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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.

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    $\begingroup$ 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 $\endgroup$
    – Masacroso
    Commented Dec 9, 2022 at 13:21
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    $\begingroup$ 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 $R\to\infty$, $$I_R=\int_0^\pi\frac{iRe^{i\phi}d\phi}{Re^{i\phi}-i\alpha}\to\pi i\,\Rightarrow\,I(\alpha)=\pi i$$ $\endgroup$
    – Svyatoslav
    Commented Dec 9, 2022 at 13:41
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    $\begingroup$ 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. $\endgroup$
    – Svyatoslav
    Commented Dec 9, 2022 at 13:41
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    $\begingroup$ @Svyatoslav why not put it into an answer. Very helpful. $\endgroup$ Commented Dec 9, 2022 at 13:57

2 Answers 2

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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}

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

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