# How to show that $\left|\int_0^\infty \cos(ax)e^{-x^4} dx \right| \le \left|\int_0^\infty \cos(bx)e^{-x^4} dx \right|$

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?

## 1 Answer

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 $[0,10]$:

$\hspace{2cm}$

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

• 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 $x^4$? – Boby Jul 1 '16 at 16:36
• @Boby: because the Fourier transform is slightly oscillating, while that does not happen in the other cases. – Jack D'Aurizio Jul 1 '16 at 16:43
• 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 $0< k\le2$ ? – Boby Jul 1 '16 at 16:46
• Interesting fact, it probably deserves a separate question. – Jack D'Aurizio Jul 1 '16 at 17:20
• Ok. I will post a new question. – Boby Jul 1 '16 at 17:22