0
$\begingroup$

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 $k=4$.

  • 0
    @SimplyBeautifulArt Integral of $f(x)$ or the integral of $\cot(xt) e^{-t^4}$?? – 2017-02-28
  • 2
    Related: http://math.stackexchange.com/questions/1847213/fourier-cosine-transforms-of-schwartz-functions-and-the-fejer-riesz-theorem – 2017-02-28
  • 0
    @JackD'Aurizio Hey Jak. Yes, I was going to add this. If you remember a while back we formulated all this together. – 2017-02-28
  • 0
    @SimplyBeautifulArt: the given function has a real zero close to $x=3.5$. – 2017-02-28
  • 0
    @JackD'Aurizio Well, if you'll excuse me... I think I'll be taking my leave on this question XD – 2017-02-28
  • 0
    @SimplyBeautifulArt No. Come on stay... – 2017-02-28

1 Answers 1

0

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); $$3.453464128$$

  • 0
    Thank you. I am however interested in providing a result that would give lower and upper bound on $x_0$ and is more analytic. – 2017-02-28
  • 0
    You might be able to get bounds by considering the ratio of those two hypergeometric functions... – 2017-02-28
  • 0
    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 $k=4$. – 2017-02-28