frac{1+b^2}{1+a^2} \\ \frac{\int_0^\infty \cos(ax) e^{-x^2} }{\int_0^\infty \cos(bx) e^{-x^2} } =e^{\frac{b^2-a^2}{4}}\end{align} Both of which are less than 1 if $a>b$. So, why is it not the same for $x^4$? $\endgroup$ – Boby Jul 1 '16 at 16:36
  • $\begingroup$ @Boby: because the Fourier transform is slightly oscillating, while that does not happen in the other cases. $\endgroup$ – Jack D'Aurizio Jul 1 '16 at 16:43
  • $\begingroup$ Could you explain more along the lines of Fourier analysis? I am very curios. Also, is it correct to say that if we exponent $x^k$ then the inequality is true for all $0< k\le2$ ? $\endgroup$ – Boby Jul 1 '16 at 16:46
  • $\begingroup$ Interesting fact, it probably deserves a separate question. $\endgroup$ – Jack D'Aurizio Jul 1 '16 at 17:20
  • $\begingroup$ Ok. I will post a new question. $\endgroup$ – Boby Jul 1 '16 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.