Referring this answer, I tried to evaluate $\int_0^\infty\frac{\sin x}{x}dx$ by using contour integral $\int_C\frac{e^{iz}-1}{z}dz$ where $C$ is a upper half circle of radius $R$ centered at $0$. By Cauchy theorem, $$0=\int_C\frac{e^{iz}-1}{z}dz = \int_{-R}^R\frac{e^{ix}-1}{x}dx-\pi i+\int_0^\pi e^{-R\sin\theta}\operatorname{exp}(iR\cos\theta)d\theta$$ and the second summand $\to 0$ as $R\to \infty$. From this, I think we conclude $$\pi i = \int_{-\infty}^\infty \frac{e^{ix}-1}{x}dx$$ but how can I conclude $\int_0^\infty\frac{\sin x}{x}dx = \frac{\pi}{2}$?
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Hint: What is the relation between $\sin x$ and $e^{ix}$? – Kolja May 30 '21 at 09:52
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1@Kolja Ok the imaginary part is the key. But $\int_{-\infty}^\infty\frac{\sin x}{x}dx = \pi$ implies $\int_0^\infty\frac{\sin x}{x}dx = \frac{\pi}{2}$ is an obvious fact? just because the integrand is even function? – one potato two potato May 30 '21 at 09:56
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That is correct ! – Kolja May 30 '21 at 10:27
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@OliverDiaz No. This is a different question – one potato two potato Jun 06 '21 at 17:39
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Take the imaginary part on both sides: $$\Im(\pi i) = \Im \left(\int_{-\infty}^\infty \frac{e^{ix}-1}{x}\,\mathrm{d}x\right)\implies\pi=\int_{-\infty}^\infty \frac{\sin(x)}{x}\,\mathrm{d}x$$ Now use the symmetry of the integrand to change bounds.
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