Dominating convergence / Monotone convergence I need some pointers trying to solve the following:

Calculate
  $$\lim_{n\to \infty}\int_\mathbb{R} e^{-nx^2+x}dx$$

This is what I have: (I assume $n\in \mathbb{N}\setminus \{ 0\})$


*

*Define $f_n: \mathbb{R} \to [0,+\infty[ =e^{-nx^2+x}$

*Each $f_n$ is Lebesgue-measureable (how do you prove this?)

*Since $f_n$ is not increasing I'll resort to dominating convergence. I start looking for a certain $g:\mathbb{R}\to \mathbb{R}$ such that $|f_n(x)|\leq g(x)$ and $\int_\mathbb{R} g <+\infty$.
I noted that $-nx^2+x$ reaches its maximum for $n=1$ such that $e^{-nx^2+x} \leq e^{-x^2+x}$ However is this integrable? And how should i do this? Should I use a different bound?
 A: Noting that for $n\ge\frac54$, $-nx^2+x\le1-x^2$, we have that $e^{-nx^2+x}$ is dominated by $e^{1-x^2}$. Then Dominated Convergence yields that
$$
\lim_{n\to\infty}\int_{\mathbb{R}}e^{-nx^2+x}\,\mathrm{d}x=\int_{\mathbb{R}}0\,\mathrm{d}x
$$

With a couple of changes of variables, we also have
$$
\begin{align}
\int_{\mathbb{R}}e^{-nx^2+x}\,\mathrm{d}x
&=\int_{\mathbb{R}}e^{-(\sqrt{n}x-1/\sqrt{4n})^2+\frac1{4n}}\,\mathrm{d}x\\
&=e^{\frac1{4n}}\int_{\mathbb{R}}e^{-(\sqrt{n}x)^2}\,\mathrm{d}x\\
&=\frac1{\sqrt{n}}e^{\frac1{4n}}\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x\\
\end{align}
$$
A: Take $g(x) = e^{-x^2 + x}$, and $\int_{\mathbb{R}} g(x)dx = \int_{-\infty}^{-1} g(x)dx + \int_{-1}^1 g(x)dx + \int_{1}^\infty g(x)dx$. You can easily show each of them converges.
$\int_{-\infty}^{-1} g(x)dx \leq \int_{-\infty}^{-1} e^{-x^2 - 1} dx =e^{-1}\cdot  \int_{-\infty} ^{-1} e^{-x^2}dx = e^{-1}\cdot \int_{1}^\infty e^{-x^2}dx \leq e^{-1}\cdot \int_{1}^\infty e^{-x}dx = e^{-2}$.
The middle term is surely convergent since the integrand is continuous on $[-1,1]$, hence integrable there.
The last term is convergent because: $\int_{1}^\infty g(x)dx = \int_{1}^2 g(x)dx + \int_{2}^\infty g(x)dx \leq \int_{1}^2 g(x)dx + \int_{2}^\infty e^{-x}dx$ since $-x^2 + x \leq -x$ on $[2, \infty)$, and observe that each of the two integrals converges as the first one converges since the integrand is continous on $[1,2]$, and the second integral converges shown earlier.
Observe that: $f_n(x) \to f(x) = 0$ if $x\neq 0$, and $f(0) = 1$, and $\int_{\mathbb{R}} f(x)dx = 0$, and the answer follows.
A: You are right, for $n \geq 1 \ e^{-n x^2+x}\leq e^{-x^2+x} = h(x)$, and you can complete the square for this function:
$$
\int_{-\infty}^{\infty}h(x) = \int_{-\infty}^{\infty}e^{-x^2 +x}dx = e^{\frac{1}{4}}\int_{-\infty}^{\infty}e^{-(x-\frac{1}{2})^2}dx = e^{\frac{1}{4}}\sqrt{\pi}
$$
since the last integral is an error function. Hence by dominated convergence theorem you can interchange limit and integration, and get a $0$.
