0
$\begingroup$

Does anyone have any advice or help on how to analytically solve the following problem?

Prove that the function $$ \operatorname{f}\left(r\right) = \int_{0}^{\infty}\operatorname{J}_{0}\left(rx\right)\, \mathrm e^{-ax^{2}}\cos\left(x\right)x\,\mathrm dx\,,\quad r \geq 0 $$ is positive beyond a certain value for $a$ positive, where the exact threshold for $a$ is the one needed to ensure that it is positive at the origin.

Some notes:

  • It is easy to calculate the exact value of $a$ beyond which $\operatorname{f}\left(0\right)$ is positive. Also, $\operatorname{f}\left(r\right) \to 0$ as $r \to \infty$.
  • So if I can prove that $\operatorname{f}$ cannot have any zeros (or local minima) it would prevent it from becoming negative.
  • I am not sure though if looking at the critical points of $\operatorname{f}$ is the right way to go.

Thanks in advance for any help :) Even help in gauging how easy/difficult this question is is welcome.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
0
$\begingroup$

For large $a$ you can replace $\cos x$ by 1 (since only the interval $x\lesssim 1/\sqrt a$ contributes), and then the integral evaluates to $$\int_0^\infty J_0(rx) e^{-ax^2} x\,dx = \frac{e^{-r^2/4 a}}{2 a},$$ which is positive for all $r$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks for the prompt answer! Would it be possible to say something about the size of a that is needed? Many thanks again for your answer, I highly appreciate it! $\endgroup$
    – Sabiske8
    Commented Jun 20 at 6:40
  • $\begingroup$ the positivity condition $a>0.2927$ can be found numerically from the value of the integral at $x=0$. $\endgroup$
    – Carlo Beenakker
    Commented Jun 20 at 7:25
  • 3
    $\begingroup$ This is, at best, heuristics. Even if it is possible to make it rigorous, it will not answer the OP question, which asks to prove the following: If $f(0)>0$ for $a>a_0$, then $f>0$ for $a>a_0$, $\endgroup$
    – Iosif Pinelis
    Commented Jun 20 at 15:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .