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Here’s problem 31(e) from [1]. Find
\begin{equation}\label{eqn:thirdOrderPole:20}
I = \int_0^\infty \frac{x^2 dx}{\lr{ a^2 + x^2 }^3 }.
\end{equation}
Again, we use the contour C illustrated in fig. 1
fig. 1. Standard above the x-axis, semicircular contour.
Along the infinite semicircle, with z = R e^{i\theta} ,
\begin{equation}\label{eqn:thirdOrderPole:40}
\Abs{ \int \frac{z^2 dz}{\lr{ a^2 + z^2 }^3 } } = O(R^3/R^6),
\end{equation}
which tends to zero. We are left to just evaluate some residues
\begin{equation}\label{eqn:thirdOrderPole:60}
\begin{aligned}
I
&= \inv{2} \oint \frac{z^2 dz}{ \lr{ a^2 + z^2 }^3 } \\
&= \inv{2} \oint \frac{z^2 dz}{ \lr{ z – i a }^3 \lr{ z + i a }^3 } \\
&= \inv{2} \lr{ 2 \pi i } \inv{2!} \evalbar{ \frac{d^2}{dz^2} \lr{ \frac{z^2}{ \lr{ z + i a }^3 } } }{z = i a}
\end{aligned}
\end{equation}
Evaluating the derivatives, we have
\begin{equation}\label{eqn:thirdOrderPole:80}
\begin{aligned}
\lr{ \frac{z^2}{ \lr{ z + i a }^3 } }’
&= \frac{ 2 z \lr{ z + i a } – 3 z^2 }{ \lr{ z + i a }^4 } \\
&=
\frac{ – z^2 + 2 i a z }
{ \lr{ z + i a }^4 },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:thirdOrderPole:100}
\begin{aligned}
\frac{d^2}{dz^2} \lr{ \frac{z^2}{ \lr{ z + i a }^3 } }
&= \lr{ \frac{ – z^2 + 2 i a z }
{ \lr{ z + i a }^4 } }’ \\
&= \frac{ \lr{ – 2 z + 2 i a }\lr{ z + i a} – 4 \lr{ – z^2 + 2 i a z }}{ \lr{ z + i a }^5 },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:thirdOrderPole:120}
\begin{aligned}
\evalbar{ \frac{d^2}{dz^2} \lr{ \frac{z^2}{ \lr{ z + i a }^3 } } }{z = i a}
&=
\frac{ \lr{ – 2 i a + 2 i a }\lr{ 2 i a} – 4 \lr{ a^2 – 2 a^2 }}{ \lr{ 2 i a }^5 } \\
&=
\frac{ 4 a^2 }{ \lr{ 2 i a }^5 } \\
&=
\inv{8 a^3 i}.
\end{aligned}
\end{equation}
Putting all the pieces together, we have
\begin{equation}\label{eqn:thirdOrderPole:140}
\boxed{
I = \frac{\pi}{16 a^3}.
}
\end{equation}
References
[1] F.W. Byron and R.W. Fuller. Mathematics of Classical and Quantum Physics. Dover Publications, 1992.