I have to resolve the following integral for a proof of theorem. It is used the residue theorem. However I am not in confidence with this argument. I would like to have more detailed step. The integral is
$$\int_{-\infty}^{\infty}\frac{{\rm d}k_{1}}{2\pi} \frac{e^{{\rm i}k_{1}\,x_{1}}}{{\rm i}k_{0} + k_{1}}$$
The integral can be solved in the comples plane in the half circle with radius $R\rightarrow \infty$ that rests on the real axis and lies in the upper half plane.
$$\int_{-\infty}^{\infty}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}=\int_{half\,circle}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}
$$
The integral on the circumnference goes to zero, infact, using the change of variable $k_1=Re^{i\theta}$, one has:
\begin{align}
&\int_{\smallfrown R}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}
\\[5mm] = &\
\int_{0}^{\pi} \frac{d\theta i Re^{i\theta}}{2\pi}\frac{e^{ix_1Rcos\theta}e^{-x_1Rsin\theta}}{ik_0+Re^{i\theta}}\rightarrow 0 \,\,\,\mbox{for}\,R\rightarrow\infty
\end{align}
And here the first question: how can it goes to zero?
At this point one can resolves using the residue theorem. The unique pole is in $k_1=-ik_0$ so to have a non zero result, one needs $k_0<0$. Why is the integral equal to zero in the case $k_0>0$?
So in the hypothesis that $x_1>0$ and $k_0<0$ one has:
$$\int_{-\infty}^{\infty}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}=ie^{k_0x_1}$$
Best Answer
For the half circle, the integral is bounded, up to a constant, by $$ \int_0^\pi e^{-x_1 R\sin \theta}\,d\theta=2\int_0^{\pi/2}e^{-x_1 R\sin\theta}\,d\theta \leq 2\int_0^{\pi/2}e^{-x_1R\theta/2}\,d\theta=\frac{2(1-e^{-x_1R\pi/4})}{x_1 R}\to0. $$ Or, even easier, you can apply Dominated Convergence.
When $k_0>0$, the pole lies outside your semicircle, so Cauchy's Theorem gives you that the integral on $[-R,R]$ is the same as the integral on the semicircle.