# Optimization problem with definite integral inequality constraints

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 $$r\in\left[1,\sqrt{d}\right]$$?)

• I get that your integral is less than $2r^{d-1}$ for all $r \ge1$. Just estimate the integrand with $r^{d-1}$. – Giorgio Metafune Aug 29 '20 at 18:14
• Just estimate the LHS by $r^{d-1}\int_{-\infty}^{\infty}\exp\{-(d-1)\frac{x^2}{2r^2}\}dx=\sqrt{2\pi}\frac{r^d}{\sqrt{d-1}}$. – fedja Aug 29 '20 at 18:22
• 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 $r\in\left[1,\sqrt{d}\right]$. – Penelope Benenati Aug 29 '20 at 18:24
• Thank you once again @fedja! This is precisely consistent with what I was expecting to obtain! – Penelope Benenati Aug 29 '20 at 18:31
• 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. – fedja Aug 29 '20 at 21:06

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 $$r \to \infty$$.

• Thank you! However I would like to prove it mathematically. – Penelope Benenati Aug 29 '20 at 17:00
• @Penelope Benenati: But this is math. – user64494 Aug 29 '20 at 17:09
• Elementary asymptotic estimates give $\sqrt{2\pi}=2.506628274631$ for the constant. – user64494 Aug 30 '20 at 0:23
• 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 $d,r\to\infty$"? – Penelope Benenati Aug 30 '20 at 8:02
• @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 $r\approx \sqrt d$. – user64494 Aug 30 '20 at 13:48