# Prove that $\frac{\sin(x)}{1+|x|}$ is not Lebesgue integrable

I would like to show that $$\frac{\sin(x)}{1+|x|}$$ is not Lebesgue integrable by showing it is not absolutely improperly Riemann integrable.

To this end, let $\epsilon >0$ be sufficiently small, say $\epsilon = \frac{\pi}{n}$ for some $n \in \mathbb{N}$ large.

Then, on $[\epsilon + k\pi,(k+1)\pi-\epsilon]$, $|\sin(x)|$ achieves a positive minimum, say $\alpha >0$.

Thus,

\begin{align*}\int_{-R\pi}^{R\pi} \left|\frac{\sin(x)}{1+|x|} \right| dx &\geq \alpha \int_{0}^{R\pi} \left|\frac{1}{1+x} \right| dx \\ & \geq \alpha \sum_{k=0}^{R-1} \int_{\epsilon + k\pi}^{(k+1)\pi-\epsilon} \left|\frac{1}{1+x} \right| dx \\ &= \alpha \sum_{k=0}^{R} \left [\log((k+1)\pi -\epsilon)) - \log(\epsilon + k\pi)\right] \end{align*}

and thus \begin{align*} \lim_{R \to \infty} \int_{-R}^{R} \left|\frac{sin(x)}{1+|x|} \right| dx &\geq \alpha \pi\left(\lim_{R \to \infty} \sum_{k=0}^{R} \left [\log((k+1)\pi -\epsilon)) - \log(\epsilon + k\pi)\right]\right) \\ &= \alpha \pi\left(\lim_{R \to \infty} \log((R+1)\pi -\epsilon))-\log(\epsilon)\right) \\ &= \infty \end{align*}

Is this a valid proof? Note the extra factor of $\pi$ due to the change of variables $y=x\pi$ in the beginning for notational convenience.

By Dirichlet's test, it is improperly Riemann integrable on $\mathbb{R}^+$. In explicit terms, the inverse Laplace transform gives $$\int_{0}^{+\infty}\frac{\sin(x)}{1+x}\,dx = \int_{0}^{+\infty}\mathcal{L}(\sin x)(s)\mathcal{L}^{-1}\left(\frac{1}{1+x}\right)(s)\,ds = \int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds$$ and by the Cauchy-Schwarz inequality $$\int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds\leq\sqrt{\frac{\pi}{8}}.$$ Similarly, for any $n\in\mathbb{N}^+$ we have $$0\leq \int_{0}^{+\infty}\frac{\sin(nx)}{1+x}\,dx \leq\frac{1}{n},\qquad 0\leq \int_{0}^{+\infty}\frac{\cos(nx)}{1+x}\,dx \leq \frac{1}{n^2}$$ and the last inequality can be used for proving $\frac{\sin x}{1+x}\not\in L^1(\mathbb{R}^+)$. Indeed $$\left|\sin x\right|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}$$ holds uniformly over any compact subset of the real line, hence $$\int_{0}^{M}\frac{\left|\sin x\right|}{1+x}\,dx = \frac{2}{\pi}\log(M+1)+O\left(\sum_{n\geq 1}\frac{1}{n^2(4n^2-1)}\right)=\frac{2}{\pi}\log M+O(1).$$
• @JohnDD: it works, but it is not really related to the improper Riemann-integrability of $\frac{\sin x}{1+x}$ and the constant appearing in the lower bound is not optimal. But of course we do not need an optimal constant for proving that $\int_{0}^{+\infty}\frac{|\sin x|}{1+x}\,dx = +\infty$. Feb 27 '18 at 17:33