As an introduction to Lebesgue integration, our professor gave us some problems of Riemann integration. One of these problems is the following function:
$$f_{n}(x) = n^2 x e^{-nx}.$$
He said the problem with this function is that:
$\lim_{n \rightarrow \infty} f_{n}(x) = 0$ for $x \in [0,1]$ And so the integration from $0$ to $1$ is also $0.$ But $\int_{0}^{1} f_{n}(x)dx = 1.$
My questions are:
1- Does the limit and the integral sign can always be interchanged in the case of Riemann integration? I do not think so, I think this is only true in case of uniformly continuous functions. Am I correct?
2- What is the reason for taking $x \in [0,1],$ is it for integrating reasons or is there a reason regarding the process of taking the limit?
3- Why did not we integrate over $n$ and not $x$?
4- How are we comparing integration over $x$ to taking the limit over $n$? Is not those are 2 very different things?
Could anyone help me in answering these questions that irritates my mind please?
EDIT:
Also, I calculated $\int_{0}^{1} f_{n}(x)dx $ but it was not 1 (I got $\frac{-n}{e^n} – \frac{1}{e^n} + 1$). am I correct? it is 1 after taking the limit as $n \to \infty.$
Best Answer
Integrating by parts, as you apparently have done, gives $\displaystyle\int_0^1 f_n(x) \, dx = 1 - (n+1)e^{-n}$ and, thus,
$$\lim_{n\to \infty} \int_0^1 f_n(x) \, dx = 1 \neq 0 = \int_0^1 \lim_{n\to \infty}f_n(x) \, dx$$
To show that $\displaystyle\lim_{n \to \infty}f_n(x) =\lim_{n \to \infty}n^2x e^{-nx} = 0$, note that $f_n(0) = 0$ and for $0 < x \leqslant 1$ we have
$$0 \tag{*}\leqslant n^2x e^{-nx} = \frac{n^2x}{e^{nx}} < \frac{n^2x}{\frac{n^3 x^3}{3!}} = \frac{6}{nx^2}$$
The inequality holds because $\displaystyle e^{nx} = 1+ nx + \frac{(nx)^2}{2!} + \frac{(nx)^3}{3!} + \ldots> \frac{(nx)^3}{3!} $.
The RHS of (*) converges to $0$ as $n \to \infty$ and, by the squeeze theorem, it follows that $n^2x e^{-nx} \to 0$.