Improper integral $\sin(x)/x $ converges absolutely, conditionally or diverges?

Question

Improper integral of $\sin(x)/x $ converges absolutely, conditionally or diverges?

We have $$\int_1^{\infty}\frac{\sin x}{x}\text{d}x$$

Integrating by parts $$u=\frac{1}{x}$$ $$\text{d}u=-\frac{1}{x^2}\text{d}x$$ $$\text{d}v=\sin x\;\text{d}x$$ $$v=-\cos x$$ $$ \begin{aligned} \int_1^{\infty} \frac{\sin x}{x} \text{d}x & = \frac{-\cos x}{x} \Big|_1^{\infty} - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \\ & = \cos 1 - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \end{aligned} $$

$\int_1^{\infty} \frac{\cos x}{x^2} \text{d}x$ converges absolutely (using the Comparison Test For Improper Integrals):

$$ \int_1^{\infty} \frac{|\cos x|}{x^2} \text{d}x < \int_1^{\infty} \frac{1}{x^2} \text{d}x $$

So $\int_1^{\infty} \frac{\sin x}{x} \text{d}x$ converges.

Now I need to find out if $\int_1^{\infty} |\frac{\sin x}{x}| \text{d}x$ converges or diverges.

Answer 1

Let $N \in \Bbb N, N > 1$, we have:

\begin{align} \int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\ &\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\ &= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx \\ &= \sum_{n=0}^{N-1} \frac{2}{\pi (n+1)} \end{align}

The last sum diverges as $N \to \infty$, and so does the original integral.

Your integral is on $[1, \infty]$, but it also diverges because $\left|\frac{\sin{x}}{x}\right|$ is continuous on $[0, 1]$. My proof is on $[0, \infty]$ because it makes managing the summation slightly easier.

Answer 2

If one's allowed to use the Absolute Divergent Theorem for Improper Integrals, then one could use what follows:

$\:|\sin x|>\frac{\:1}{\sqrt2}\:,\:\:\:\forall x\in\left]\pi(j+\frac{1}{4}),\pi(j+\frac{3}{4})\right[=I_j\:,\:\:\:j\in\mathbf N.$

$\:\:\forall x\in I_j,\: \forall q\in\:]0,1]\:$ we have $$\sum_{j\in\mathbf N}\int_{I_j}{{\text{d}x}\over x^q}\:\le\:\int_1^\infty{{\text{d}x}\over x^q}=\infty\:,$$by comparison.

So with $\{I_j\}$ being an increasing sequence and $|I_j|=\pi/2\:$ with $\:\lim_{j\in\mathbf N}\:\pi(j+3/4)=\infty,$ $$\int_{1}^\infty{|\sin x|\over x^q}\:\text{d}x=\infty$$

Answer 3

I would like to provide an intuitive graphic explanation.

Since $\left|\frac{\sin(x)}{x}\right|$ is continuous on $[0, \infty)$, thus it suffices to show the absolute divergence of the tail, namely $\int_{\pi}^{\infty} \left|\frac{\sin(x)}{x}\right| dx$.

Consider bounding this integral below by the infinite sum of area of triangles, namely the n-th triangle has width $\pi$ and height $\frac{1}{(n +1/2) \pi}$, thus the total area (of triangles) = $$\sum_{n = 1}^{\infty} \frac{1}{n + 1/2} = \infty$$ By a comparison test to the harmonic series $\sum_{k = 2}^{\infty} \frac{1}{k}$, thus the original integral diverges.