Is it true that \begin{align} \int_0^A (A-x) \cos(x) \ f(x) dx \ge 0 \end{align} for all $A\ge 0$, if for $f(x)$ monotone decreasing, non-negative and bounded.

This question came up in the discussion here where it was also shown that for $A \le \pi$ this is true.


For $f(x)=e^{-tx}$ we have that

\begin{align} \int_0^a (a-x) \cos (x) e^{-t x} \, dx=\frac{e^{-a t} \left(\left(t^2-1\right) \cos (a)-2 t \sin (a)+e^{a t} \left(a \left(t^3+t\right)-t^2+1\right)\right)}{\left(t^2+1\right)^2} \end{align}

the expression $\left(t^2-1\right) \cos (a)-2 t \sin (a)+e^{a t} \left(a \left(t^3+t\right)-t^2+1\right)$ apprears to be postive for all $a>0$ and $t>0$.

  • $\begingroup$ It might be interesting to prove it first for any function of the form $f(x)=e^{-\kappa x}$ or $f(x)=\frac{1}{x+\kappa}$ with $\kappa>0$. $\endgroup$ – Jack D'Aurizio Dec 13 '16 at 0:06
  • $\begingroup$ @JackD'Aurizio Done with $e^{-tx}$ case. $\endgroup$ – Boby Dec 13 '16 at 0:20
  • $\begingroup$ By the properties of the Fejér kernel, if $\cos(x)\omega(|x|)$ has a non-negative Fourier cosine transform, then $$\int_{0}^{a}(a-x)\cos(x)\omega(x)\,dx \geq 0.$$ That gives an alternative answer to your previous question: both $\cos(x)$ and $\frac{1+x^2}{1+b^2 x^2}$ have non-negative Fourier cosine transform, so does their product, so $\int_{0}^{a}(a-x)\cos(x)\frac{1+x^2}{1+b^2 x^2}\,dx \geq 0$ for any $b\in\mathbb{R}$ and $a\in\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Dec 13 '16 at 0:43
  • $\begingroup$ It follows that if the even extension of $f(x)$ has a non-negative Fourier cosine transform, the given inequality holds. The cases $\omega(x)=e^{-\kappa x}$ and $f(x)=\frac{1}{\kappa^2+x^2}$ are in this sense the dual of each other. $\endgroup$ – Jack D'Aurizio Dec 13 '16 at 0:54

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