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I know that $\int_{0}^{+\infty}\frac{\sin(x)}{x}\,dx=\dfrac \pi 2$, by constructing the 2 variable integration $$\int_{0}^{+\infty}\sin(x)\int_{0}^{+\infty}\dfrac{1}{e^{xy}}\,dy\,dx$$ But this seems to be an ad hoc solution to me.

If I only need to prove it converges to a finite number, is there any general method?

Thanks a lot.

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    @Dr.MV Because evaluating a series when only asked for convergence is overkill apparently. – 2017-02-28
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    You are missing a $dx$. – 2017-02-28
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    This is a duplicate of http://math.stackexchange.com/q/390810/269764 but my phone won't let me mark to close it. Would someone please do so? – 2017-02-28
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    @BrevanEllefsen Perhaps you ought to refresh the page. – 2017-02-28
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    @BrevanEllefsen That "duplicate" is different and the single answer in the link does not address the conditional convergence. – 2017-02-28
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    1+: Thank you for sharing this solution, didn't know one could derive the integral so easily. – 2017-02-28

2 Answers 2

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To show that the improper Riemann integral $\int_0^\infty \frac{\sin(x)}{x}\,dx$ exists, we can use the Abel-Dirichlet test. There, we only need to establish that there exists a number $M$ such that for any $L>0$

$$\left|\int_0^L \sin(x)\,dx\right|\le M$$

Inasmuch as $\frac{1}{x}$ decreases to $0$ monotonically, then the test guarantees that the integral $\int_0^\infty \frac{\sin(x)}{x}\,dx$ exists.

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    So we reduce to the corresponding series. I see~ – 2017-02-28
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Here are two proofs.

For the first, we write $$ \int_0^\infty \frac{\sin x}{x} \ dx = \sum_{n=0}^\infty \int_{n \pi}^{ (n + 1) \pi} \frac{\sin x}{x} \ dx \ .$$ Now define $$ x_n = \int_{n \pi}^{ (n + 1) \pi} \frac{\sin x}{x} \ dx \ ,$$ and note that the terms $x_n$ alternate in sign since $\sin x$ alternates in sign. We have $$ - \frac{1}{ n} \leq x_n \leq \frac{1}{n} \ ,$$ so $\lim_{n \to \infty} x_n = 0$. Finally, note that $$|x_n| = \int_{n \pi}^{(n + 1) \pi } \frac{|\sin x|}{x} \ dx = \int_{(n + 1) \pi}^{(n + 2) \pi } \frac{|\sin x|}{x - \pi} \ dx \geq \int_{(n + 1) \pi}^{(n + 2) \pi } \frac{|\sin x|}{x} \ dx = |x_{n+1}| \ , $$ so $|x_n|$ is monotonically decreasing. We can thus conclude that the series $$ \sum_{n=0}^\infty x_n$$ satisfies the conditions of Leibniz's test for alternating series, and thus converges.

For the second, using continuity, one can show that $$ \int_0^1 \frac{\sin x}{x} \ dx < \infty \ .$$ Now for the remaining integral, we have $$ \int_1^\infty \frac{\sin x}{x} \ dx = - \frac{\cos x}{x} |_{1}^\infty - \int_1^\infty \frac{\cos x}{x^2} \ dx = \cos 1 - \int_1^\infty \frac{\cos x}{x^2} \ dx \ .$$ Now note that $$ \left| \int_1^\infty \frac{\cos x}{x^2} \ dx \right| \leq \int_1^\infty \frac{1}{x^2} \ dx < \infty \ .$$ So, the integral $$ \int_1^\infty \frac{\sin x}{x} \ dx < \infty \ .$$

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    These are not general methodologies, rather specific to this problem. – 2017-02-28
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    I'm not sure if your proof of the first part is valid, since this proves only that for *this specific* splitting of the integral (i.e. in evenly $\pi$ long integrals) the limit exists. You actually need to show that for any arbitrary splitting the series converges. That means for *any* sequence $b_k$ with $\lim_{k \to \infty} b_k=\infty$ the series $\sum_{k=0}^\infty \int_{b_{k-1}}^{b_k} sin(x)/x \,dx$ converges. – 2018-02-19