Day: January 21, 2025

A contour integral with a third order pole.

January 21, 2025 math and physics play , , ,

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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.

Another real integral using contour integration.

January 21, 2025 math and physics play , , ,

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Here’s (31(d)) from [1]. Find
\begin{equation}\label{eqn:fourPoles:20}
I = \int_0^\infty \frac{dx}{1 + x^4} = \inv{2}\int_{-\infty}^\infty \frac{dx}{1 + x^4}.
\end{equation}
This one is easy conceptually, but a bit messy algebraically. We integrate over the contour \( C \) illustrated in fig. 1.

fig. 1. Standard above the x-axis, semicircular contour.

We want to evaluate
\begin{equation}\label{eqn:fourPoles:40}
2 I = \oint_C \frac{dz}{1 + z^4},
\end{equation}
because the semicircular part of the integral is \( O(R^{-3}) \), which tends to zero in the \( R \rightarrow \infty \) limit.

The poles are at the points
\begin{equation}\label{eqn:fourPoles:60}
\begin{aligned}
z^4
&= -1 \\
&= e^{i \pi + 2 \pi i k},
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:fourPoles:80}
\begin{aligned}
z
&= e^{i \pi/4 + \pi i k/2},
\end{aligned}
\end{equation}
These are the points \( z = (\pm 1 \pm i)/\sqrt{2} \), two of which are enclosed by our contour. Specifically
\begin{equation}\label{eqn:fourPoles:100}
\begin{aligned}
2 I
&= \oint_C \frac{dz}{
\lr{ z – \frac{1 + i}{\sqrt{2}} }
\lr{ z – \frac{-1 + i}{\sqrt{2}} }
\lr{ z – \frac{1 – i}{\sqrt{2}} }
\lr{ z – \frac{-1 – i}{\sqrt{2}} }
} \\
&= \oint_C \frac{dz}{
\lr{ z – \frac{1 + i}{\sqrt{2}} }
\lr{ z – \frac{-1 + i}{\sqrt{2}} }
\lr{ \lr{z + \frac{i}{\sqrt{2}}}^2 – \inv{2} }
} \\
&=
\evalbar{
\frac{ 2 \pi i }
{
\lr{ z – \frac{-1 + i}{\sqrt{2}} }
\lr{ \lr{z + \frac{i}{\sqrt{2}}}^2 – \inv{2} }
}
}{z = \frac{1 + i}{\sqrt{2}}}
+
\evalbar{
\frac{ 2 \pi i }
{
\lr{ z – \frac{1 + i}{\sqrt{2}} }
\lr{ \lr{z + \frac{i}{\sqrt{2}}}^2 – \inv{2} }
}
}
{z = \frac{-1 + i}{\sqrt{2}} } \\
&=
\evalbar{
\frac{(2 \pi i )(2 \sqrt{2})}
{
\lr{ z’ + 1 – i }
\lr{ \lr{z’ + i}^2 – 1 }
}
}{z’ = 1 + i}
+
\evalbar{
\frac{(2 \pi i )(2 \sqrt{2})}
{
\lr{ z’ – 1 – i }
\lr{ \lr{z’ + i}^2 – 1 }
}
}
{z’ = -1 + i}
\\
&=
\frac{2 \pi i \sqrt{2}}
{
\lr{2 i + 1}^2 – 1 }
–
\frac{2 \pi i \sqrt{2}}
{ \lr{2 i – 1}^2 – 1 }
\\
&=
\frac{\pi i \sqrt{2}}
{
2 (-1 + i)
}
+
\frac{\pi i \sqrt{2}}
{ 2(1 + i) }
\\
&=
\lr{ -1 – i }
\frac{\pi i}
{
2 \sqrt{2}
}
+
\lr{ 1 – i }
\frac{\pi i}
{ 2 \sqrt{2} }
\\
&=
\frac{\pi}
{ \sqrt{2} }
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:fourPoles:120}
\boxed{
I = \frac{\pi}{2 \sqrt{2}}.
}
\end{equation}

References

[1] F.W. Byron and R.W. Fuller. Mathematics of Classical and Quantum Physics. Dover Publications, 1992.