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The fact that $\int_{0}^{\infty} \frac{\sin{x} dx}{x} $ converges has been proven multiple times for example: here

However I am wondering what is the way of proving it using Cauchy-Bolzano criterion.

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John Lennon
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    If you only want to know that the integral converges, you can break $(0,\infty)$ into $(0,\pi/2)$, $(\pi/2, 3\pi/2)$..... and argue similarly as in alternating series test for series, –  Mar 27 '15 at 07:50
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    @John: In my opinion, the easiest way to prove convergence is to integrate by parts: this leads to the integral $\int \frac{\cos x}{x^2},dx$ which converges absolutely. – Greg Martin Mar 27 '15 at 07:55

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The criterion would be: for every $\epsilon$, there exists $T$ such that for every $U > T$, we have $\big|\int_T^U \frac{\sin x}x\,dx\big| < \epsilon$. This can be proved by integrating by parts: \begin{align*} \bigg| \int_T^U \frac{\sin x}x\,dx \bigg| &= \bigg| {-}\frac{\cos x}x \bigg|_T^U - \int_T^U \frac{\cos x}{x^2}\,dx \bigg| \\ &\le \frac1T + \frac1U + \int_T^U \frac{1}{x^2}\,dx < \frac3T. \end{align*} So choosing $T = \frac3\epsilon$ suffices.

Greg Martin
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