<s>[INST] Evaluate $\operatorname{lim}_{n\to\infty} \int_0^\infty \frac{n^2x}{1+x^2} e^{-n^2x^2} dx$ Evaluate the following limit for $a=0$ and for $a>0$
$$\lim_{n\to\infty} \int_a^\infty \frac{n^2x}{1+x^2} e^{-n^2x^2} dx.$$
For $a>0$, I can use dominated convergence theorem as $\frac{n^2x}{1+x^2} e^{-n^2x^2}\leq \frac{1}{x(1+x^2)}\in L^1(a,\infty)$, then the limit can be put into the integral to get the limit as $0$. But I'm not sure how to approach the case $a=0$. Can you please give me some hint?
 [/INST] Just for your curiosity.
$$\frac{n^2x}{1+x^2} e^{-n^2x^2}=e^{n^2}n^2 x\frac{e^{-n^2(1+x^2)}} {1+x^2 }$$
$$I=\int\frac{n^2x}{1+x^2} e^{-n^2x^2}\,dx=\frac 12e^{n^2}n^2\int \frac{e^{-n^2(1+x^2)}} {1+x^2 }\,d(1+x^2)$$ I suppose that the change of variable $t=1+x^2$ is clear and it makes
$$I=\frac{1}{2} e^{n^2} n^2 \,\text{Ei}\left(-n^2 \left(1+x^2\right)\right)$$ where appears the exponential integral function.
$$J=\int_0^\infty\frac{n^2x}{1+x^2} e^{-n^2x^2}\,dx=\frac{1}{2} e^{n^2} n^2\, \Gamma \left(0,n^2\right)$$ the expansion of which being
$$J=\frac{1}{2}-\frac{1}{2 n^2}+O\left(\frac{1}{n^4}\right)$$
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