$$lim_{n \to \infty} \int_0^1(1 - e^{\frac{-x^2}{n}})x^{-1/2}dx$$

I want to use the Dominated Convergence Theorem to solve this.

Let $f_n = (1 - e^{\frac{-x^2}{n}})x^{-1/2}$.

Step 1: Determining convergence of $f_n$

Fix $x$ to be some constant number.

Now, bringing the limit inside the integral, we have $lim_{n \to \infty}(1 - \frac{1}{e^{\frac{k}{n}}})t$ where $k, t$ are constants.

As $n \to \infty, \ \frac{1}{e^{\frac{k}{n}}} \to 1$

So we get $(1 - 1)t = 0$

So $f_n$ converges to the zero function $f_0$. Note that it doesn't converge at $x = 0$ as we have a division by zero but on $(0, 1]$ it converges to $f_0$ and the measure of $\{0\}$ is $0$ so we have convergence almost everywhere.

Step 2: Determining a dominating function $g$

Now, when $x \in (0, 1)$, for all $n$ we have,

$$e^{\frac{1}{n}} > 1$$

$$\implies e^{\frac{x^2}{n}} \ge 1$$

$$\implies 0 < \frac{1}{e^{\frac{x^2}{n}}} \le 1$$

$$\implies -1 \le - \frac{1}{e^{\frac{x^2}{n}}} < 0$$

$$\implies 0 \le 1 - \frac{1}{e^{\frac{x^2}{n}}} < 1$$

$$\implies \mid 1 - \frac{1}{e^{\frac{x^2}{n}}}\mid < 1$$

$$\implies \mid (1 - \frac{1}{e^{\frac{x^2}{n}}})x^{-1/2}\mid < x^{-1/2}$$

Let $g = x^{-1/2}$

$g$ is integrable as $\int_0^\pi x^{-1/2} = 2\sqrt{x} \mid_0^\pi = 2\sqrt\pi$

So we have a sequence of integrable functions $f_n$ that converges a.e. to the function $f_0$. Let $g = \frac{1}{\sqrt{x}}$. By the dominated convergence theorem as $\mid f_n \mid \le g$ on $(0, 1)$ for all $n$ we have that $$\int_0^1 f_0 = lim_{n \to \infty} \int_0^1 f_n $$

And the integral of the zero function $f_0$ is $0$ so we have that

$$lim_{n \to \infty} \int_0^1(1 - e^{\frac{-x^2}{n}})x^{-1/2}dx = 0$$

Have I made any mistakes here? In particular, I have an issue with the way I brought the limit inside the integral in step 1. I couldn't apply the Monotone Convergence Theorem as we have decreasing sequence, so is there some other way to justify bringing it inside the integral?


2 Answers 2


Your argument is ok. In step 1 there are no integrals; you are just considering the limit of $f_n$.

  • 1
    $\begingroup$ Ah yes, I see that now upon re-reading the definition of LDCT. $\endgroup$
    – sonicboom
    Nov 24, 2013 at 20:24

Everything is ok. Step 1 could be the last step also. When you interchange the integral and the limit, with the help of DCT, that's when you need what is $\lim\limits_{n\to\infty} f_n(x)$.


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