Hfrxdjt

Consider the following definite real integral:
$$I = int_0^infty dx frace^-ix - e^ixx$$

Using the $textSi(x)$ function, I can solve it easily,
$$I = -2i int_0^infty dx frace^-ix - e^ix-2ix = -2i int_0^infty dx fracsinxx = -2i lim_x to infty textSi(x) = -2i left(fracpi2right) = - i pi,$$
simply because I happen to know that $mathrmSi(x)$ 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:
$$I= int_0^infty dx frace^-ixx - int_0^infty dx frac e^ixx = colorred-int_-infty^0 dx frace^ixx - int_0^infty dx frac e^ixx
= -int_-infty^infty dx frace^ixx $$
Then I define $$I_epsilon := -int_-infty^infty dx frace^ixx-ivarepsilon$$ 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
$$lim_x to +iinfty frace^ixx-ivarepsilon = 0$$ and it contains the simple pole at $x_0 = i varepsilon$. Using the residue theorem with the contour enclosing $x_0$ $$I_varepsilon = -2 pi i , textRes_x_0 left( frace^ixx-ivarepsilonright) = -2 pi i left( frace^ix1 right)Biggrvert_x=x_0=ivarepsilon=-2 pi i , e^-varepsilon.$$
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 $int_0^infty fracmathrme^-mathrmi x - mathrme^mathrmi xx ,mathrmdx$ with two divergent integrals, $int_0^infty fracmathrme^-mathrmi xx ,mathrmdx$ and $int_0^infty fracmathrme^mathrmi xx ,mathrmdx$. (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 $pm pi mathrmi$, not $pm 2 pi mathrmi$.

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.