Help with a contour integration

Question

I've been trying to derive the following formula

$$\int_\mathbb{R} \! \frac{y \, dt}{|1 + (x + iy)t|^2} = \pi$$ for all $x \in \mathbb{R}, y > 0$. I was thinking that the residue formula is the way to go (and would prefer a solution by this method), but I keep getting stuck either proceeding with the function as is and choosing the correct contour or finding a substitution which makes things easier. I would greatly appreciate some help on how best to proceed. Thanks in advance.

Answer 1

Let $z = x + iy$ $$ f(w) = \frac{y}{(1+zw)(1+\overline{z}w)} $$ Let $\gamma$ be the contour consisting of the real axis from $-R$ to $R$ followed by the semi-circle in the upper half plane from $R$ to $-R$.
$f$ has a one simple pole in the upper half plane at $w = -\frac{\overline{z}}{|z|^2} = -\frac{1}{z}$.
The residue of $f$ at this pole is $\frac{1}{2i}$. Apply the Residue Theorem and let $R \to \infty$. In the limit, the integral on the portion of $\gamma$ in the upper half plane goes to zero using the simple estimate given by the max of the modulus of $f$ on the curve multiplied by the length of the curve.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{% ? \equiv \int_{\mathbb{R}}{y\,\dd t \over \verts{1 + (x + \ic y)t}^{2}}}$

\begin{align} \color{#0000ff}{\large ?} &= y\int_{-\infty}^{\infty}{\dd t \over \pars{1 + xt}^{2} + \pars{yt}^{2}} = y\int_{-\infty}^{\infty}{\dd t \over \pars{x^{2} + y^{2}}t^{2} + 2xt + 1} \\[3mm]&= {y \over x^{2} + y^{2}} \int_{-\infty}^{\infty} {\dd t \over t^{2} + 2\bracks{x/\pars{x^{2} + y^{2}}}t + 1/\pars{x^{2} + y^{2}}} \\[3mm]&= {y \over x^{2} + y^{2}} \int_{-\infty}^{\infty} {\dd t \over \bracks{t + x/\pars{x^{2} + y^{2}}}^{2} - x^{2}/\pars{x^{2} + y^{2}}^{2} + 1/\pars{x^{2} + y^{2}}} \\[3mm]&= {y \over x^{2} + y^{2}} \int_{-\infty}^{\infty} {\dd t \over t^{2} + y^{2}/\pars{x^{2} + y^{2}}^{2}} = {y \over x^{2} + y^{2}}\,{1 \over \verts{y}/\pars{x^{2} + y^{2}}} \int_{-\infty}^{\infty} {\dd t \over t^{2} + 1} \\&=\color{#0000ff}{\large\sgn\pars{y}\pi} \end{align}

Answer 3

Note that

$$|1+(x+i y)t|^2=1+2 x t +(x^2+y^2) t^2$$

The integral is therefore a straightforward application of the residue theorem, if you want. That is, evaluate

$$\int_{-\infty}^{\infty} \frac{dt}{1+2 x t +(x^2+y^2) t^2}$$

The poles are at $t_{\pm}=(-x \pm i y)/(x^2+y^2)$. If we close in the upper half plane with a semicircle of radius $R$, and let $R\to\infty$, the integral about the circular arc vanishes as $\pi/((x^2+y^2)R)$, and by the residue theorem, the integral is

$$i 2 \pi \frac1{2 (x^2+y^2) t_++2 x} = \frac{i 2 \pi}{2 (-x+i y)+2 x} = \frac{\pi}{y}$$

as claimed.