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?


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.

  • $\begingroup$ 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

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