Find $\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{2x^2+y^2<n}\Big(1-\frac{2x^2+y^2}{n^2}\Big)^{n^2}x^{2m}dxdy$ Find

$$
\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{2x^2+y^2&lt;n}\Big(1-\frac{2x^2+y^2}{n^2}\Big)^{n^2}x^{2m}dxdy
$$

My main (so far) problem is that I do not know what I should do with $n$ which defines set over I integrate. To move limit inside I should remove it. Is it correct to move it inside anyway and just assume that integral is defined over $\mathbb{R^2}$?
Define $M_n:=\{(x,y) : 2x^2+y^2&lt;n\}$
and $$f_n(x,y):=\chi_{M_n}(x,y)\left(1-\frac{2x^2+y^2}{n^2}\right)^{n^2}x^{2m}.$$
Note that $f_n\geq0$ converges monotonically increasing to $f(x,y):=e^{-2x^2-y^2}x^{2m}.$ With Levi you may interchange the limit with integral and then use Fubini:
\begin{align}
&amp;\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{2x^2+y^2&lt;n}\left(1-\frac{2x^2+y^2}{n^2}\right)^{n^2}x^{2m}dxdy \\
&amp;=\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{\Bbb R^2}\chi_{M_n}(x,y)\left(1-\frac{2x^2+y^2}{n^2}\right)^{n^2}x^{2m}dxdy \\
&amp;=\sum_{m=0}^\infty\int_{\Bbb R^2}e^{-2x^2-y^2}x^{2m}dxdy \\
&amp;=\sum_{m=0}^\infty\int_{\Bbb R^2}e^{-2x^2}x^{2m}e^{-y^2}dxdy \\
&amp;=\sum_{m=0}^\infty\int_{\Bbb R}e^{-2x^2}x^{2m}dx\int_{\Bbb R}e^{-y^2}dy \\
&amp;=\sqrt{\pi}\sum_{m=0}^\infty\int_{\Bbb R}e^{-2x^2}x^{2m}dx.
\end{align}
Now some kind of Gamma function is left. Try to proceed from here.

Update: Well, when I transform $\int_{\Bbb R}e^{-2x^2}x^{2m}dx$ further, I'm getting something like $\approx\left(\frac{1}{2}\right)^{m+\frac{1}{2}}\Gamma(m+\frac{1}{2})$, which doesnt converge to $0$. Thus, I'm having the notion that the series doesnt exist.

Update2: Okay, an easier way of seeing this: For the summands we obtain:
\begin{align}
\int_{\Bbb R}e^{-2x^2}x^{2m}dx
\geq \int_{[1,\infty)}e^{-2x^2}x^{2m}dx
\geq \int_{[1,\infty)}e^{-2x^2}dx=:\delta&gt;0.
\end{align}
This positive $\delta$ is independent of $m$ and thus the series over them goes to infinity.
Let $D_n$ be the set $\{(x,y)\in\Bbb R^2:2\,x^2+y^2&lt;n\}$ and $\chi_n\colon\Bbb R^2\to\Bbb R$ its characteristic function. Then for any function $f(x,y)$ we have
$$
\int_{D_n}f(x,y)\,dxdy=\int_{\Bbb R^2}\chi_n(x,y)\,f(x,y)\,dxdy.
$$