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Consider the following definite real integral: $$I = \int_{0}^\infty dx \frac{e^{-ix} - e^{ix}}{x}$$

Using the $\text{Si}(x)$ function, I can solve it easily, $$I = -2i \int_{0}^\infty dx \frac{e^{-ix} - e^{ix}}{-2ix} = -2i \int_{0}^\infty dx \frac{\sin{x}}{x} = -2i \lim_{x \to \infty} \text{Si}(x) = -2i \left(\frac{\pi}{2}\right) = - i \pi,$$ simply because I happen to know that $\mathrm{Si}(x)$ asymptotically approaches $\pi/2$.

However, if I try to calculate it using the residue theorem, I get the wrong answer, off by a factor of $2$ and I'm not sure if I understand why. Here's the procedure: $$I= \int_{0}^\infty dx \frac{e^{-ix}}{x} - \int_{0}^\infty dx \frac{ e^{ix}}{x} = \color{red}{-\int_{-\infty}^0 dx \frac{e^{ix}}{x}} - \int_{0}^\infty dx \frac{ e^{ix}}{x} = -\int_{-\infty}^\infty dx \frac{e^{ix}}{x} $$ Then I define $$I_\epsilon := -\int_{-\infty}^\infty dx \frac{e^{ix}}{x-i\varepsilon}$$ for $\varepsilon > 0$ so that$$I=\lim_{\varepsilon \to 0^+} I_\varepsilon.$$ Then I complexify the integration variable and integrate over a D-shaped contour over the upper half of the complex plane. I choose that contour because $$\lim_{x \to +i\infty} \frac{e^{ix}}{x-i\varepsilon} = 0$$ and it contains the simple pole at $x_0 = i \varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_\varepsilon = -2 \pi i \, \text{Res}_{x_0} \left( \frac{e^{ix}}{x-i\varepsilon}\right) = -2 \pi i \left( \frac{e^{ix}}{1} \right)\Bigg\rvert_{x=x_0=i\varepsilon}=-2 \pi i \, e^{-\varepsilon}.$$ Therefore, $$I=\lim_{\varepsilon \to 0^+} \left( -2 \pi i \, e^{-\varepsilon} \right) = -2\pi i.$$

However, that is obviously wrong. Where exactly is the mistake?

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You've replaced the converging integral $\int_0^\infty \frac{\mathrm{e}^{-\mathrm{i} x} - \mathrm{e}^{\mathrm{i} x}}{x} \,\mathrm{d}x$ with two divergent integrals, $\int_0^\infty \frac{\mathrm{e}^{-\mathrm{i} x}}{x} \,\mathrm{d}x$ and $\int_0^\infty \frac{\mathrm{e}^{\mathrm{i} x}}{x} \,\mathrm{d}x$. (That something divergent has been introduced is evident in your need to sneak up on a singularity at $0$ that was not in the original integral.)

Also, notice that your D-shaped contour does not go around your freshly minted singularity at $x = 0$. The singularity lands on your contour. See the Sokhotski–Plemelj theorem to find that the multiplier for the residue of the pole is $\pm \pi \mathrm{i}$, not $\pm 2 \pi \mathrm{i}$.

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  • $\begingroup$ Ah, of course! And thank you for the additional info, very useful. $\endgroup$ – PhysSE is Cancer Mar 29 at 23:45
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You cannot shift the pole from the integration contour at will. Imagine that you shift it in the lower complex half-plane. Then instead of $-2\pi i$ you would obtain for the integral the value $0$!

The correct way to handle the pole is to take the half of its residue value, which is equivalent to bypassing the pole along a tiny semicircle around it (observe that the result does not depend on the choice between upper and lower semicircle).

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There is a problem at the very first step. You cannot split the integral because both integrals are divergent.

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