# Is $\lim_{n\to\infty}\int_0^1f_n(x)=\int_0^1f(x)$ in this case?

Let $$f_n(x)=\begin{cases}\frac{e^{x^2}}{x^2} &x\in\left[1/n,1\right]\\ 0 &x\in(-\infty,1/n)\cup(1,+\infty)\end{cases}.$$

This converges to $$f(x)=\begin{cases}\frac{e^{x^2}}{x^2} &x\in\left(0,1\right]\\ 0 &x\in(-\infty,0]\cup(1,+\infty)\end{cases}$$ pointwise on $$\Bbb R$$ and uniformly on $$(-\infty,-\varepsilon]\cup[\varepsilon,+\infty)$$ for every $$\varepsilon>0$$.

I'm then asked to find $$\lim_{n\to\infty}\int_0^1f_n(x)dx$$, and I think the following holds: $$\lim_{n\to\infty}\int_0^1f_n(x)dx=\lim_{n\to\infty}\int_0^{1/n}0+\lim_{n\to\infty}\int_{1/n}^1\frac{e^{x^2}}{x^2}dx=\lim_{n\to\infty}\int_{1/n}^1\frac{e^{x^2}}{x^2}=\int_{0}^1\frac{e^{x^2}}{x^2}=+\infty.$$Am I correct?

• I edited out a few typos in the definition of $f$ – Learner Nov 12 '18 at 13:19

## 3 Answers

You are correct. Also, you can use Dominated convergence Theorem with the Dominant function $$f$$ observing that $$|f_n(x)|\leq |f(x)|$$ $$\forall n$$ $$\forall x$$ since $$\frac{e^{x^2}}{x^2}\geq 0 \hspace{2mm}\forall x\in (0,1].$$ Thus, by the Dominated convergence theorem,

$$\lim_{n\rightarrow \infty}\int f_n=\int \lim_{n\rightarrow \infty}f_n =\int f =\int_{(0,1]}\frac{e^{x^2}}{x^2}.$$

Since $$+\infty=\int_{(0,1]}\frac{1}{x^2}\leq \int_{(0,1]}\frac{e^{x^2}}{x^2},$$

we have $$\lim_{n\rightarrow \infty}\int f_n=\int_{(0,1]}\frac{e^{x^2}}{x^2}=+\infty.$$

Yes it is true. This follows from Beppo Levi's monotone convergence theorem for Lebesgue integral here: https://en.wikipedia.org/wiki/Monotone_convergence_theorem

$$f_n(x) \ge \frac{1}{x^2}$$ on $$[1/n,1]$$ and by positivity of Riemann integral $$\int_0^1 f_n(x) dx = \int_{1/n}^1 \frac{e^{x^2}}{x^2} dx \ge \int_{1/n}^1 \frac{1}{x^2} dx = n-1.$$

The result follows.