math and physics play

More residue calculus: sinc squared, fractional exponent, log, pie contour

January 26, 2025 math and physics play , , , ,

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Sinc squared.

This is problem 31(g) from [1]. Find
\begin{equation}\label{eqn:sincSquared:20}
I = \int_{-\infty}^\infty \frac{\sin^2 x}{x^2} dx.
\end{equation}

We will use the same upper half plane semicircular contour, enclosing the second order pole at the origin. This time we write
\begin{equation}\label{eqn:sincSquared:40}
\sin^2 x = \frac{ 1 – \cos\lr{ 2x } }{2},
\end{equation}
allowing us to write
\begin{equation}\label{eqn:sincSquared:60}
\begin{aligned}
I
&= \inv{2} \int_{-\infty}^\infty \frac{1 – \cos\lr{ 2 x } }{x^2} dx \\
&= \inv{2} \textrm{Re} \int_{-\infty}^\infty \frac{1 – e^{2 i x } }{x^2} dx.
\end{aligned}
\end{equation}
This has exactly the structure required to apply Jordan’s lemma, and conclude that the integral over the infinite semicircular part of the contour is zero.

We can proceed to compute the residues
\begin{equation}\label{eqn:sincSquared:80}
\begin{aligned}
I
&=
\inv{2} \textrm{Re} \oint \frac{1 – e^{2 i z } }{z^2} dz \\
&=
\inv{2} \textrm{Re} \frac{ \pi i }{1!} \evalbar{ \lr{1 – e^{2 i z } }’ }{ z = 0 }.
\end{aligned}
\end{equation}
Because we are sneaking around the pole at the origin with a half semicircle, we multiply the residue by \( \pi i \), not \( 2 \pi i \). This leaves
\begin{equation}\label{eqn:sincSquared:100}
I = \textrm{Re} \frac{\pi i}{2} \evalbar{ (- 2 i) e^{2 i z } }{ z = 0 },
\end{equation}
or
\begin{equation}\label{eqn:sincSquared:640}
\boxed{
I = \pi.
}
\end{equation}

An fractional exponent integral.

Next, let’s do 31(h). Given \( 0 < a < 1 \), we want to evaluate \begin{equation}\label{eqn:sincSquared:120} I = \int_0^\infty \frac{x^{2 a – 1}}{x^2 + b^2} dx. \end{equation} Let’s start with a \( x = b u \), where we assume that \( b > 0 \). This gives us
\begin{equation}\label{eqn:sincSquared:140}
\begin{aligned}
I
&= \int_0^\infty \frac{ b^{2 a – 1} u^{2 a – 1}}{b^2 \lr{ u^2 + 1 } } b du \\
&= b^{2(a – 1)} \int_0^\infty \frac{ u^{2 a – 1}}{u^2 + 1 } du.
\end{aligned}
\end{equation}
We now want to solve the slightly simpler integral, let’s call it
\begin{equation}\label{eqn:sincSquared:160}
J = \int_0^\infty \frac{ u^{2 a – 1}}{u^2 + 1 } du.
\end{equation}

Let’s now look at
\begin{equation}\label{eqn:sincSquared:180}
\begin{aligned}
K
&= \int_{-\infty}^0 \frac{ u^{2 a – 1}}{u^2 + 1 } du \\
&= -\int_{\infty}^0 \frac{ \lr{ – v } ^{2 a – 1}}{v^2 + 1 } dv \\
&= \lr{-1}^{2 a – 1} J.
\end{aligned}
\end{equation}
However
\begin{equation}\label{eqn:sincSquared:200}
\begin{aligned}
\lr{-1}^{2 a – 1}
&=
e^{i \pi \lr{ 2 a – 1 } } \\
&=
– e^{2 i \pi a },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:sincSquared:220}
\begin{aligned}
J
&= \inv{ 1 – e^{2 i \pi a } } \int_{-\infty, \infty} \frac{ u^{2 a – 1}}{u^2 + 1 } du \\
&= \inv{ 1 – e^{2 i \pi a } } \oint \frac{ z^{2 a – 1}}{z^2 + 1 } dz,
\end{aligned}
\end{equation}
that is, assuming the integral on the upper half plane semicircle is zero.

On that semicircle, with \( z = R e^{i\theta} \),
\begin{equation}\label{eqn:sincSquared:240}
\begin{aligned}
\oint \frac{ z^{2 a – 1}}{z^2 + 1 } dz
&= \int_0^\pi \frac{ \lr{ R e^{i \theta} }^{2 a – 1} }{ R^2 e^{2 i \theta} + 1 } R e^{i \theta} d\theta \\
&\rightarrow \int_0^\pi R^{2 a – 1 + 1 – 2} e^{i \theta (2 a – 1) -i \theta} d\theta.
\end{aligned}
\end{equation}
Near \( a \approx 0 \) this is \( O(R^{-2}) \), which tends to zero. At the upper tend of the a range, where \( a = 1 – \epsilon \), this is
\begin{equation}\label{eqn:sincSquared:260}
O(R^{2(1 – \epsilon) – 2}) = O(1/R^{2 \epsilon}),
\end{equation}
which also tends to zero. We have only one enclosed pole, at \( z = i \), so
\begin{equation}\label{eqn:sincSquared:280}
\begin{aligned}
I
&= \frac{b^{2(a-1)}}{ 1 – e^{2 i \pi a } } ( 2 \pi i ) \evalbar{ \frac{z^{2 a – 1}}{z + i} }{z = i} \\
&= \frac{\pi b^{2(a-1)}}{ 1 – e^{2 i \pi a } } i^{2 a – 1}.
\end{aligned}
\end{equation}

But
\begin{equation}\label{eqn:sincSquared:300}
\begin{aligned}
i^{2 a – 1}
&= \lr{ e^{i \pi/2} }^{2 a – 1} \\
&= e^{i \pi a} e^{-i \pi/2} \\
&= -i e^{i \pi a},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:sincSquared:320}
\begin{aligned}
I
&= \frac{-i \pi b^{2(a-1)} e^{i \pi a}}{ 1 – e^{2 i \pi a } } \\
&= \frac{-i \pi b^{2(a-1)} }{ e^{-i \pi a} – e^{i \pi a } },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:sincSquared:660}
\boxed{
I = \frac{\pi b^{2(a-1)} }{ 2 \sin\lr{ \pi a } }.
}
\end{equation}

A log integral over the positive x-axis.

Problem 31(i) is
\begin{equation}\label{eqn:sincSquared:340}
I = \int_0^\infty \frac{\ln x}{x^2 + b^2} dx.
\end{equation}
Let’s see what happens to this integral over the \( x < 0 \) range.
\begin{equation}\label{eqn:sincSquared:360}
\begin{aligned}
\int_{-\infty}^0 \frac{\ln x}{x^2 + b^2} dx
&=
-\int_{\infty}^0 \frac{\ln(- u)}{u^2 + b^2} d u \\
&=
I + \int_0^\infty \frac{\ln e^{i \pi}}{u^2 + b^2} du \\
&=
I + \frac{i \pi}{2} \int_{-\infty}^\infty \inv{u^2 + b^2} du.
\end{aligned}
\end{equation}

This means that we have
\begin{equation}\label{eqn:sincSquared:380}
\int_{-\infty}^\infty \frac{\ln x}{x^2 + b^2} dx = 2 I + \frac{i \pi}{2} \int_{-\infty}^\infty \inv{u^2 + b^2} du,
\end{equation}
or
\begin{equation}\label{eqn:sincSquared:480}
\begin{aligned}
I
&= \inv{2} \int_{-\infty}^\infty \frac{\ln x – i \pi/2}{x^2 + b^2} dx \\
&= \inv{2} \int_{-\infty}^\infty \frac{\ln(-i x)}{x^2 + b^2} dx \\
&= \inv{2} \oint \frac{\ln(-i z)}{z^2 + b^2} dz \\
&= (\pi i) \evalbar{ \frac{\ln(-i z)}{z + i b} }{z = i b},
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:sincSquared:680}
\boxed{
I = \frac{\pi \ln b}{2 b}.
}
\end{equation}

A pie contour.

Skipping ahead to 31(k), we want to find
\begin{equation}\label{eqn:sincSquared:500}
I = \int_0^\infty \frac{dx}{x^3 + a^3}.
\end{equation}
Some googling shows that we can evaluate integrals with \( x^n + b^n \) denominators, using a pie shaped contour, with slice sizes of \( 2 \pi /n \),
as plotted in fig. 1.

fig. 1. Pie shaped contour for cubic integral.

Integrating (backwards) along the \( 2 \pi/3 \) slice line, with \( \alpha = e^{\pi i/3} \), and \( z = \alpha^2 u \), we have
\begin{equation}\label{eqn:sincSquared:520}
\begin{aligned}
\int_0^\infty \frac{ \alpha^2 du }{ \alpha^6 u^3 + a^3 }
&=
\alpha^2 \int_0^\infty \frac{ du }{ u^3 + a^3 } \\
&= \alpha^2 I,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:sincSquared:540}
I = \inv{ 1 – \alpha^2 } \oint \frac{dz}{z^3 + a^3}.
\end{equation}
We can factor the denominator as
\begin{equation}\label{eqn:sincSquared:560}
z^3 + a^3 = \lr{ z – \alpha a }\lr{ z + a }\lr{ z – a/\alpha },
\end{equation}
where only the \( z = \alpha a \) pole is enclosed. The residue calculation is
\begin{equation}\label{eqn:sincSquared:580}
\begin{aligned}
I
&= \frac{ 2 \pi i }{1 – \alpha^2 } \evalbar{ \inv{\lr{z + a}\lr{ z – a/\alpha } } }{z = a\alpha} \\
&= \frac{ 2 \pi i }{a^2} \inv{ \lr{1 – \alpha^2 } \lr{ \alpha + 1 } \lr{ \alpha – 1/\alpha } } \\
\end{aligned}
\end{equation}
We now want to expand the denominator. I had trouble simplifying this by hand, and it took me a few tries to get it right:
\begin{equation}\label{eqn:sincSquared:600}
\begin{aligned}
-\lr{1 – \alpha^2 } \lr{ \alpha + 1 } \lr{ \alpha – 1/\alpha }
&=
-\frac{\lr{\alpha^2 -1}^2 \lr{ \alpha + 1 }}{\alpha} \\
&=
-\lr{ e^{2 \pi i/3} – 1}^2 \lr{ 1 + e^{i\pi/3} } e^{-i\pi/3} \\
&=
-\lr{ e^{4 \pi i/3} + 1 – 2 e^{ 2 \pi i /3} } \lr{ 1 + e^{-i\pi/3} } \\
&=
-\lr{ e^{4 \pi i/3} + 1 – 2 e^{ 2 \pi i /3} + e^{3 \pi i/3} + e^{-i\pi/3} – 2 e^{ \pi i /3} } \\
&=
-\lr{ e^{4 \pi i/3} – 2 e^{ 2 \pi i /3} + e^{-i\pi/3} – 2 e^{ \pi i /3} } \\
&=
-\lr{ \inv{2} \lr{ -1 – \sqrt{3} i } – \lr{-1 + \sqrt{3} i} + \inv{2}\lr{ 1 – \sqrt{3} i } – \lr{ 1 + \sqrt{3} i} } \\
&=
-\lr{ -\inv{2} – 1 – \inv{2} – 1 } \sqrt{3} i \\
&=
3 \sqrt{3} i.
\end{aligned}
\end{equation}
Putting the pieces together
\begin{equation}\label{eqn:sincSquared:620}
\boxed{
I = \frac{2 \pi}{3 \sqrt{3} a^2}.
}
\end{equation}

References

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

A weighted sinc function integral.

January 25, 2025 math and physics play , , , , ,

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Here’s another real integral problem from [1] (31(f)). Find
\begin{equation}\label{eqn:weightedSinc:20}
I = \int_0^\infty \frac{\sin x dx}{x\lr{ x^2 + a^2 }}.
\end{equation}
Both Mathematica and the text state that the answer is
\begin{equation}\label{eqn:weightedSinc:40}
I = \frac{\pi}{2 a^2} \lr{ 1 – e^{-a}}.
\end{equation}

My initial attempt to evaluate this using contour integral techniques gets the wrong answer. I’m going to post both my wrong solution method, and why the initial method was wrong.

Then I’ll follow up with the corrected method. The mistake, and it’s identification, is probably more interesting than the solution.

The contour.

Before delving into the residue calculations, it’s first helpful to note that the integrand is an even function, so we may transform it first into an integral over the entire real axis:
\begin{equation}\label{eqn:weightedSinc:60}
I = \inv{2} \int_{-\infty}^\infty \frac{\sin x dx}{x\lr{ x^2 + a^2 }}.
\end{equation}
The obvious choice for the contour is illustrated in fig. 1, where \( R \rightarrow \infty \), and \( \rho \rightarrow 0 \).

fig. 1. Contour with two semicircles.

We have to figure out if the circular contour integrals are zero, in the limit. Let’s start with the small contour, with \( z = \rho e^{i \theta} \)
\begin{equation}\label{eqn:weightedSinc:80}
\begin{aligned}
I_\rho
&= \inv{2} \oint \frac{\sin z dz}{z \lr{ z^2 + a^2 }} \\
&= \inv{2} \int_\pi^{2 \pi} \frac{\sin\lr{ \rho e^{i\theta} } \rho i e^{i \theta} d\theta}{\rho e^{i\theta} \lr{ \rho^2 e^{2 i \theta} + a^2 }} \\
&\approx \inv{2} \int_\pi^{2 \pi} \frac{\rho e^{i\theta} d\theta}{\lr{ \rho^2 e^{2 i \theta} + a^2 }} \\
&= \frac{\rho}{2} \int_\pi^{2 \pi} \frac{e^{i\theta} d\theta}{\lr{ \rho^2 e^{2 i \theta} + a^2 }} \\
&\rightarrow \frac{\rho}{2 a^2} \int_\pi^{2 \pi} e^{i\theta} d\theta \\
&\rightarrow 0.
\end{aligned}
\end{equation}

The wrong way.

To complete the problem, I did a similar argument showing that the limit along the infinite semicircle was zero, and then proceeded to evaluate the residues. Let’s see what we get if this is done. We are trying to evaluate

\begin{equation}\label{eqn:weightedSinc:100}
I = \inv{2} \oint \frac{\sin z dz}{z \lr{ z + ia } \lr{ z – ia } },
\end{equation}
where we have a “half-enclosed” pole at \( z = 0 \), and fully enclosed pole at \( z = i a \). The residues are

\begin{equation}\label{eqn:weightedSinc:120}
\begin{aligned}
I
&= \inv{2} \lr{ 2 \pi i } \lr{ \evalbar{ \frac{\sin z}{z \lr{ z + ia } } }{z = i a} + \evalbar{ \inv{2} \frac{\sin z}{\lr{ z^2 + a^2 }} }{z = 0} } \\
&= \inv{2} \lr{ 2 \pi i } \frac{e^{i^2 a} – e^{-i^2 a}}{2i \lr{ ia } \lr{ 2 ia } } \\
&= \frac{ \pi }{4 a^2} \lr{e^{a} – e^{-a}}.
\end{aligned}
\end{equation}

This has a similar structure to the actual answer, but is wrong. I couldn’t find any obvious mistake in the residue calculation, so I was scratching my head for a while about what I did wrong. This confusion was only compounded by figuring out a different way to evaluate this, which did yield the right answer.

The right way.

The key to getting the right answer is to notice that
\begin{equation}\label{eqn:weightedSinc:65}
\inv{2} \int_{-\infty}^\infty \frac{\cos x dx}{x\lr{ x^2 + a^2 }} = 0,
\end{equation}
so we can, instead compute
\begin{equation}\label{eqn:weightedSinc:70}
I = \inv{2} \textrm{Im} \int_{-\infty}^\infty \frac{e^{ix} dx}{x\lr{ x^2 + a^2 }}.
\end{equation}
It’s easy to repeat the argument that after the \( x \rightarrow z \) substitution, this is also zero on the small semicircle, in the \( \rho \rightarrow 0 \) limit. On the large semicircle, the integrand is now in the perfect form to apply Jordan’s lemma. Recall that lemma was:

Lemma 1.1: Jordan’s Lemma.

Given \( R(z) \rightarrow 0 \), then for \( \alpha > 0 \) on a upper half-plane semicircular arc,
\begin{equation*}
\lim_{\Abs{z} \rightarrow \infty} \oint R(z) e^{i \alpha z} dz = 0.
\end{equation*}

We have \( \alpha = 1 \), which is greater than zero, and the rest of the integrand is \( O(R^{-3}) \), so it’s zero. Now we can just compute the residues

\begin{equation}\label{eqn:weightedSinc:140}
\begin{aligned}
\inv{2} \oint \frac{e^{iz} dz}{z\lr{ z^2 + a^2 }}
&=
\lr{ \pi i } \lr{ \evalbar{ \frac{e^{iz}}{z \lr{ z + i a }} }{z = i a} + \inv{2} \evalbar{ \frac{e^{iz}}{z^2 + a^2 }}{z = 0} } \\
&=
\lr{ \pi i } \lr{ \frac{e^{-a}}{ia \lr{ 2 i a }} + \inv{2} \frac{1}{a^2}} \\
&=
\frac{\pi i}{2 a^2} \lr{ 1 – e^{-a} }.
\end{aligned}
\end{equation}
Taking the imaginary part of this integral, we have the solution.

What went wrong?

I don’t think there is anything wrong with the residue calculation for “the wrong way”, but it was not correct to argue that the integral along the infinite semicircular arc was zero in that case.

I’d made that argument in the following fashion, looking for the limit of
\begin{equation}\label{eqn:weightedSinc:160}
\begin{aligned}
\int \frac{\sin z}{z \lr{ z^2 + a^2 } } dz
=
\int \frac{e^{iz}}{2 i z \lr{ z^2 + a^2 } } dz –
\int \frac{e^{-iz}}{2 i z \lr{ z^2 + a^2 } } dz
\end{aligned}
\end{equation}
We can apply Jordan’s lemma to the first integral, but not the second. To see why specifically, consider an explicit \( z = R e^{i \theta } \) parameterization of that integral
\begin{equation}\label{eqn:weightedSinc:180}
\begin{aligned}
\Abs{\int \frac{e^{-iz}}{2 i z \lr{ z^2 + a^2 } } dz}
&=
\inv{2} \Abs{ \int_0^\pi \frac{e^{-i R e^{i\theta} }}{R e^{i\theta} \lr{ R^2 e^{2 i \theta} + a^2 } } i R e^{i \theta} d\theta } \\
&=
\inv{2} \Abs{ \int_0^\pi \frac{e^{-i R e^{i\theta} }}{\lr{ R^2 e^{2 i \theta} + a^2 } } d\theta } \\
&=
\inv{2} \Abs{ \int_0^\pi \frac{ e^{-i R \cos\theta} e^{R \sin\theta} }{\lr{ R^2 e^{2 i \theta} + a^2 } } d\theta }.
\end{aligned}
\end{equation}
Not only is this not zero, it’s very obviously infinite for \( R \rightarrow \infty \), as the real exponential will dominate (at least for some angles.) In particular, note that along the imaginary axis, say \( z = i u\), we have \( \sin z = i \sinh u \), which blows up as \( u \rightarrow \infty \). That sine is not well behaved, so we have to use the imaginary part trick, to convert it to an exponential that will submit properly to Jordan’s lemma.

The moral of the story is that if we incorrectly identify that a portion of the contour integral is zero, when it isn’t, the rest of the results that follow are garbage.

References

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

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.

Evaluating some real trig integrals with unit circle contours.

January 19, 2025 math and physics play , , , , ,

[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.