Evaluate $ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$ I am trying to evaluate the following limit:
$$ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$$
Now seems we can use the fact that $$ \int_\mathbb{R}\frac{1}{1+x^2}dx = \pi$$ since this is the definite integral of $\arctan(x)$.
I tried finding a clever way to re-write the denominator but I am not getting anywhere.
 A: This problem is a straightforward application of the dominated convergence theorem. For all $n \ge 1$, we have $\dfrac{1}{1+x^2e^{nx^2}} \le \dfrac{1}{1+x^2}$, and $\displaystyle\int_{-\infty}^{\infty}\dfrac{1}{1+x^2}\,dx = \pi < \infty$.
Hence, we can use the dominated convergence theorem to say that $$\lim_{n \to \infty}\left[\int_{-\infty}^{\infty}\dfrac{1}{1+x^2e^{nx^2}}\,dx\right] = \int_{-\infty}^{\infty}\left[\lim_{n \to \infty}\dfrac{1}{1+x^2e^{nx^2}}\right]\,dx$$ provided that $\displaystyle\lim_{n \to \infty}\dfrac{1}{1+x^2e^{nx^2}}$ exists for all $x \in \mathbb{R}$. Can you evaluate this limit for all $x \in \mathbb{R}$? Then all you have to do is integrate that limit over $\mathbb{R}$.
A: $\lim\limits_{n\rightarrow +\infty}\frac{1}{1+x^2e^{nx^2}}=0$ if $x\neq 0$ and $1$ if $x=0$, the dominated convergence tells you that $$\lim\limits_{n\rightarrow +\infty}\int_{-\infty}^{+\infty}\frac{dx}{1+x^2e^{nx^2}}=0$$
(using the domination $\frac{1}{1+x^2e^{nx^2}}\leqslant\frac{1}{1+x^2}$)
A: One does not need any measure theory. It suffices to study $\int_0^\infty \frac{dx}{1+x^2e^{nx^2}}$.  Observe that for $x>0$,
\begin{align}
0< \frac1{1+x^2e^{nx^2}}
&=\frac{e^{-nx^2}}{e^{-nx^2}+x^2}
\le \frac{e^{-nx^2}}{x^2}
\end{align}
Also, $\frac1{1+x^2e^{nx^2}}\le 1$.
So for any $\epsilon>0$,
\begin{align}
0&\le \int_0^\infty \frac{dx}{1+x^2e^{nx^2}} 
\\&
\le \int_0^\epsilon dx + \int_{\epsilon}^\infty \frac{e^{-nx^2}}{x^2}dx 
\\&
= \epsilon + {\sqrt n}\int_{\epsilon\sqrt n}^\infty \frac{e^{-y^2}}{y^2}dy 
\\&
\le \epsilon + \frac1{\epsilon^2\sqrt n}\int_{\epsilon\sqrt n}^\infty e^{-y^2}dy   
\\&
\le \epsilon + \frac1{\epsilon^2\sqrt n}\int_{0}^\infty e^{-y^2}dy    \end{align}
Its well known that $\int_{0}^\infty e^{-x^2}dx $ is finite, but a good enough elementary bound follows from $e^t \ge 1+t$, or rather $e^{-t} \le \frac1{1+t}$, and the fact that $\int_0^\infty \frac{dy}{1+y^2} =\frac\pi2$ which is assumed in the OP. In addition, the integrand $\frac1{1+x^2e^{nx^2}}$ is decreasing as $n\to\infty$. So, the desired limit exists, and
$$0 \le \lim_{n\to\infty} \int_0^\infty \frac{dx}{1+x^2e^{nx^2}} \le \epsilon$$
Since $\epsilon$ was arbitrary, we conclude that $ \lim_{n\to\infty} \int_0^\infty \frac{dx}{1+x^2e^{nx^2}}=0$.
