# 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.

• this integral doesn't exists in the common sense, just as a principal value integral. This is because the integrals $\int_{0}^{\infty }\frac1{x-i\alpha }dx$ and $\int_{-\infty }^0 \frac1{x-i\alpha }dx$ both diverge Commented Dec 9, 2022 at 13:21
• If we understand the integral in the principal value sense as $\displaystyle I(\alpha)=\lim_{R\to\infty}\int_{-R}^R\frac{dx}{x-i\alpha}$, it can be evaluated. For $\alpha>0$ we close the contour in the upper half of complex plane. Denoting $I_R$ the integral along the half-circle of the radius $R$, $$\oint=I(\alpha)+I_R=2\pi iRes_{z=i\alpha}\frac{1}{z-i\alpha}=2\pi i$$ and, leading $R\to\infty$, $$I_R=\int_0^\pi\frac{iRe^{i\phi}d\phi}{Re^{i\phi}-i\alpha}\to\pi i\,\Rightarrow\,I(\alpha)=\pi i$$ Commented Dec 9, 2022 at 13:41
• For $\alpha$<0 we close the contour in the lower half-plane and integrate in the negative direction: $I(\alpha<0)=-\pi i$. For $\alpha=0$ we can define the integral as $\lim_{R\to\infty}\lim_{r\to0}\int_{-R}^{-r}+\int_r^R$ and get zero. Commented Dec 9, 2022 at 13:41
• @Svyatoslav why not put it into an answer. Very helpful. Commented Dec 9, 2022 at 13:57

In fact $$\begin{eqnarray} \mathrm{P.V.}\int_{-\infty }^{\infty }\frac1{x-i\alpha }\,d x&=&\lim_{R\to \infty }\int_{-R}^R \frac1{x-i\alpha }\,d x\\&=&\lim_{R\to \infty }\int_{-R}^R \frac{x+i\alpha}{x^2+\alpha^2 }\,dx\\ &=&\lim_{R\to \infty }\left(\frac12\log(x^2+\alpha^2)-i\arctan\left(\frac x\alpha \right)\right)\Biggr|_{-R}^{R}\\ &=&\pi i \text{ sign}(\alpha). \end{eqnarray}$$

• Sorry, I can only choose one answer, through I like your answer most. Commented Dec 10, 2022 at 2:03
• There is a typo: the last second line, first term should be log. Commented Dec 10, 2022 at 2:15

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.∎