Show that: $$ p.v.\int_{-\infty}^\infty\frac{1}{(\omega' - \omega_0)^2 + a^2}\frac{1}{\omega' - \omega}d\omega' = \frac{\pi}{a}\frac{\omega - \omega_0}{(\omega - \omega_0)^2 + a^2} $$

Where $p.v.$ is the principal value of the integral.

I suspect it has something to do with the "integral analog" of cauchy-riemann equations (below) which came from Cauchy Integral formula. $$ f(x_0) = \frac{1}{\pi i}p.v.\int_{-\infty}^\infty\frac{u(x) + iv(x)}{x-x_0}dx $$

One can separate the values of $u(x_0)$ and $v(x_0)$ separating real and imaginary parts from above formula.

$$ u(x_0) = \frac{1}{\pi}p.v\int_{-\infty}^\infty\frac{v(x)}{x-x_0}dx \\ v(x_0) = -\frac{1}{\pi}p.v\int_{-\infty}^\infty\frac{u(x)}{x-x_0}dx $$

However.. I am really stuck from here.... Indeed, the answer seems really close to these relations... But, I have no idea how this helps to solve integral, because, integration under $v(x)$ will give $u(x_0)$ and so forth.

Any help? How to solve?


$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\mathrm{P.V.}\int_{-\infty}^{\infty} {1 \over \pars{\omega' - \omega_{0}}^2 + a^{2}}\, {1 \over \omega' - \omega}\,\dd\omega'} = \mathrm{P.V.}\int_{-\infty}^{\infty} {1 \over \pars{\omega' - \Omega}^2 + a^{2}}\, \,{\dd\omega' \over \omega'}\,,\quad\Omega \equiv \omega_{0} - \omega \end{align}

\begin{align} &\color{#f00}{\mathrm{P.V.}\int_{-\infty}^{\infty} {1 \over \pars{\omega' - \omega_{0}}^2 + a^{2}}\, {1 \over \omega' - \omega}\,\dd\omega'} \\[5mm] = &\ \int_{0}^{\infty} \bracks{{1 \over \pars{\omega' - \Omega}^2 + a^{2}} - {1 \over \pars{-\omega' - \Omega}^2 + a^{2}}}\,{\dd\omega' \over \omega'} \\[5mm] = &\ 4\Omega\int_{0}^{\infty} {\dd\omega' \over \bracks{\pars{\omega' - \Omega}^2 + a^{2}} \bracks{\pars{\omega' + \Omega}^2 + a^{2}}} = \\[5mm] = &\ 2\Omega\int_{-\infty}^{\infty} {\dd\omega' \over \bracks{\pars{\omega' - \Omega}^2 + a^{2}} \bracks{\pars{\omega' + \Omega}^2 + a^{2}}} = 2\Omega\bracks{{\pi\,\mathrm{sgn}\pars{a} \over 2a\pars{a^{2} + \Omega^{2}}}} \label{1}\tag{1} \\[5mm] & = \color{#f00}{-\,{\pi \over \verts{a}} {\omega - \omega_{0} \over \pars{\omega - \omega_{0}}^{2} + a^{2}}} \end{align}

The integral, in expression \eqref{1}, is evaluated by 'standard means'. For example, by a contour integration in the complex plane or by partial fraction expansion.


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