Is $\frac{\sin(x)}{1 + x^2}$ Lebesgue integrable?

Question

Is the following function Lebesgue measurable? $$f(x):= \frac{\sin x}{1 + x^2} $$The problem confuses me a bit, since it doesn't state where it wants it to be Lebesgue integrable. I figure that since $f$ is continuous on any interval $[a,b]$, further it is bounded by $1$, then it is Riemann-integrable (since the points where it is discontinuous is a Lebesgue null set), and since it is Riemann-integrable, it's definitely Lebesgue-integrable.

Is this correct? If yes, how do I extend "Lebesgue integrable on $[a,b]$" to just "Lebesgue integrable"?

Answer 1

For all $x\in\mathbb R$, $$0\leq \left|\frac{\sin(x)}{1+x^2}\right|\leq \left|\frac{1}{1+x^2}\right|=\frac{1}{1+x^2}.$$

Since $x\mapsto \frac{1}{1+x^2}$ is Lebesgue integrable (even Riemann integrable), $x\mapsto \frac{\sin(x)}{1+x^2}$ is Lebesgue integrable (and even Riemann integrable).

Answer 2

It is Lebesgue integrable on the entire line. Observe that $$\left| \frac{\sin x}{x^2 + 1} \right| \le \frac 1{x^2 + 1}$$ for all $x$. The integral of the latter function can be computed using Riemann integrals and the monotone convergence theorem.

Answer 3

You may actually prove: $$ I=\int_{0}^{+\infty}\frac{\sin x}{x^2+1}\,dx \leq \sqrt{\frac{2}{3}\,\log 2}\tag{1}$$ by using integration by parts and the Cauchy-Schwarz inequality:

$$ I = 2\int_{0}^{+\infty}\frac{s(1-\cos s)}{(1+s^2)^2}\,ds\leq 2\sqrt{\int_{0}^{+\infty}\frac{(1-\cos(s))^2}{s^3}\,ds\cdot \int_{0}^{+\infty}\frac{s^5}{(s^2+1)^4}\,ds}.\tag{2}$$

On the other hand, $\int_{0}^{+\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}$ and $\left|\sin x\right|\leq 1$. Moreover, $\frac{\sin x}{x^2+1}$ is an improperly Riemann-integrable function over $\mathbb{R}^+$ by Dirichlet's test, since $\sin x$ has a bounded primitive and $\frac{1}{x^2+1}$ is a decreasing function with limit zero.