I am interested in estimating the position of the first positive zero of the following function \begin{align} f(x)= \int_0^\infty \cos(tx) e^{-t^4} dt. \end{align}

For the story of this question please see Q1, Q2, Q3.

Numerical simulation seem to suggest that the zero is around $3.4$.

The end goal of this question is to find a way of estimating zeros of a more general function \begin{align} f(x)= \int_0^\infty \cos(tx) e^{-t^k} dt. \end{align}

for $k>2$. However, as can be seen from Q1 the function can behave differently, depending if $k$ is odd or even. To reduce complexity, I therefore, decided to focus on a specific case of $k=4$.

  • $\begingroup$ @SimplyBeautifulArt Integral of $f(x)$ or the integral of $\cot(xt) e^{-t^4}$?? $\endgroup$ – Boby Feb 28 '17 at 21:32
  • 2
    $\begingroup$ Related: math.stackexchange.com/questions/1847213/… $\endgroup$ – Jack D'Aurizio Feb 28 '17 at 21:33
  • $\begingroup$ @JackD'Aurizio Hey Jak. Yes, I was going to add this. If you remember a while back we formulated all this together. $\endgroup$ – Boby Feb 28 '17 at 21:34
  • $\begingroup$ @SimplyBeautifulArt: the given function has a real zero close to $x=3.5$. $\endgroup$ – Jack D'Aurizio Feb 28 '17 at 21:35
  • $\begingroup$ @JackD'Aurizio Well, if you'll excuse me... I think I'll be taking my leave on this question XD $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 21:36

In Maple:

F:= int(cos(t*x)*exp(-t^4),t=0..infinity) assuming x>0; $$ 1/4\,\sqrt {2}\sqrt {\pi} \left( {\frac {\sqrt {\pi}}{\Gamma \left( 3/ 4 \right) }{\mbox{$_0$F$_2$}(\ ;\,1/2,3/4;\,{\frac {{x}^{4}}{256}})}}- 1/4\,{\frac {\sqrt {2}\Gamma \left( 3/4 \right) {x}^{2}}{\sqrt {\pi}} {\mbox{$_0$F$_2$}(\ ;\,5/4,3/2;\,{\frac {{x}^{4}}{256}})}} \right) $$

fsolve(F,x=0..4); $$3.453464128$$

  • $\begingroup$ Thank you. I am however interested in providing a result that would give lower and upper bound on $x_0$ and is more analytic. $\endgroup$ – Boby Feb 28 '17 at 21:50
  • $\begingroup$ You might be able to get bounds by considering the ratio of those two hypergeometric functions... $\endgroup$ – Robert Israel Feb 28 '17 at 22:05
  • $\begingroup$ Sure. But I was hoping for an approach that can be extended to any $k$. We are kind of lucky that we can come up with a closed form expression of an integral for $k=4$. $\endgroup$ – Boby Feb 28 '17 at 22:10

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