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

Question

Calculate

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

We use the Monotone Convergence Theorem:

If $f_n \to f$ is a sequence of monotonically increasing non-negative measurable functions, then

$$ \lim_{n\to\infty}\int_{\Omega} f_n \;dx = \int_{\Omega } f \;dx $$

Notice that $e^{x-nx^2}\to e^{-\infty}=0 $ as $n\to\infty$ but the functions are decreasing so we consider the functions $$\frac{1}{e^{x-nx^2}}\to \infty$$

So the integral is $\infty$? Is this correct?

EDIT: According to wolframalpha, it should be $0$ not $\infty$. Not sure what I'm doing wrong.

Answer 1

Your sequence $f_n$ is monotonically decreasing to $0$, and $f_1\in L^1(\mathbb{R})$. So you can use the dominated convergence theorem (since $0 < f_n \leq f_1$ for every $n\geq 1$) to conclude that the limit is $0$.

Answer 2

Just for verification, we can actually compute the integral for each $n$: $$ \begin{align} \int_{\mathbb{R}}e^{x-nx^2}\,\mathrm{d}x &=\frac{e^{\frac1{4n}}}{\sqrt{n}}\int_{\mathbb{R}}e^{-\frac1{4n}+\frac x{\sqrt{n}}-x^2}\,\mathrm{d}x\\ &=\frac{e^{\frac1{4n}}}{\sqrt{n}}\int_{\mathbb{R}}e^{-\left(x-\frac1{2\sqrt{n}}\right)^2}\,\mathrm{d}x\\ &=\frac{e^{\frac1{4n}}}{\sqrt{n}}\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x\\[3pt] &=e^{\frac1{4n}}\sqrt{\frac\pi{n}} \end{align} $$ which tends to $0$.

Answer 3

You can also avoid the DCT and only use elementary inequalities. Over $\mathbb{R}^+$ the function $x-nx^2$ attains its maximum at $x=\frac{1}{2n}$ and the value of such maximum is $\frac{1}{4n}$. We have

$$ \int_{-\infty}^{+\infty}\exp\left(x-nx^2\right)\,dx = e^{\frac{1}{4n}}\int_{-\infty}^{+\infty}e^{-nx^2}\,dx\leq \frac{K}{\sqrt{n}} $$ so the wanted limit is clearly zero.