Uniform boundedness of integral?

Question

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).

Answer 1

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 $\sup_{b\le0}I(b)J(b)<\infty$.

Answer 2

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);

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