Is the integral
$$
t^2\left(\iint\_{\mathbb{R}^2}|\xi|e^{\frac{-a\xi^4t}{\xi^2+b}}\,d\xi\_1\,d\xi\_2\right)$$ bounded when $t\rightarrow\infty$? Here


* $\xi=(\xi\_1,\xi\_2)\in\mathbb{R}^2$,
* $|\xi|=\sqrt{\xi\_1^2+\xi\_2^2}$,
* $a$ and $b$ are fixed positive numbers.



Whether the integral $t^2(\iint_{\mathbb{R}^2}|\xi|e^{\frac{-a\xi^4t}{\xi^2+b}} \,d\xi_1 \,d\xi_2)$ is bounded?
Assuming that by $\xi^2$ and $\xi^4$ you meant, respectively, $|\xi|^2$ and $|\xi|^4$, switching to polar coordinates, and making the substitution $r=ut^{-1/4}$, we have
$$
I(t):=t^2\left(\iint\_{\mathbb{R}^2}|\xi|e^{\frac{-a\xi^4t}{\xi^2+b}}\,d\xi\_1\,d\xi\_2\right)
=2\pi t^2\int\_0^\infty dr\,r^2\exp\Big(-\frac{tar^4}{r^2+b}\Big)
=2\pi t^{5/4}J(t),$$
where
$$J(t):=\int\_0^\infty du\,u^2\exp\Big(-\frac{au^4}{t^{-1/2}u^2+b}\Big).$$
By dominated convergence, as $t\to\infty$,
$$J(t)\to J(\infty):=\int\_0^\infty du\,u^2\exp\Big(-\frac{au^4}{b}\Big)\in(0,\infty).$$
So,
$$
I(t)\sim2\pi J(\infty) t^{5/4}\to\infty$$
as $t\to\infty$. In particular, $I(t)$ is unbounded.


