\(\S\S\) 7.3 Q[1] \(\int_0^\infty\frac{1}{(x^2+p^2)^2} dx \)

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Full details are written out for this problem. This includes a proof that certain integrals vanish when Darboux theorem is used. In solutions to all other similar problems of this section, some of these details, being repetitive in nature, are suppressed. It is hoped that, if required, the reader will be able to supply the details by consulting this solution with full details.

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 \((-\infty, \infty)\).
    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}
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 \((-R,R)\).
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}
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}
Therefore, we have
\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}
and the required integral is
\[ I = \frac{1}{2}\oint_{AOBCA} \frac{dz}{(z^2+p^2)^2} = \frac{\pi}{4p^3} \]










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