Does contour integral over $\mathbb{R}$ give a step function?

Question

I have the following integral $$ \int \frac{dx}{2\pi i} \frac{1}{(x+ia)^2+b^2} $$ where $x$ is a real variable and we integrate in the real axis from $-\infty$ to $+\infty$. I am also given the fact that $$ \int dx \frac{1}{(x + ia)^2+b^2}=\frac{\pi}{|b|}\theta(|b|-|a|) $$ integrating again in the real line and where $\theta$ is the step function. I want to know why the latter integral gives me this theta function and if I am right that the first integral would then give me $$ \frac{-i}{|b|}\theta(|b|-|a|) $$

Answer 1

Let $I(a,b)$ be integral of interest given by

$$I(a,b)=\int_{-\infty}^\infty \frac{1}{(x+ia)^2+b^2}\,dx \tag 1$$

We assume that $b\ne0$ else the integral is trivially $0$ for $a\ne 0$.

Note that the function $f(z)$, defined as $f(z)=\frac{1}{(z+ia)^2+b^2}$ has first-order poles at $z=-i(a\pm b)$ with corresponding residues given by

$$\text{Res}\left(\frac{1}{(z+ia)^2+b^2},z=-i(a\pm b)\right)=\mp \frac{1}{2ib} \tag 2$$

Next, $C$ be the closed contour comprised of the real line segment from $-R$ to $R$ and the semi-circle in the upper-half plane, with radius $R$ and centered at the origin. Define $J(a,b)$ as the closed-contour integral around $C$ of $f(z)$. Then, we can write

$$\begin{align} J(a,b)&=\oint_C \frac{1}{(z+ia)^2+b^2}\,dz\\\\ &=\int_{-R}^R \frac{1}{(x+ia)^2+b^2}\,dx+\int_0^{\pi}\frac{1}{(Re^{i\phi}+ia)^2+b^2}\,iRe^{i\phi}\,d\phi \tag 3 \end{align}$$

In the limit as $R\to \infty$, the first integral on the right-hand side of $(3)$ tends to $I(a,b)$ as given in $(1)$ while the second integral on the right-hand side of $(2)$ tends to $0$.

Moreover, the residue theorem guarantees that $J(a,b)$ is $2\pi i$ times the residues from the poles enclosed in $C$ as given in $(2)$. We have four cases to examine.

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Case 1: Neither pole is in the upper-half plane.

Here, if $\text{Re}(a-b)>0$ and $\text{Re}(a+b)>0$, then no pole is enclosed and $I(a,b)=0$.

Note that this corresponds to $|\text{Re}(a)|>|\text{Re}(b)|$.

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Case 2: Both poles are enclosed in the upper-half plane.

Here, if $\text{Re}(a-b)<0$ and $\text{Re}(a+b)<0$, then both poles are enclosed. The sum of the residues is $0$ and $I(a,b)=0$.

Note that this also corresponds to $|\text{Re}(a)|>|\text{Re}(b)|$.

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Case 3: Only the pole at $z=-i(a-b)$ is enclosed in the upper-half plane.

Here, if $\text{Re}(a-b)<0$ and $\text{Re}(a+b)>0$, then only the pole at $z=-i(a-b)$ is enclosed. The residue from this pole is $\frac{1}{2ib}$ and $I(a,b)=\frac{\pi}{b}$.

Note that this corresponds to $|\text{Re}(b)|>|\text{Re}(a)|$.

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Case 4: Only the pole at $z=-i(a+b)$ is enclosed in the upper-half plane.

Here, if $\text{Re}(a-b)>0$ and $\text{Re}(a+b)<0$, then only the pole at $z=-i(a+b)$ is enclosed. The residue from this pole is $-\frac{1}{2ib}$ and $I(a,b)=-\frac{\pi}{b}$.

Note that this also corresponds to $|\text{Re}(b)|>|\text{Re}(a)|$.

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Putting all $4$ cases together, we see that if $|\text{Re}(a)|>|\text{Re}(b)|$, $I(a,b)=0$, while if $|\text{Re}(b)|>|\text{Re}(a)|$, then $I(a,b)=\pm \frac{\pi}{b}$. In addition, for Case $3$, we see that $\text{Re}(b)>0$, while for Case $4$, $\text{Re}(b)<0$. Thus, we can finally write

$$\bbox[5px,border:2px solid #C0A000]{I(a,b)=\text{sgn}\left(\text{Re}(b)\right)\,\left(\frac{\pi}{b}\right)\,\theta\left(|\text{Re}(b)|-|\text{Re}(a)|\right)} \tag 4$$

where

$$\text{sgn}(x)=\begin{cases}1&,x>0\\\\0&,x=0\\\\1&,x< 0 \end{cases}$$

and

$$\theta(x)=\begin{cases}1&,x>0\\\\0&,x\le 0 \end{cases}$$

For real-valued $a$ and $b$, $(4)$ simplifies to

$$\bbox[5px,border:2px solid #C0A000]{I(a,b)=\left(\frac{\pi}{|b|}\right)\,\theta\left(|b|-|a|\right)} $$

as was to be shown!