Integration using residue theorem

Question

Can you find, using residue theorem, ($\epsilon >0$), the value of this integral ($I$)?

\begin{equation} I=\lim_{\epsilon->0^{+}} \int_{-\infty}^{\infty} \frac{dw}{w+i\epsilon} \end{equation}

Thank you very much!

Finally I have proved that its value is $I=-i\pi$ using a semi-circular contour $C=C_{1}+C_{2}$,

$C_{1}: z=x, -R<x<R$

$C_{2}: z=Re^{i\theta}, 0<\theta<\pi$

Supposing that the residue of integrand at $z=-i\epsilon$ is not in $C$ because $\epsilon>0$, we have

$$\lim_{\epsilon->0^{+}} \int_{C} \frac{dz}{z+i\epsilon}=I+\lim_{R->\infty}\int_{0}^{\pi}\frac{iRe^{i\theta} d\theta}{Re^{i\theta}+0}=2\pi i Res(f(z), z\in C)=0$$

and $$ \lim_{R->\infty}\int_{0}^{\pi}\frac{iRe^{i\theta} d\theta}{Re^{i\theta}+0}=\int_{0}^{\pi} id\theta=i\pi $$ Finally, $$ I=-\lim_{R->\infty}\int_{0}^{\pi}\frac{iRe^{i\theta} d\theta}{Re^{i\theta}+0}=-i\pi$$

Answer 1

By Cauchy's theorem (no residue involved ($\varepsilon>0$)), \begin{gather*} \forall R>0:\int_{-R}^R\frac{dw}{w+i\varepsilon} =-i\int_0^{\pi}\frac{Re^{i\varphi}} {Re^{i\varphi}+i\varepsilon}\,d\varphi\to-i\pi, \quad \varepsilon\to0.\\ \lim_{\varepsilon\to0}\left[p.v.\int_{-\infty}^\infty\frac{dw}{w+i\varepsilon}\right]=\lim_{R\to\infty}\left[\lim_{\varepsilon\to0} \int_{-R}^R\frac{dw}{w+i\varepsilon}\right]=-i\pi. \end{gather*}