A contour integral with a third order pole.

January 21, 2025 math and physics play No comments , , ,

[Click here for a PDF version of this post]

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 No comments , , ,

[Click here for a PDF version of this post]

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.

Evaluating some real trig integrals with unit circle contours.

January 19, 2025 math and physics play No comments , , , , ,

[Click here for a PDF version of this post]

Here are a couple of problems from [1].

A sine integral.

This is problem (31(a)). For \( a > b > 0 \), find
\begin{equation}\label{eqn:unitCircleContourIntegrals:20}
I = \int_0^{2 \pi} \frac{d\theta}{a + b \sin\theta}.
\end{equation}
We can proceed by making a change of variables \( z = e^{i\theta} \), for which \( dz = i z d\theta \). Also let \( \alpha = a/b \), so
\begin{equation}\label{eqn:unitCircleContourIntegrals:40}
\begin{aligned}
I
&= \inv{b} \oint_{\Abs{z} = 1} \frac{-i dz}{z} \inv{\alpha + (1/2i)\lr{z – 1/z}} \\
&= \frac{2}{b} \oint_{\Abs{z} = 1} \frac{dz}{2 i z \alpha + z^2 – 1} \\
&= \frac{2}{b} \oint_{\Abs{z} = 1} \frac{dz}{\lr{ z + i \alpha + i\sqrt{\alpha^2 – 1}}\lr{ z + i \alpha – i\sqrt{\alpha^2 – 1}}}.
\end{aligned}
\end{equation}
Clearly the mixed sign factor represents the pole that falls within the unit circle, so we have only one residue to include
\begin{equation}\label{eqn:unitCircleContourIntegrals:60}
\begin{aligned}
I
&= \frac{2}{b} 2 \pi i \evalbar{ \inv{ z + i \alpha + i\sqrt{\alpha^2 – 1}} }{ z = -i \alpha + i \sqrt{ \alpha^2 – 1 } } \\
&= \frac{4 \pi i}{b} \inv{ 2 i \sqrt{\alpha^2 – 1}} \\
&= \frac{2 \pi}{\sqrt{a^2 – b^2}}.
\end{aligned}
\end{equation}

Sines and cosines upstairs and downstairs.

This is problem (31(b)). Given \( a > b > 0 \) (again), this time we want to find
\begin{equation}\label{eqn:unitCircleContourIntegrals:80}
I = \int_0^{2 \pi} \frac{\sin^2 \theta d\theta}{a + b \cos\theta}.
\end{equation}
We’d like to make the same \( z = e^{i \theta} \) substitution, but have to prepare a bit. We rewrite the sine
\begin{equation}\label{eqn:unitCircleContourIntegrals:100}
\begin{aligned}
\sin^2 \theta
&= \inv{2} \lr{ 1 – \cos(2\theta) } \\
&= \inv{2} – \inv{4}\lr{ e^{2 i \theta} + e^{-2 i \theta} },
\end{aligned}
\end{equation}
so, again setting \( \alpha = a/b \), we have
\begin{equation}\label{eqn:unitCircleContourIntegrals:120}
\begin{aligned}
I
&= \inv{b} \oint_{\Abs{z} = 1} \lr{ \inv{2} – \inv{4}\lr{ z^2 + 1/z^2 } } \frac{-i dz}{z} \inv{\alpha + (1/2)\lr{ z + 1/z} } \\
&= \frac{-i}{2 b} \oint_{\Abs{z} = 1} \lr{ 2 – z^2 – \inv{z^2} } \frac{dz}{ 2 \alpha z + z^2 + 1 } \\
&= \frac{-i}{2 b} \oint_{\Abs{z} = 1} dz \frac{ 2 z^2 – z^4 – 1 }{ z^2 \lr{ 2 \alpha z + z^2 + 1} } \\
&= \frac{-i}{2 b} \oint_{\Abs{z} = 1} dz \frac{ 2 z^2 – z^4 – 1 }{ z^2 \lr{ z + \alpha + \sqrt{ \alpha^2 – 1} }\lr{ z + \alpha – \sqrt{ \alpha^2 – 1} } }.
\end{aligned}
\end{equation}
The enclosed poles are at \( z = 0 \) (a second order pole) and \( z = -\alpha + \sqrt{ \alpha^2 – 1} \), so the integral is
\begin{equation}\label{eqn:unitCircleContourIntegrals:140}
\begin{aligned}
I
&= \lr{ 2 \pi i } \lr{ \frac{-i}{2 b} } \lr{
\evalbar{ \lr{ \frac{ 2 z^2 – z^4 – 1 }{ 2 \alpha z + z^2 + 1 } }’ }{z = 0}
+
\evalbar{ \frac{ 2 z^2 – z^4 – 1 }{ z^2 \lr{ z + \alpha + \sqrt{ \alpha^2 – 1} } } }{ z = -\alpha + \sqrt{ \alpha^2 – 1} }
}
\end{aligned}
\end{equation}
The derivative residue simplifies to
\begin{equation}\label{eqn:unitCircleContourIntegrals:160}
\begin{aligned}
\evalbar{ \lr{ \frac{ 2 z^2 – z^4 – 1 }{ 2 \alpha z + z^2 + 1 } }’ }{z = 0}
&=
\evalbar{ \frac{ 4 z – 4 z^3 }{2 \alpha z + z^2 + 1} – \frac{ 2 z^2 – z^4 – 1}{\lr{ 2 \alpha z + z^2 + 1 }^2 }\lr{ 2 \alpha + 2 z } }{z = 0} \\
&= 2 \alpha,
\end{aligned}
\end{equation}
whereas the remaining residue is
\begin{equation}\label{eqn:unitCircleContourIntegrals:180}
\evalbar{ -\frac{ \lr{z^2 – 1}^2 }{ z^2 \lr{ 2 \sqrt{ \alpha^2 – 1} } } }{ z = -\alpha + \sqrt{ \alpha^2 – 1} }
=
\evalbar{ -\lr{z – \inv{z}}^2 \inv{ 2 \sqrt{ \alpha^2 – 1} } }{ z = -\alpha + \sqrt{ \alpha^2 – 1} },
\end{equation}
but
\begin{equation}\label{eqn:unitCircleContourIntegrals:220}
\begin{aligned}
\inv{z}
&= \inv{ -\alpha + \sqrt{ \alpha^2 – 1 } } \frac{ \lr{ \alpha + \sqrt{ \alpha^2 – 1 }} }{ \lr{ \alpha + \sqrt{ \alpha^2 – 1 } } } \\
&= \frac{ \alpha + \sqrt{ \alpha^2 – 1 } }{ -\alpha^2 + \lr{ \alpha^2 – 1} } \\
&= -\lr{ \alpha + \sqrt{ \alpha^2 – 1 } },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:unitCircleContourIntegrals:240}
\begin{aligned}
z – \inv{z}
&= -\alpha + \sqrt{ \alpha^2 – 1 } + \alpha + \sqrt{ \alpha^2 – 1 }
&= 2 \sqrt{ \alpha^2 – 1 },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:unitCircleContourIntegrals:260}
\begin{aligned}
\evalbar{ -\frac{ \lr{z^2 – 1}^2 }{ z^2 \lr{ 2 \sqrt{ \alpha^2 – 1} } } }{ z = -\alpha + \sqrt{ \alpha^2 – 1} }
&= – \frac{ \lr{ 2 \sqrt{ \alpha^2 – 1 } }^2 }{ 2 \sqrt{ \alpha^2 – 1} } \\
&= – 2 \sqrt{ \alpha^2 – 1 },
\end{aligned}
\end{equation}
for a final answer of
\begin{equation}\label{eqn:unitCircleContourIntegrals:200}
\begin{aligned}
I
&= \frac{2 \pi}{b} \lr{ \alpha – \sqrt{\alpha^2 – 1} } \\
&= \frac{2 \pi}{b^2} \lr{ a – \sqrt{a^2 – b^2} }.
\end{aligned}
\end{equation}

Another cosine integral.

Last problem of this sort (31 (c)), was to find, again with \( a > b > 0 \)
\begin{equation}\label{eqn:unitCircleContourIntegrals:280}
I = \int_0^{2 \pi} \frac{ d\theta} {\lr{ a + b \cos \theta }^2 }.
\end{equation}
Making our \( z = e^{i \theta} \) substitution, and setting \( \alpha = a/b \), we have
\begin{equation}\label{eqn:unitCircleContourIntegrals:300}
\begin{aligned}
I &= \inv{b^2} \oint_{\Abs{z} = 1} \frac{ -i dz/z} {\lr{ \alpha + (1/2)\lr{ z + 1/z } }^2 } \\
&= \frac{-4 i}{b^2} \oint_{\Abs{z} = 1} \frac{ z dz}{\lr{ 2 \alpha z + z^2 + 1 }^2 } \\
&= \frac{-4 i}{b^2} \oint_{\Abs{z} = 1} \frac{ z dz}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^2\lr{ z + \alpha – \sqrt{\alpha^2 – 1} }^2}.
\end{aligned}
\end{equation}
Again, only this mixed sign pole will be within the unit circle, so
\begin{equation}\label{eqn:unitCircleContourIntegrals:320}
\begin{aligned}
I
&= \lr{\frac{-4 i}{b^2} }\lr{ 2 \pi i }
\lr{
\evalbar{ \lr{ \frac{z}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^2} }’ }{z = -\alpha + \sqrt{\alpha^2 – 1} }
}
\end{aligned}
\end{equation}

That derivative is
\begin{equation}\label{eqn:unitCircleContourIntegrals:340}
\begin{aligned}
\lr{ \frac{z}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^2} }’
&=
\inv{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^2} – \frac{2 z}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^3} \\
&= \frac{z + \alpha + \sqrt{\alpha^2 – 1} – 2 z}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^3} \\
&= \frac{-z + \alpha + \sqrt{\alpha^2 – 1}}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^3}.
\end{aligned}
\end{equation}
Evaluating it at our pole \( z = -\alpha + \sqrt{\alpha^2 – 1} \), we have
\begin{equation}\label{eqn:unitCircleContourIntegrals:360}
\begin{aligned}
\frac{-z + \alpha + \sqrt{\alpha^2 – 1}}{\lr{ z + \alpha + \sqrt{\alpha^2 – 1} }^3}
&=
\frac{ \alpha – \sqrt{\alpha^2 – 1} + \alpha + \sqrt{\alpha^2 – 1}}{\lr{ -\alpha + \sqrt{\alpha^2 – 1} + \alpha + \sqrt{\alpha^2 – 1} }^3} \\
&= \frac{ 2 \alpha }{\lr{ 2 \sqrt{\alpha^2 – 1} }^3 } \\
&= \inv{4} \frac{ \alpha }{\lr{ \alpha^2 – 1}^{3/2} },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:unitCircleContourIntegrals:380}
\begin{aligned}
I
&= \frac{8 \pi}{b^2} \inv{4} \frac{ \alpha }{\lr{ \alpha^2 – 1}^{3/2} } \\
&= \frac{2 \pi a }{b^3 \lr{ \alpha^2 – 1}^{3/2} },
\end{aligned}
\end{equation}
but \( b^3 = \lr{ b^2}^{3/2} \), for
\begin{equation}\label{eqn:unitCircleContourIntegrals:400}
I = \frac{ 2 \pi a }{ \lr{ a^2 – b^2 }^{3/2} }.
\end{equation}

References

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

What will be the value of k to satisfy this integral equation

January 19, 2025 math and physics play No comments , ,

[Click here for a PDF version of this post]

Another problem from x/twitter [1]:

Find \( k \), where
\begin{equation}\label{eqn:trigProp:20}
\int_0^{2 \pi} \sin^4 x dx = k \int_0^{\pi/2} \cos^4 x dx.
\end{equation}
I initially misread the integration range in the second integral as \( 2 \pi \), not \( \pi/2 \), in which case the answer is just 1 by inspection. However, solving the stated problem, is not much more difficult.

Since sine and cosine are equal up to a shift by \( \pi/2 \)
\begin{equation}\label{eqn:trigProp:40}
\sin(u + \pi/2) = \frac{e^{i(u + \pi/2)} – e^{-i(u + \pi/2)}}{2i} = \frac{e^{i u} + e^{-i u}}{2} = \cos u,
\end{equation}
we can make an \( x = u + \pi/2 \) substitution in the sine integral.

Observe that \( \cos^4 x = \Abs{\cos x}^4 \), but the area under \( \Abs{\cos x} \) is the same for each \( \pi/2 \) interval. This is shown in fig. 1.

fig. 1. Plot of |cos x|

Of course, the area under \( \cos^4 x \), will also have the same periodicity, but those regions will be rounded out by the power operation, as shown in fig. 2.

fig. 2. Plot of cos^4 x.

Since the area under \( \cos^4 x \) is the same for each \( \pi/2 \) wide interval, we have
\begin{equation}\label{eqn:trigProp:60}
\boxed{
k = 4.
}
\end{equation}

References

[1] CalcInsights. What will be the value of k to satisfy this integral equation, 2025. URL https://x.com/CalcInsights_/status/1880932308108341443. [Online; accessed 19-Jan-2025].

A fun ellipse related integral.

January 18, 2025 math and physics play No comments , , ,

[Click here for a PDF version of this post]

Motivation.

This was a problem I found on twitter ([2])

Find
\begin{equation}\label{eqn:ellipicalIntegral:20}
I = \int_0^\pi \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}.
\end{equation}

I posted my solution there (as a screenshot), but had a sign wrong. Here’s a correction.

Solution.

Let’s first assume we aren’t interested in the \( a^2 = b^2 \), nor either of the \( a = 0, b = 0\) cases (if either of \( a, b \) is zero, then the integral is divergent.)

We can make a couple simple transformations to start with
\begin{equation}\label{eqn:ellipicalIntegral:40}
\begin{aligned}
\cos^2 x &= \frac{\cos(2x) + 1}{2} \\
\sin^2 x &= \frac{1 – \cos(2x)}{2},
\end{aligned}
\end{equation}
and then \( u = 2 x \), for \( dx = du/2 \)
\begin{equation}\label{eqn:ellipicalIntegral:60}
\begin{aligned}
I
&= \int_0^{2\pi} 2 \frac{du/2}{a^2 \lr{ 1 + \cos u } + b^2 \lr{1 – \cos u}} \\
&= \int_0^{2\pi} \frac{du}{ \lr{ a^2 – b^2 } \cos u + a^2 + b^2 }.
\end{aligned}
\end{equation}
There is probably a simple way to evaluate this integral, but let’s try it the fun way, using contour integration. Following examples from [1], let \( z = e^{i u} \), where \( dz = i z du \), and \( \alpha = \lr{ a^2 + b^2 }/\lr{ a^2 – b^2 } \), for
\begin{equation}\label{eqn:ellipicalIntegral:80}
\begin{aligned}
I
&= \oint_{\Abs{z} = 1} \frac{dz/(i z)}{ \lr{ a^2 – b^2 } \lr{ z + \inv{z}}/2 + a^2 + b^2 } \\
&= \frac{2}{i \lr{ a^2 – b^2 }} \oint_{\Abs{z} = 1} \frac{dz}{ z \lr{ z + \inv{z} + 2 \alpha} } \\
&= \frac{2}{i \lr{ a^2 – b^2 }} \oint_{\Abs{z} = 1} \frac{dz}{ z^2 + 2 \alpha z + 1 } \\
&= \frac{2}{i \lr{ a^2 – b^2 }} \oint_{\Abs{z} = 1} \frac{dz}{ \lr{ z + \alpha – \sqrt{\alpha^2 – 1}}\lr{ z + \alpha + \sqrt{\alpha^2 – 1}} }.
\end{aligned}
\end{equation}

There is a single enclosed pole on the real axis. For \( a^2 > b^2 \) where \( \alpha > 0 \) that pole is at \( z = -\alpha + \sqrt{ \alpha^2 – 1} \), so the integral is
\begin{equation}\label{eqn:ellipicalIntegral:100}
\begin{aligned}
I
&= 2 \pi i \frac{2}{i \lr{ a^2 – b^2 }} \evalbar{ \frac{1}{ z + \alpha + \sqrt{\alpha^2 – 1 } }}{z = -\alpha + \sqrt{ \alpha^2 – 1}} \\
&= \frac{4 \pi}{ a^2 – b^2 } \frac{1}{ 2 \sqrt{\alpha^2 – 1 } } \\
&= \frac{4 \pi }{ 2 \sqrt{\lr{a^2 + b^2}^2 – \lr{a^2 – b^2}^2 } } \\
&= \frac{2 \pi }{ \sqrt{4 a^2 b^2 } } \\
&= \frac{\pi }{ \Abs{ a b } },
\end{aligned}
\end{equation}
and for \( a^2 < b^2 \) where \( \alpha < 0 \), the enclosed pole is at \( z = -\alpha – \sqrt{ \alpha^2 – 1} \), where
\begin{equation}\label{eqn:ellipicalIntegral:120}
\begin{aligned}
I
&= 2 \pi i \frac{2}{i \lr{ a^2 – b^2 }} \evalbar{ \frac{1}{ z + \alpha – \sqrt{\alpha^2 – 1 } }}{z = -\alpha – \sqrt{ \alpha^2 – 1}} \\
&= \frac{4 \pi}{ a^2 – b^2 } \frac{1}{ -2 \sqrt{\alpha^2 – 1 } } \\
&= \frac{4 \pi}{ b^2 – a^2 } \frac{1}{ 2 \sqrt{\alpha^2 – 1 } } \\
&= \frac{4 \pi }{ 2 \sqrt{\lr{a^2 + b^2}^2 – \lr{b^2 – a^2}^2 } } \\
&= \frac{2 \pi }{ \sqrt{4 a^2 b^2 } } \\
&= \frac{\pi }{ \Abs{ a b } }.
\end{aligned}
\end{equation}
Observe that this also holds for the \( a = b \) case.

References

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

[2] CalcInsights. A decent integral problem to try out, 2025. URL https://x.com/CalcInsights_/status/1880110549146431780. [Online; accessed 18-Jan-2025].