I am trying to find $\lim_{n \to \infty} \int_{(0,\infty)} \frac{n \sin (x/n)}{x(1+x^2)}dx$ by applying the Dominated Convergence Theorem for Lebesgue Integrals.

First, considering $(f_n): f_n=\frac{n \sin (x/n)}{x(1+x^2)}=\frac{\sin(x/n)}{(x/n)} \frac{1}{1+x^2},$ it then follows that $\lim_{n \to \infty}\frac{\sin(x/n)}{(x/n)}\frac{1}{1+x^2}=\frac{1}{1+x^2}\lim_{n \to \infty}\frac{\sin(x/n)}{(x/n)}$. And thus substituting with $u=\frac{x}{n}$, we get that as $n \to \infty$, $u \to 0$. Obviously, $\lim_{u \to 0} \frac{\sin u }{u}=1 \Rightarrow \lim_{n \to \infty} f_n = \frac{1}{1+x^2}=f(x).$ Hence $f_n$ converges pointwise to $f$.

Then the requirement is that $(f_n)$ is in $L_1$. I go on to argue that since on $(0,\infty)$ both $\frac{\sin(x/n)}{(x/n)}$ and $\frac{1}{1+x^2}$ are continuous, then both functions are measurable. And because the measurable functions form an algebra, then $f_n$, their product, is also measurable. How do we argue however that they are Lebesgue integrable? I know, for example, that the Lebesgue integral $\int_{(0,\infty)}\frac{\sin x}{x}$ does not exist.

Also, I reckon we are going to use $g(x)=\frac{1}{1+x^2} \in L_1$ to dominate $f_n$ such that $|f_n|\leq g$ for all $n$. I argue as follows: on $(0,\infty)$ we have $|\sin (x/n)|\leq|x/n|$ for any $n$. Hence $|f_n|=\left|\frac{\sin(x/n)}{(x/n)}\frac{1}{1+x^2} \right|\leq \frac{1}{1+x^2}$. So, once again we get to the question of the function being in $L_1$.

Finally, it is straightforward to apply DCT and get $\lim_{n \to \infty}\int f_n= \int f=\left.\tan^{-1}(x)\right|^{\infty}_{0}=\frac{\pi}{2}$. However, application of the Riemann integral bothers me somewhat, as we know that for the bounded Riemann integrable function, the proper Riemann integral is equivalent to the Lebesgue version. Does this application hold?


It is not necessary to check that each $f_n$ is Lebesgue integrable. To apply Dominated Convergence, you only need to check that $g$ is Lebesgue integrable. [Anyway, once you have checked that $g$ is Lebesgue integrable, it follows immediately from your inequality $|f_n| \leq |g|$ that each $f_n$ is also Lebesgue integrable.]

So how do you show that $g$ itself is Lebesgue integrable? Perhaps you could first verify that $|g|\chi_{(0,N)}$ is Lebesgue integrable for each $N \in \mathbb N$. [As you correctly pointed out, the Lebesgue integral of $|g|\chi_{(0,N)}$ is equal to the Riemann integral of $|g|$ on the interval $(0,N)$.] Then apply Monotone convergence to deduce that $|g|$ is Lebesgue integrable (with integral equal to $\pi / 2$).

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    $\begingroup$ P.S. Since your $g$ is equal to your $f = \lim_{n \to \infty}f_n$, the same method can be used to show that $\int_{(0,\infty)}f = \pi / 2$. $\endgroup$ – Kenny Wong May 6 '17 at 11:51
  • $\begingroup$ Oh, that was pretty graceful as an approach. Thank you, Sir. :) $\endgroup$ – theheroneeded May 6 '17 at 13:17

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