Question: Evaluate the integrals $$I(t) = \int_{-\infty}^\infty \frac{e^{itx}}{(x+i)^2}dx, \ \ \ -\infty < t < \infty.$$
My attempt: I have seen the answer to this question on this site and elsewhere, and I understand most of the algebra/calc steps given in those solutions, but I'm very confused about something. So, to evaluate the integral when $t > 0$, we use the basic semi-circle in the upper half plane contour, noting that the integral of $$f(z) = \frac{e^{itz}}{(z+i)^2}$$ is $0$ over that contour, and thus the integral over the real line must be $0$. That makes sense to be, but I am confused as to why we can't just do the same thing when $t < 0$. I can see that then we would have: $$f(z) = \frac{e^{-i|t|z}}{(z+i)^2},$$ but isn't this function still holomorphic in the semi-circle contour? And it seems like the bound $$\frac{|e^{-i|t|z}|}{|z+i|^2} \le \frac{1}{(R+1)^2}$$ still applies over the semi-circle contour, which should give the same result, that the integral over the real line is $0$.
I know I'm just being dumb here, but this is clearly a huge conceptual gap in my knowledge and I am hoping that someone can help explain where I'm going wrong. Thanks!
Note: It might possibly have to do with switching up sines and cosines when $t$ is negative in the Euler's formula? Not sure though.
For all $t\geq 1$ the integral will be zero according to Cauchy’s theorem which tells us that the integral of $f(z)$ around any simple closed curve that doesn’t enclose any singular points is zero. And we have two equal poles $z = -i $. Now for $t\leq -1$ we have $$\int \limits^{+\infty }_{-\infty }\frac{e^{itz}}{\left( z+i\right) ^{2}} dz=2\pi i \sum Res\left( f\left( z\right) , z_{0}\right) $$. Now we will work in the lower half plane. So our residue will be $$2\pi i\left( \lim \limits_{z\to -i}\left( \frac{1}{1! } \frac{d}{dz} \left( z+i\right) ^{2}\frac{e^{-itz}}{\left( z+i\right) ^{2}} \right) \right) =2\pi i \left( \lim \limits_{z\to -i}\frac{d}{dz} e^{-itz}\right) =2\pi i\times -ite^{t}=2\pi te^{t}$$. So the General form of your integral for all $t\leq -1$ $$\int \limits^{+\infty }_{-\infty }\frac{e^{itz}}{\left( z+i\right) ^{2}} dz=\boxed { 2\pi te^{t}}$$