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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, $$I(b)\sim-b^{-3}e^{b^4}/4$$ 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 $\sup_{b\le0}I(b)J(b)<\infty$.

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

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 $$I(y):=\int_0^ye^{s^4}\,ds,\quad J(b):=\int_{-\infty}^b e^{-s^4}\,ds.$$ 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 $\sup_{b\le0}I(|b|)J(b)<\infty$.

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, $$I(b)\sim-b^{-3}e^{b^4}/4$$ 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 $\sup_{b\le0}I(b)J(b)<\infty$.

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

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 $$I(y):=\int_0^ye^{s^4}\,ds,\quad J(b):=\int_{-\infty}^b e^{-s^4}\,ds.$$ 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 $\sup_{b\le0}I(|b|)J(b)<\infty$.