Limit of a sequence of integrals Determine
$$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n}$$
Since $e^{x} \geq t+1$ we know that $e^{x^2}-1 \geq 2t +t^2 \geq t^2$ if $t\geq 0$
So due to symmetry, we should be able to use DCT?
But my solution is $$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n} = \int_{-\infty}^\infty \lim _{n \rightarrow \infty} \frac{dx}{n(e^{x^2}-1) +1/n} = 0$$. Which is the wrong answere.
Can someone help me out here?
We have not startet with variable change in measure theory, so please do not use that.
 A: We start from the the inequality $ e^y \le 1+y +\frac {3y^2} {2}$ if $ y \ge 0,\, y \le 1$. This follows from the Taylor's theorem 
  in the Lagrange form. Next, we obtain with help of Maple
  $$J_n:=\int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \ge \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{nx^2+3nx^4/2 +1/n}=$$ $$ \frac {1}{\sqrt {{n}^{2}-6}\sqrt { \left( \sqrt {{n}^{2}-6}-n
 \right) n}\sqrt { \left( \sqrt {{n}^{2}-6}+n \right) n}}$$ $$-2\,n\sqrt {3} \left( \mathop{\rm arctanh} \left( {\frac {\sqrt {3n}}{\sqrt { \left( \sqrt {{n}^{2}-6}-n \right) n}}} \right) \sqrt {
 \left( \sqrt {{n}^{2}-6}+n \right) n}+$$
 $$ \left.\sqrt { \left( \sqrt{{n}^{2}-6
}-n \right) n}\arctan \left( {\frac {\sqrt {n}\sqrt {3}}{\sqrt {
 \left( \sqrt {{n}^{2}-6}+n \right) n}}} \right)  \right)\to \pi,\, n \to \infty.(1)
 $$
On the other hand,
  $$J_n= \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \le \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{nx^2 +1/n} =\frac 1 n \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{x^2 +1/n^2}=2\arctan(\sqrt{n}) \to \pi,\,n \to \infty .(2)$$ 
  It remains to estimate the tails: $$\int_{-\infty}^{-\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n}+ \int_{\frac 1 {\sqrt{n}}}^{\infty} \frac{dx}{n(e^{x^2}-1) +1/n}\le 2\int_{\frac 1 {\sqrt{n}}}^{\infty} \frac{dx}{nx^2} =\frac 2 {\sqrt{n}} \to 0,\, n \to \infty .(3)$$  Combining (1)-(3), we deduce that the limit under consideration equals $\pi$.
