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Case $1$

In order to evaluate the Dirichlet integral $$I_1=\int\limits_{-\infty}^{+\infty}\frac{\sin x}{x}dx=\pi,$$ we can consider another integral $$I_1^\prime=\oint_C\frac{e^{iz}}{z}.$$ Now, we can choose a semicircular contour C of radius $R$ to close in the upper half-plane and make another small semicircular indentation around $z=0$ (of radius $\varepsilon$) to avoid the pole at $z=0$. In this problem, the little semicircle about $z=0$ does contribute and it cannot be neglected.

Case $2$

Now consider the integral $$I_2=\int\limits_{-\infty}^{+\infty}\frac{e^{ix}}{x-x_0}=2\pi i e^{ix_0}$$ with $x_0>0$. Let us make it a closed contour integral by regarding $x$ to be complex, choosing a large semicircle enclosed in the upper half-plane and a smaller semicircular indentation (of radius $\varepsilon$) in the lower half-plane about $x=x_0$. If we want to do this using residue theorem, picking up the residue at $x=x_0$, the small semicircular contribution does not matter.

Why does the smaller semicircular contribution matter in the first case but not in the second case?

Solidification
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    at a glance it is beacuse in $I_1$ there is no singularity at zero in the original integral, while the complex integral $I_1'$ has a singularity there, so the small circle compensates for that; $I_2$ already has a singularity, so the big circle already accoounts for it – Conrad Jul 28 '20 at 02:08
  • Thanks, but how could you know (without evaluating) that the small semicircular contribution around $x=x_0$ vanishes in the limit $\varepsilon\to 0$ for the second case? Don't you need to check that explicitly? Sorry if I am asking a blunt question. – Solidification Jul 28 '20 at 02:16
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    I thought a little more and I believe we are both wrong - first $I_1=\pi$ (the residue gives $2\pi$ and the small indentation takes $\pi$ away) but the second integral has to be interpreted in some way (the integrand is not absolutely convergent at $x_0$) and I think the Cauchy PV is actually $\pi i e^{ix_0}$ so indeed the small semicircle contributes too - assuming one interpretes $I_2$ that way of course - so it may be a matter of defintions if something else is meant by $I_2$ ($I_1$ being convergent there is no ambiguity) – Conrad Jul 28 '20 at 03:04
  • You are right! $I_1$ has value $\pi$. Will correct it. – Solidification Jul 28 '20 at 03:10
  • what is your meaning for $I_2$? – Conrad Jul 28 '20 at 03:11
  • I am not sure what you mean by 'meaning'. One usually uses a $i\varepsilon$ prescription or indent the contour to avoid the pole. My mathematics is kind of lousy. – Solidification Jul 28 '20 at 03:15
  • the standard way to interpret $I_2$ is as Cauchy PV (principal value) so the limit $\epsilon \to 0, \epsilon >0$ of the integral up to $x_0-\epsilon$ and then from $x_0+\epsilon$ on; this way $I_2=\pi i e^{ix_0}$; – Conrad Jul 28 '20 at 03:18
  • @Conrad You may want to look at this question https://math.stackexchange.com/questions/4106170/with-this-contour-specified-how-can-the-integral-int-limits-infty-infty – Solidification Apr 21 '21 at 09:16

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You shouldn't ever "neglect" a piece of the contour: what you may do is estimate its contribution and show that it goes to $0$.

Robert Israel
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