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

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"?

• Probably it means "Lebesgue integrable on $\mathbb R$". Sep 28 '15 at 13:33
• Out of context one would expect the domain to be $\mathbb{R}$ (so that in fact the domain could be whatever measurable set you like). In this case it is Lebesgue integrable on $\mathbb{R}$, so that seems to support the conclusion.
– Ian
Sep 28 '15 at 13:33
• Are you asking about measurability or integrability? It is both measurable and Lebesgue integrable, anyway, since $|f|$ is a non-negative Lipschitz function bounded by $\frac{1}{1+x^2}$ that is clearly integrable over $\mathbb{R}$. Sep 28 '15 at 13:33
• @JackD'Aurizio It's not nonnegative (maybe you meant $|f|$)
– Ian
Sep 28 '15 at 13:34
• @Ian: sure, fixed. Thanks. Sep 28 '15 at 13:36

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.

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).

• Why is $x \mapsto \frac{1}{1 + x^2}$ Lebesgue integrable? You should know I have yet to meet a single Lebesgue-integrable function in my studies other than the very trivial ones. Sep 28 '15 at 13:36
• Because $x\mapsto \frac{1}{1+x^2}$ is Riemann integrable and Riemann integrable $\implies$ Lebesgue integrable !
– Surb
Sep 28 '15 at 13:37
• ... allowing me to conclude it is Lebesgue integrable on every $[a,b]$, which is the only theorem I can refer to. Does it also hold for the entire real line? Sep 28 '15 at 13:38
• I don't understand your question. If it's integrable on $\mathbb R$, it's of course integrable on all $[a,b]$.
– Surb
Sep 28 '15 at 13:41
• No, I meant that using the theorems at $\textbf{my}$ disposal, I can only conclude it is integrable on $[a,b]$ (which doesn't imply integrability on all of $\mathbb{R}$). So what I'm asking is, does Riemann-integrability on all of $\mathbb{R}$ imply Lebesgue integrability on all of $\mathbb{R}$, and the theorem in my book just happens to not state that? (Obviously that's what your answer implies, I'm just curious why it's not mentioned in my book?) Sep 28 '15 at 13:42

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.