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).



Uniform boundedness of integral?
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 $\sup\_{b\le0}I(b)J(b)<\infty$.


