Kdjykuj

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 $$I_epsilon := -int_{-infty}^infty dx frac{e^{ix}}{x-ivarepsilon}$$ 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
$$lim_{x to +iinfty} frac{e^{ix}}{x-ivarepsilon} = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , text{Res}_{x_0} left( frac{e^{ix}}{x-ivarepsilon}right) = -2 pi i left( frac{e^{ix}}{1} right)Biggrvert_{x=x_0=ivarepsilon}=-2 pi i , e^{-varepsilon}.$$
Therefore,
$$I=lim_{varepsilon to 0^+} left( -2 pi i , e^{-varepsilon} right) = -2pi i.$$

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

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 $pm 2 pi mathrm{i}$.

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 $-2pi i$ 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).

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