How to prove convergence of $\int_0^{\infty}{\frac{sin(x)}{x}dx}$ without evaluating it How to prove convergence of $\int_0^{\infty}{\frac{sin(x)}{x}dx}$ without evaluating it. Since $\lim_{x \to 0}\frac{sin(x)}{x}=1$, there is no problem around zero. But how to show the convergence as the limit goes to $\infty$?
Using integration by parts we can write $\int_1^{\infty}{\frac{sin(x)}{x}dx}=\lim_{c \to \infty}\int_1^{c}{\frac{sin(x)}{x}dx}=\lim_{c \to \infty}[\frac{-cos(x)}{x}]_1^{c}-\lim_{c \to \infty}\int_1^c\frac{cos(x)}{x^2}$
The second term can be managed by comparison since $\int_1^c\frac{cos(x)}{x^2} \leq \int_1^c\frac{1}{x^2} $ which is convergent.
But how do you manage $\lim_{c \to \infty}[\frac{-cos(x)}{x}]_1^{c}=\lim_{c \to \infty}[\frac{-cos(c)}{c}] + cos(1)$?
 A: We know that $\lim_{x \to \infty} \frac{\sin{x}}{x} = 0$. Because the sine oscillates between being positive and negative, we can divide the integral up into 'chunks', on each of which the function is either wholly positive or wholly negative. If the integral exists, then this will be equal to the series of these 'chunks'. Then, by the alternating series theorem, the series (which equals the integral should it exist) converges - the terms go to zero monotonically, for sine is bounded and $\frac{1}{x}$ goes to zero monotonically.
A: Integration by parts gives
$$\int_0^N\frac{\sin x}{x}\,dx
=\left[\frac{1-\cos x}{x}\right]_0^N+\int_0^N\frac{1-\cos x}{x^2}\,dx
=\frac{1-\cos N}{N}+\int_0^N\frac{1-\cos x}{x^2}\,dx.$$
So
$$\lim_{N\to\infty}\int_0^N\frac{\sin x}{x}\,dx
=\lim_{N\to\infty}\int_0^N\frac{1-\cos x}{x^2}\,dx.$$
This last is a convergent integral.
A: About your last question in your post:
$$\lim_{x\to\infty}-\frac{\cos c}c=-\lim_{c\to\infty}\,\frac1c\cdot\cos c=0$$
since the last is the limit of a function whose limit zero times a bounded one.
A: $\int_0^\infty \frac {sin x}{x} dx = \int_0^\pi \frac {sin x}{x} dx + \int_\pi^{2\pi} \frac {sin x}{x} dx +\cdots$
or $\sum_\limits{i=0}^{\infty} \int_{i\pi}^{(i+1)\pi} \frac {sin x}{x} dx$
$\frac {\sin x}{x}\le 1$ for all $x$
$\int_0^\pi \frac {sin x}{x} dx < \pi$
$|\int_{i\pi}^{(i+1)\pi} \frac {\sin x}{x} dx|< \frac {\pi}i$
$\lim_\limits {i\to\infty} \frac {1}{i} = 0$
$|\int_{i\pi}^{(i+1)\pi} \frac {\sin x}{x} dx| < |\int_{(i+1)\pi}^{(i+2)\pi} \frac {\sin x}{x} dx|$
The sign is alternating.
We have passed the alternating series test.
