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Question 1: Compute \(\int_0^\infty \frac{dx}{(x^2+p^2)^2}\) using the method of contour integration.
Solution: Let the given integral be denoted by \(I\). We first transform the given integral into an integral over full range \((-\infty, \infty)\).
\begin{equation}
I = \int_0^\infty \frac{dx}{(x^2+p^2)^2} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{(x^2+p^2)^2}\label{EQ1}
\end{equation}
Since along the real axis \(z=x, -\infty < x < \infty\), and \(dz-dx\) we can write the real integral as a contour integral along the real line which is in turn is equal to a \(\lim_{R\to\infty}\) of the contour integral taken over the interval \((-R,R)\).
\begin{equation}
I = \frac{1}{2}\int _\text{real line} \frac{dz}{(z^2+p^2)^2} = \frac{1}{2}\ \int_{AOB}\frac{dz}{(z^2+p^2)^2}.\label{EQ2}
\end{equation}
We need to close the contour so that the integration can be done using the Cauhcy residue theorem. The contour for this type of problems can be closed by adding a semi circular contour BCA. The interal along BCA will vanish as \(R \to \infty\). This is proved by a simple application of Darboux theorem. We note that along the semicircle BCA
\begin{eqnarray} |z^2+ p^2| & > & |R^2-p^2|, \qquad \because |\alpha+ \beta| \ge ||\alpha|-|\beta||\\ \therefore \quad \frac{1}{(z^2+p^2)^2} &\le& \frac{1}{(R^2-p^2)^2} \\ \int_\text{BCA} \frac{dz}{(z^2+p^2)^2} &\le& \frac{\pi R }{(R^2-p^2)^2}\end{eqnarray}
Therefore in the limit \( R \to \infty\), we get
\begin{equation} \lim_{R\to\infty} \int_\text{BCA} \frac{dz}{(z^2+p^2)^2} \longrightarrow 0 .\'end{equation} The \eqref{EQ2} can now be written as
\begin{eqnarray} I &=& \frac{1}{2} \lim_{R\to\infty} \big( \int_AOB \frac{dz}{(z^2+p^2)^2}+ \int BCA\frac{dz}{(z^2+p^2)^2} \big) \\
&=& \frac{1}{2} \lim_{R\to\infty} \oint_{AOBCA} \frac{dz}{(z^2+p^2)^2}\end{eqnarray}This integral can be computed easily by applying Cauchy residue theorem. The answer will be \(2\pi i \) \times sum of
residues at poles enclosed by the contour. Only the double pole at \(z=ip\) is enclosed inside the contour AOBCA. The residue at
this pole is given by \begin{eqnarray}Res \left[\frac{1}{(z^2+p^2)^2}\right]_{z=ip}
&=& \frac{d}{dz} \left. \frac{(z-ip)^2}{(z^2+p^2)^2}\right|_{z=ip}\\
&=& \frac{d}{dz} \frac{1}{(z+ip)^2}\Big|_{z=ip}
= \frac{-2}{(z+ip)^3} \Big|_{z=ip}\\
&=&\frac{-2}{-8ip^3} = \frac{1}{4ip^3}
\end{eqnarray} Therefore, we have
\begin{equation}
\lim_{R\to\infty} \oint_{AOBCA} \frac{dz}{(z^2+p^2)^2} = (2\pi i) \frac{1}{4ip^3} = \frac{\pi}{2 p^3}
\end{equation}and the required integral is
\[ I = \frac{1}{2}\oint_{AOBCA} \frac{dz}{(z^2+p^2)^2} = \frac{\pi}{4p^3} \]
