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 $\sup_{b\le0}I(|b|)J(b)<\infty$, where 
$$I(y):=\int_0^ye^{s^4}\,ds,\quad J(b):=\int_{-\infty}^b e^{-s^4}\,ds.$$
By l'Hospital's rule,
$$I(y)\sim y^{-3}e^{y^4}/4$$
as $y\to\infty$, and 
$$J(b)\sim -b^{-3}e^{-b^4}/4$$
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 $\sup_{b\le0}I(|b|)J(b)<\infty$.