0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Well, the result should be zero. I think $e^{x-nx^2}\rightarrow0$ is enough to say that is it zero, I don't know why the last consequence is in $\endgroup$ – Evgeny Nov 2 '17 at 17:51
  • 1
    $\begingroup$ How do you think the functions $$\frac{1}{e^{x-nx^2}}$$ are rrelated to your question? $\endgroup$ – Did Nov 2 '17 at 17:52
  • $\begingroup$ Why do you put it in the denominator? Seems like the answer is that the integral is zero. $\endgroup$ – jdods Nov 2 '17 at 17:52
  • $\begingroup$ @Did Cause I want a monotonically increasing function so that I can use the Monotone Convergence Theorem $\endgroup$ – berrygreen Nov 2 '17 at 17:55
4
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ Yes, I know that I can use the Dominated Convergence Theorem, but in this context it was implied that the question should be done using the Monotone Convergence Theorem. Can you help me figure out how to accomplish that? $\endgroup$ – berrygreen Nov 2 '17 at 17:57
  • $\begingroup$ @berrygreen: Monotone convergence is usually used for increasing functions, so it may be possible, but might be a bit contrived to apply it here. $\endgroup$ – robjohn Nov 2 '17 at 18:00
  • 1
    $\begingroup$ @berrygreen: you can use the version of the monotone convergence theorem that says that, if $f_1 \geq f_2 \geq \cdots$, $f_1 \in L^1$ and $f_n \to f$ a.e., then $\lim_n \int f_n = \int f$. $\endgroup$ – Rigel Nov 2 '17 at 18:05
2
$\begingroup$

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 $0$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.