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I have perhaps a very simple question where I lack some inutition at the moment: Is the expression $$\sup_{\alpha < 0, \lambda \in \mathbb N}\int_{-\infty}^{\alpha} e^{-\lambda t^4} \ dt \int_{\alpha}^0 e^{\lambda t^4} \ dt $$

finite or not?-My main concern is $\alpha$ close to zero since $t^4$ is not strongly convex (this is something that would've bailed me out I think, so for Gaussians I think I could show this boundedness).

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2 Answers 2

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Let $a:=\alpha$ and $x:=\lambda$. Using now the substitution $t=x^{-1/4}s$ and introducing $b:=x^{1/4}a$, we see that the problem is to show that $\sup_{b\le0,x\ge1}x^{-1/2}I(b)J(b)=\sup_{b\le0}I(b)J(b)<\infty$, where $$I(b):=\int_b^0 e^{s^4}\,ds=\int_0^{-b}e^{s^4}\,ds,\quad J(b):=\int_{-\infty}^b e^{-s^4}\,ds$$ for $b\le0$. By l'Hospital's rule, $$I(b)\sim-b^{-3}e^{b^4}/4,\quad J(b)\sim -b^{-3}e^{-b^4}/4$$ as $b\to-\infty$. Thus, $I(b)J(b)\to0$ as $b\to-\infty$. Also, the functions $I$ and $J$ are continuous and hence locally bounded. Thus, indeed $\sup_{b\le0}I(b)J(b)<\infty$.

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  • $\begingroup$ Do you take into account that $dt=\lambda^{1/4}\,ds$? I don't see $\lambda^{1/4}$ in the above. $\endgroup$
    – user64494
    Jul 27, 2020 at 18:13
  • $\begingroup$ Sorry, I don't see any denominator in the above. Could you elaborate your comment? TIA. $\endgroup$
    – user64494
    Jul 27, 2020 at 18:40
  • $\begingroup$ @user64494 : Oops! I have now inserted the missing factor $x^{-1/4}x^{-1/4}=x^{-1/2}$. Since this factor is bounded for $x\ge1$, it does not affect the conclusion. $\endgroup$ Jul 27, 2020 at 18:58
  • $\begingroup$ Iosif Pinelis, can you do the right edit, in particular, defining $y$? I think, $y=b$ in your notation. $\endgroup$
    – user64494
    Jul 27, 2020 at 19:06
  • $\begingroup$ @user64494 : $y$ is a dummy variable, and thus can be replaced by any symbol. Later, we substitute $|b|$ for $y$. I only did not want to use $b$ in place of $y$, because $b<0$ and hence $b\ne|b|$. $\endgroup$ Jul 27, 2020 at 19:28
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Mathematica 12.0 does the job by

Integrate[Exp[\[Lambda]*t^4],{t, \[Alpha], 0},  Assumptions->\[Alpha]<0 && \[Lambda] >= 1]*
Integrate[Exp[-\[Lambda]*t^4],{t,-Infinity,\[Alpha]},Assumptions->\[Alpha]<0&&\[Lambda]>=1]

$$-\frac{(-1)^{3/4} \alpha E_{\frac{3}{4}}\left(\alpha ^4 \lambda \right) \left(\Gamma \left(\frac{1}{4},-\alpha ^4 \lambda \right)-\Gamma \left(\frac{1}{4}\right)\right)}{16 \sqrt[4]{\lambda }} $$

NMaximize[{%,\[Alpha]<0&&\[Lambda]>=1},\[Alpha],\[Lambda]},Method-> "DifferentialEvolution"]

$$\{0.209323,\{\alpha \to -0.476784,\lambda \to 1.\}\}$$

Addition. Maple confirms it by

DirectSearch:-Search((alpha, lambda) -> int(exp(-lambda*t^4), t = -infinity .. alpha, numeric)*int(exp(lambda*t^4), t = alpha .. 0, numeric), {-100 <= alpha, 1 <= lambda, alpha <= 0, lambda <= 100}, maximize);

$$[ 0.209323347704846, \left[ \begin {array}{c} - 0.476781454615864297 \\ 1.00000000002946488\end {array} \right] ,117] $$

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  • $\begingroup$ You know that NMaximize does not guarantee a true global maximum, right? $\endgroup$ Jul 27, 2020 at 17:22
  • $\begingroup$ @Iosif Pinelis: Maple produces the same. $\endgroup$
    – user64494
    Jul 27, 2020 at 17:49
  • $\begingroup$ Does that Maple numerical routine guarantee a true global maximum? Does the coincidence (up to some, possibly high, accuracy) of the Mathematica and Maple results guarantee a true global maximum? $\endgroup$ Jul 27, 2020 at 18:33
  • $\begingroup$ @Iosif Pinelis: Thank you for your interest to the answer. Of course, neither DirectSearch nor NMaximize guarantees a true global maximum for all cases. The default accuracy of the Search command is $10^{−6}$ both for maximum value and variables. The coincidence of the results produced by two independent CASes suggests that the upper bound under consideration does equal ≈0.2093, no more and no less $\endgroup$
    – user64494
    Jul 27, 2020 at 19:21

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