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Question: How can we prove that there exists a real constant $c\ge 1$ such that the following inequality holds for all integers $d>1$ and all real numbers $r\in\left[1,\sqrt{d}\right]$?

$$\int_{-1}^1 \left(\sqrt{r^2-x^2}\right)^{d-1} dx\le c\cdot \frac{r^d}{\sqrt{d-1}}$$


(Furthermore, is it also possible to find an upper bound for the minimum value of $c\in [1,\infty)$ such that the above inequality holds for all $d>1$ and all $r\in\left[1,\sqrt{d}\right]$?)

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  • $\begingroup$ I get that your integral is less than $2r^{d-1}$ for all $r \ge1$. Just estimate the integrand with $r^{d-1}$. $\endgroup$ Aug 29, 2020 at 18:14
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    $\begingroup$ Just estimate the LHS by $r^{d-1}\int_{-\infty}^{\infty}\exp\{-(d-1)\frac{x^2}{2r^2}\}dx=\sqrt{2\pi}\frac{r^d}{\sqrt{d-1}}$. $\endgroup$
    – fedja
    Aug 29, 2020 at 18:22
  • $\begingroup$ Thank you @GiorgioMetafune. This way, whan $r$ is small (say constant), we get $c$ depending on $d$, and we do not get that the inequality holds for a constant $c$ and for all $d>1$ and all $r\in\left[1,\sqrt{d}\right]$. $\endgroup$ Aug 29, 2020 at 18:24
  • $\begingroup$ Thank you once again @fedja! This is precisely consistent with what I was expecting to obtain! $\endgroup$ Aug 29, 2020 at 18:31
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    $\begingroup$ Looks like just constant number of times, i.e., there is a lower bound of the same type in the range you are interested in. $\endgroup$
    – fedja
    Aug 29, 2020 at 21:06

1 Answer 1

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Mathematica answers

NMaximize[{Integrate[(r^2 - x^2)^(d/2 - 1/2), {x, -1, 1}, 
Assumptions -> d > 1 && r >= 1]/r^d*Sqrt[d - 1], 
r >= 1 && r <= Sqrt[d] && d > 1 && d \[Element] Integers},{r, d}] 

$$\{2.43959,\{r\to 1.01254,d\to 28\}\} $$ and

NMaximize[{Integrate[(r^2 - x^2)^(d/2 - 1/2), {x, -1, 1}, 
Assumptions -> d > 1 && r >= 1]/r^d*Sqrt[d - 1], 
 r >= 1 && r <= Sqrt[d] && d > 1}, {r, d}, AccuracyGoal -> 4,PrecisionGoal -> 4]

$$ \{2.50662,\{r\to 149.294,d\to 611671.\}\}$$

Addition. The command of Maple confirms it by

restart;Digits := 20;DirectSearch:-Search((d, r) -> int((r^2 - x^2)^(1/2*d - 1/2),x = -1 .. 1, numeric, epsilon = 1/1000)*sqrt(d - 1)/r^d, {1 <= d, 1 <= r, r <= sqrt(d)}, maximize);

$$[ 2.5066284493892445574, \left[ \begin {array}{c} 280247431.41221862419\\ 1059.2785342279385942 \end {array} \right] ,317] $$ It seems the supremum is attained as $d \to \infty$ and $r \to \infty$.

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  • $\begingroup$ Thank you! However I would like to prove it mathematically. $\endgroup$ Aug 29, 2020 at 17:00
  • $\begingroup$ @Penelope Benenati: But this is math. $\endgroup$
    – user64494
    Aug 29, 2020 at 17:09
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    $\begingroup$ Elementary asymptotic estimates give $\sqrt{2\pi}=2.506628274631$ for the constant. $\endgroup$
    – user64494
    Aug 30, 2020 at 0:23
  • $\begingroup$ Do you mean "asymptotic" w.r.t. $d,r\to\infty$? if so, how can you first mathematically prove that "the supremum is attained as $d,r\to\infty$"? $\endgroup$ Aug 30, 2020 at 8:02
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    $\begingroup$ @Penelope Benenati: I think the asymptotics of the integral under consideration as $d\to\infty$ can be found by the Laplace's method. That asymptotics depends on $r$. According to numeric calculations, its maximum should be attained for $r\approx \sqrt d$. $\endgroup$
    – user64494
    Aug 30, 2020 at 13:48

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