Find whether $\int_{1}^{\infty} \frac{sin(x)}{x} dx$ is converging or not

Question

Find whether $\int_{1}^{\infty} \frac{sin(x)}{x} dx$ is converging or not.

I tried to use comparison test or limit comparison test but could't find a suitable function. How can I determine what type of a function to use when using these tests?

Answer 1

Consider the area of each lobe of the function to be the sequence $a_n$

The area alternates in sign, and due to $1/x$, it decreases monotonically.

By alternating series test, it will converge.

Answer 2

This, like the analogous series $\sum \limits _{n=1} ^\infty \frac {\sin n} n$, is the prototypical application of the Abel-Dirichlet test. It's a subtle problem, no other convergence test among the usual ones would do the job. Please note that if the question had been about $\int \limits _1 ^\infty \frac {\sin x} {x^p} \Bbb d x$ with $p > 1$, then it would have been much easier.

The Abel-Dirichlet test tells us that if we have two functions $\alpha, f : [a,\infty) \to [0,\infty)$ such that $\alpha$ decreases to $0$ and there exist $M>0$ with $|\int \limits _a ^b f (x) \Bbb d x| \leq 0 \space \forall b \geq a$, then $\int \limits _a ^\infty \alpha(x) f(x) \Bbb d x$ converges.

With the notations above, take $a=1, \space \alpha = \frac 1 x, \space f = \sin x$. That $\frac 1 x$ satisfies the conditions on $\alpha$ is trivial to check; concerning $f$, note that $|\int \limits _1 ^b \sin x \Bbb d x| = |- \cos b + \cos 1| \leq |\cos b| + |\cos 1| \leq 2$, so choose $M=2$ and Abel and Dirichlet do the rest for you.

Answer 3

Bound it as following:

$$\left|\int_{n\pi}^{(n+1)\pi} \frac{\sin{x}}{x}dx\right|<\left|\frac{1}{x}\Bigg|_{n\pi}\int_{n\pi}^{(n+1)\pi}\sin x\,dx\right|=\frac1{n\pi}$$ $$\left|\int_{n\pi}^{(n+1)\pi} \frac{\sin{x}}{x}dx\right|>\left|\frac{1}{x}\Bigg|_{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}\sin x\,dx\right|=\frac1{(n+1)\pi}$$

where I used that $1/x$ is monotonously decreasing (therefore the biggest on the lower bound.

EDIT:

The entire integral can be written as:

$$\int_1^\infty \frac{\sin x}{x}\,dx=-\int_0^1\frac{\sin x}{x}\,dx+\sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x}\,dx$$

For this specific choice of bounds, we can factor out the alternating sign and write: $$\sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x}\,dx=\sum_{n=0}^\infty (-1)^n\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x}\,dx\right|$$ Now you have an alternating series, which is known to converge if the absolute value is monotonously decreasing. Which it is, because we bounded it very efficiently.

Comment: if you argue that $\int_0^1$ is also not proven to converge, just write the integral as $\int_1^\pi+\sum_{n=1}$ instead.