complex integration along real axis I am reading P. Coleman's many-body physics book. It gives a integral which is odd for me,
$$
\int_{-\infty}^{\infty}\frac{{\rm d}x}{x - {\rm i}\alpha} =
{\rm i}\pi\operatorname{sgn}(\alpha),
\qquad \alpha \in \mathbb{R}.
$$
My question is how to calculate it. Sorry for my stupid question, at least it looks like.
 A: In fact
\begin{eqnarray}
\mathrm{P.V.}\int_{-\infty }^{\infty }\frac1{x-i\alpha }\,d x&=&\lim_{R\to \infty }\int_{-R}^R \frac1{x-i\alpha }\,d x\\&=&\lim_{R\to \infty }\int_{-R}^R \frac{x+i\alpha}{x^2+\alpha^2 }\,dx\\
&=&\lim_{R\to \infty }\left(\frac12\log(x^2+\alpha^2)-i\arctan\left(\frac x\alpha \right)\right)\Biggr|_{-R}^{R}\\
&=&\pi i \text{ sign}(\alpha).
\end{eqnarray}
A: As a principal value, we have that
$$
\begin{align*}
\mathrm{P.V.}\int_{-\infty }^{\infty }\frac1{x-i\alpha }\,d x:=&\lim_{R\to \infty }\int_{-R}^R \frac1{x-i\alpha }\,d x\\=&\lim_{R\to \infty }(\log (R-i\alpha )-\log (-R-i\alpha ))\\
=&i\lim_{R\to \infty }(\arg(R-i\alpha )-\arg(-R-i\alpha ))
\end{align*}
$$
where I used that $\log z=\log|z|+i\arg(z)$. Now, by a simple trigonometric observation, we can see that
$$
\arg(R-i\alpha )-\arg(-R-i\alpha )=\begin{cases}
-2\left(\frac{\pi}{2}-\arg(R-i\alpha )\right),&\alpha <0\\
2\left(\frac{\pi}{2}-\arg(R-i\alpha )\right),&\alpha \geqslant 0
\end{cases}
$$
and that $\lim_{R\to \infty }\arg(R-i\alpha )=0$ for any chosen $\alpha \in \mathbb{R}$, therefore
$$
\mathrm{P.V.}\int_{-\infty }^{\infty }\frac1{x-i\alpha }\,d x=i\pi \operatorname{sign}(\alpha )
$$
∎

A more elementary approach is the following: just note that
$$
\int_{-R}^R \frac1{x-i\alpha }\,d x=\int_{-R}^0 \frac1{x-i\alpha }\,d x+\int_{0}^{R}\frac1{x-i\alpha }\,d z=-\int_{R}^0 \frac1{-x-i\alpha }\,d x+\int_{0}^{R}\frac1{x-i\alpha }\,d z\\
=\int_{0}^R \left[\frac1{x-i\alpha }-\frac1{x+i\alpha }\right]dx=\int_{0}^R \frac{2i\alpha }{x^2+\alpha ^2}dx
$$
Therefore
$$
\mathrm{P.V.}\int_{-\infty }^{\infty }\frac1{x-i\alpha }\,d x=\int_{0}^\infty \frac{2i\alpha }{x^2+\alpha ^2}dx
$$
Now the last integral can be easily computed, and it have the expected answer.∎
