absolute value

Correcting the errors: Green’s functions for the 1D Helmholtz and Laplacian operators.

September 28, 2025 math and physics play , , , , , , , , , , , , , , , ,

[Click here for a PDF version of this post, and others in this series]

The following recent posts explored 1D Green’s functions for the Helmholtz and Laplacian operators.  There was a sign error (wrong residue sign for a negatively oriented contour) that I made near the beginning that caused a lot of trouble.  Having found the error, I’ve now reworked all that exploratory content into a more coherent form.  That reworked content can be found in today’s blog post below (or in the PDF above, which includes all of this, plus the 2D and 3D derivations.)

Part 1. Green’s functions for the Helmholtz (wave equation) operator in various dimensions.

 

A trilogy in five+ parts: Confirming an error in the derived 1D Helmholtz Green’s function.

 

A trilogy in six+ parts: 1D Laplacian Green’s function

A trilogy in 7+ parts: A better check of the 1D Helmholtz Green’s function.

 

 

Helmholtz Green’s function in 1D.

Evaluating the integral.

For the one dimensional case, we want to evaluate
\begin{equation}\label{eqn:helmholtzGreensV2:200}
G(r) = -\inv{2 \pi} \int \inv{p^2 – k^2} e^{j p r} dp,
\end{equation}
an integral which is unfortunately non-convergent. Since we are dealing with delta functions, it is not surprising that we have convergence problems. The technique used in the book is to displace the pole slightly by a small imaginary amount, and then take the limit.

That is
\begin{equation}\label{eqn:helmholtzGreensV2:220}
G(r) = \lim_{\epsilon \rightarrow 0} G_\epsilon(r),
\end{equation}
where
\begin{equation}\label{eqn:helmholtzGreensV2:240}
\begin{aligned}
G_\epsilon(r)
&= -\inv{2 \pi} \int_{-\infty}^\infty \inv{p^2 – \lr{k + j \epsilon}^2} e^{j p r} dp \\
&= -\inv{2 \pi} \int_{-\infty}^\infty \frac{e^{j p r}}{\lr{ p – k -j \epsilon}\lr{p + k + j \epsilon}} dp.
\end{aligned}
\end{equation}
Our contours, for \( \epsilon > 0 \), are illustrated in fig 1.

fig 1. Contours for 3D Green’s function evaluation

For \( r > 0 \) we can use an upper half plane infinite semicircular contour integral, with \( R \rightarrow \infty \).

The residue calculation for this contour gives
\begin{equation}\label{eqn:helmholtzGreensV2:260}
\begin{aligned}
G_\epsilon(r)
&= -\frac{(+2 \pi j)}{2 \pi} \evalbar{\frac{e^{j p r}}{p + k + j \epsilon} }{p = k + j \epsilon} \\
&= -j \frac{e^{j \lr{k + j \epsilon} r}}{2\lr{k + j \epsilon}} \\
&= -j \frac{e^{j k r} e^{-\epsilon r}}{2\lr{k + j \epsilon}} \\
&\rightarrow -\frac{j}{2k} e^{j k r}.
\end{aligned}
\end{equation}

For \( r < 0 \) we use the lower half plane infinite semicircular contour For this contour, we find \begin{equation}\label{eqn:helmholtzGreensV2:280} \begin{aligned} G_\epsilon(r) &= -\frac{2 \pi j}{(-2 \pi)} \evalbar{\frac{e^{j p r}}{p – k – j \epsilon} }{p = -k – j \epsilon} \\ &= -j \frac{e^{-j \lr{k + j \epsilon} r}}{2\lr{k + j \epsilon}} \\ &= -j \frac{e^{-j k r} e^{\epsilon r}}{2\lr{k + j \epsilon}} \\ &\rightarrow -\frac{j}{2k} e^{-j k r}. \end{aligned} \end{equation} Combining both results, our Green’s function, after a positive pole displacement \( \epsilon > 0 \), is
\begin{equation}\label{eqn:helmholtzGreensV2:300}
G(r) = \frac{1}{2 j k} e^{j k \Abs{r}}.
\end{equation}

Similarly, should we pick \( \epsilon < 0 \), the same sort of calculation yields an incoming wave solution \begin{equation}\label{eqn:helmholtzGreensV2:2240} G(r) = -\frac{1}{2 j k} e^{-j k \Abs{r}}. \end{equation} Allowing for either, we have Green’s functions for both the incoming and outgoing wave cases \begin{equation}\label{eqn:helmholtzGreensV2:2260} \boxed{ G_{\pm}(x – x’) = \pm \frac{1}{2 j k} e^{ \pm j k \Abs{x – x’}}. } \end{equation} With two Green’s functions, we can also make a linear combination. Specifically \begin{equation}\label{eqn:helmholtzGreensV2:2280} \begin{aligned} G(r) &= \inv{2}\lr{ G_{+}(r) + G_{-}(r) } \\ &= \inv{4 j k}\lr{ e^{ j k \Abs{r}} – e^{ – j k \Abs{r}} } \\ &= \inv{2 k} \sin\lr{ k \Abs{r} } \end{aligned} \end{equation} This real valued Green’s function is plotted in fig. 2.

fig. 2. Green’s function for 1D Helmholtz operator.

The convolution is now fully specified, providing a specific solution to the non-homogeneous equation \begin{equation}\label{eqn:helmholtzGreensV2:400} U(x) = \inv{2k} \int_{-\infty}^\infty \sin( k\Abs{r} ) V(x + r) dr. \end{equation}

The general solution may also include any solutions to the homogeneous Helmholtz equation \begin{equation}\label{eqn:helmholtzGreensV2:2300} U(x) = A e^{j k x} + B e^{-j k x} + \inv{2k} \int_{-\infty}^\infty \sin( k\Abs{r} ) V(x + r) dr. \end{equation}

A strictly causal solution.

We can split the convolution kernel a “causal” part, where only the spatially-“past” values of \( V \) contribute, and an “acausal” part \begin{equation}\label{eqn:helmholtzGreensV2:2320} U(x) = \inv{2k} \int_{-\infty}^0 \sin( k\Abs{r} ) V(x + r) dr + \inv{2k} \int_{0}^\infty \sin( k\Abs{r} ) V(x + r) dr. \end{equation} In a sense, we are averaging causal and acausal portions of the convolution. Suppose that we form a convolution with a built in cut-off, so that values of \( V(x’), x’ > x \) do not contribute to \( U(x) \). That is
\begin{equation}\label{eqn:helmholtzGreensV2:440}
f(x) = \int_{-\infty}^x \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’.
\end{equation}
Here the one-half factor has been dropped, since we are no longer performing a QFT like average of causal and acausal terms.

Intuition suggests this should be a solution to the Helmholtz equation, but let’s test that guess. We start with the identity
\begin{equation}\label{eqn:helmholtzGreensV2:460}
\frac{d}{dx} \int_a^x g(x, x’) dx’
=
\evalbar{g(x, x’) }{x’ = x} + \int_a^x \frac{\partial g(x,x’)}{dx} dx’.
\end{equation}
Taking the first derivative of \( f(x) \), we find
\begin{equation}\label{eqn:helmholtzGreensV2:480}
\begin{aligned}
\frac{df}{dx}
&= \evalbar{ \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) }{x’ = x} + k \int_{-\infty}^x \frac{\cos\lr{k \lr{x – x’}}}{k} V(x’) dx’ \\
&= k \int_{-\infty}^x \frac{\cos\lr{k \lr{x – x’}}}{k} V(x’) dx’,
\end{aligned}
\end{equation}
where we have, somewhat lazily, treated the infinite limit as a constant. Effectively, this requires that the forcing function \( V(x) \) is zero at \( -\infty \). Taking the second derivative, we have
\begin{equation}\label{eqn:helmholtzGreensV2:500}
\begin{aligned}
\frac{d^2f}{dx^2}
&=
\evalbar{ k \frac{\cos\lr{k \lr{x – x’}}}{k} V(x’) }{x’ = x} – k^2 \int_{-\infty}^x \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’ \\
&= V(x) – k^2 f(x),
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreensV2:520}
\frac{d^2}{dx^2} f(x) + k^2 f(x) = V(x).
\end{equation}
This verifies that \ref{eqn:helmholtzGreensV2:440} is also a specific solution to the wave equation, as expected and desired.

It appears that the general solution is likely of the following form
\begin{equation}\label{eqn:helmholtzGreensV2:380b}
U(x) =
A’ \cos\lr{ k x } +
B’ \sin\lr{ k x }
+\alpha \int_{-\infty}^x \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’
+(1-\alpha)\int_x^\infty \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’,
\end{equation}
with \( \alpha \in [0,1] \).

It’s pretty cool that we can completely solve the 1D forced wave equation, for any forcing function, from first principles. Yes, I took liberties that would make a mathematician cringe, but we are telling a story, and leaving the footnotes to somebody else.

Verification of the 1D Helmholtz Green’s function.

Let’s show that the outgoing Green’s function has the desired delta function semantics. That is
\begin{equation}\label{eqn:helmholtzGreensV2:2060}
\lr{ \spacegrad^2 + k^2 } G(x, x’) = \delta(x – x’),
\end{equation}
where
\begin{equation}\label{eqn:helmholtzGreensV2:2080}
G(x, x’) = \frac{e^{j k \Abs{x – x’}}}{2 j k}.
\end{equation}
Making a \( r = x – x’ \) change of variables gives
\begin{equation}\label{eqn:helmholtzGreensV2:2120}
\spacegrad^2 e^{j k \Abs{x – x’}} = \frac{d^2}{dr^2} e^{j k \Abs{r}}
\end{equation}

The function \( \Abs{r} \) formally has no derivative at the origin, but we may use the physics trick, rewriting the absolute in terms of the Heaviside theta function
\begin{equation}\label{eqn:helmholtzGreensV2:2340}
\Abs{r} = r \Theta(r) – r \Theta(-r).
\end{equation}
We then use the delta function identification for the derivative
\begin{equation}\label{eqn:helmholtzGreensV2:2360}
\Theta'(r) = \delta(r).
\end{equation}
In particular
\begin{equation}\label{eqn:helmholtzGreensV2:2380}
\begin{aligned}
\frac{d}{dr} \Abs{r}
&= \Theta(r) – \Theta(-r) + r \delta(r) – (-1)\delta(-r) \\
&= \Theta(r) – \Theta(-r) + 2 r \delta(r),
\end{aligned}
\end{equation}
using the symmetric property of the delta function \( \delta(-r) = \delta(r) \). The delta function contribution to this derivative is actually zero, as seen when we operate with \( r \delta(r) \) against a test function
\begin{equation}\label{eqn:helmholtzGreensV2:2400}
\begin{aligned}
\int r \delta(r) f(r) dr
&=
\evalbar{r f(r)}{r = 0} \\
&= 0.
\end{aligned}
\end{equation}
We’ve now found that
\begin{equation}\label{eqn:helmholtzGreensV2:2420}
\frac{d}{dr} \Abs{r} = \Theta(r) – \Theta(-r) = \mathrm{sgn}(r),
\end{equation}
the sign function. The derivative of the sign function is
\begin{equation}\label{eqn:helmholtzGreensV2:2440}
\begin{aligned}
\mathrm{sgn}'(r)
&= \lr{ \Theta(r) – \Theta(-r) }’ \\
&= \delta(r) -(-1)\delta(-r) \\
&= 2 \delta(r).
\end{aligned}
\end{equation}

The derivatives are
\begin{equation}\label{eqn:helmholtzGreensV2:2140}
\begin{aligned}
\frac{d}{dr} e^{j k \Abs{r} }
&=
j k e^{j k \Abs{r} } \frac{d\Abs{r}}{dr} \\
&=
j k e^{j k \Abs{r} } \mathrm{sgn}(r).
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreensV2:2160}
\frac{d^2}{dr^2} e{j k \Abs{r}}
=
\lr{ j k \mathrm{sgn}(r) }^2 e^{j k \Abs{r} } + 2 j k e^{j k \Abs{r} } \delta(r).
\end{equation}

We can identify \( e^{j k \Abs{r} } \delta(r) = \delta(r) \), just as we identified \( r \delta(r) = 0 \), by application to a test function. That is
\begin{equation}\label{eqn:helmholtzGreensV2:2180}
\begin{aligned}
\int e^{j k \Abs{r} } \delta(r) f(r) dr
&=
\evalbar{e^{j k \Abs{r} } f(r)}{r = 0} \\
&=
f(0) \\
&=
\int \delta(r) f(r) dr.
\end{aligned}
\end{equation}
With that identification
\begin{equation}\label{eqn:helmholtzGreensV2:2200}
\spacegrad^2 e^{j k \Abs{r} } = -k^2 e^{j k \Abs{r} } + 2 j k \delta(r),
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreensV2:2220}
\boxed{
\lr{ \spacegrad^2 + k^2 } \frac{e^{j k \Abs{x – x’} }}{2 j k} = \delta(x – x’).
}
\end{equation}

Verifying the Green’s function with convolution.

Avoiding the physics tricks, we may use a limiting argument to validate our Green’s function.

We first want to show that at points \( x’ \ne x \) the Helmholtz operator applied to the Green’s function is zero:
\begin{equation}\label{eqn:helmholtzGreensV2:1220}
\lr{ \spacegrad^2 + k^2} G(x,x’) = 0,
\end{equation}
Since we are avoiding the origin
\begin{equation}\label{eqn:helmholtzGreensV2:1240}
\lr{ k^2 + \frac{d^2}{dx^2} } e^{j k \Abs{x – x’}}.
\end{equation}
We expect that this will be zero. Making a change of variables \( r = x’ – x \), we want to evaluate
\begin{equation}\label{eqn:helmholtzGreensV2:1260}
\lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}},
\end{equation}
assuming that we are omitting a neighbourhood around \( r = 0 \) where the absolute value causes trouble. For \( r > 0 \)
\begin{equation}\label{eqn:helmholtzGreensV2:1280}
\begin{aligned}
\frac{d}{dr} e^{j k \Abs{r}}
&= \frac{d}{dr} e^{j k r} \\
&= j k e^{j k r},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreensV2:1300}
\begin{aligned}
\frac{d^2}{dr^2} e^{j k \Abs{r}}
&= \frac{d}{dr} j k e^{j k r} \\
&= (j k)^2 e^{j k r}.
\end{aligned}
\end{equation}
Similarly, for \( r < 0 \), we have
\begin{equation}\label{eqn:helmholtzGreensV2:1320}
\frac{d^2}{dr^2} e^{j k \Abs{r}} = (-jk)^2 e^{j k \Abs{r}}.
\end{equation}
In both cases, provided we are in a neighbourhood that omits \( r \ne 0 \), we have
\begin{equation}\label{eqn:helmholtzGreensV2:1340}
\lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}} = 0,
\end{equation}
as desired.

The takeaway is that we have
\begin{equation}\label{eqn:helmholtzGreensV2:1360}
\begin{aligned}
\lr{ \spacegrad^2 + k^2 }\int_{-\infty}^\infty G(x, x’) V(x’) dx’
&=
\int_{\Abs{x’ – x} \le \epsilon} V(x’) \lr{ \frac{d^2}{d{x’}^2} + k^2 } G(x, x’) dx’ \\
&=
\int_{-\epsilon}^\epsilon V(x + r) \lr{ \frac{d^2}{dr^2} + k^2 } G(r) dr \\
\end{aligned}
\end{equation}
for some arbitrarily small value of \( \epsilon \). Observe that after bringing the operator into the integral, we also made a change of variables, first to \( x’ \) for the Laplacian, and then to \( r = x’ – x \).

We’d like the 1D equivalent of Green’s theorem to reduce this, so let’s work that out first.
\begin{equation}\label{eqn:helmholtzGreensV2:1380}
\begin{aligned}
\int dx\, v \frac{d^2 u}{dx^2} – \int dx\, u \frac{d^2 v}{dx^2}
&=
\int dx\,
\lr{
\frac{d}{dx} \lr{
v \frac{du}{dx}
}
– \frac{dv}{dx} \frac{du}{dx}
}
–
\int dx\,
\lr{
\frac{d}{dx} \lr{
u \frac{dv}{dx}
}
–
\frac{du}{dx} \frac{dv}{dx}
}
\\
&=
\int dx\,
\frac{d}{dx}
\lr{
v \frac{du}{dx}
}
–
\int dx\,
\frac{d}{dx} \lr{
u \frac{dv}{dx}
}
\\
&=
v \frac{du}{dx}
–
u \frac{dv}{dx},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreensV2:1400}
\boxed{
\int_a^b dx\, v \frac{d^2 u}{dx^2}
=
\int_a^b dx\, u \frac{d^2 v}{dx^2}
+
\evalrange{v \frac{du}{dx}}{a}{b}
–
\evalrange{u \frac{dv}{dx}}{a}{b}.
}
\end{equation}

This gives us
\begin{equation}\label{eqn:helmholtzGreensV2:1420}
\begin{aligned}
\lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’
&=
\int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr \\
&\quad +
\evalrange{ V(x + r) \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
–
\evalrange{ e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon}.
\end{aligned}
\end{equation}
If we can assume that \( V \) and it’s first and second derivatives are all continuous over this small interval, then the first integral is approximately
\begin{equation}\label{eqn:helmholtzGreensV2:1440}
\begin{aligned}
\int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr
&\sim
\lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) }
\int_{-\epsilon}^\epsilon e^{j k \Abs{r}} dr \\
&=
\frac{2 j}{k} \lr{ 1 – e^{ j k \epsilon} }
\lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) } \\
&\rightarrow 0.
\end{aligned}
\end{equation}
Similarly, with \( dV/dr \) continuity condition, that last term is also zero. We are left, for \( \epsilon \) sufficiently small, we are left with
\begin{equation}\label{eqn:helmholtzGreensV2:1460}
\lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’
=
V(x)
\evalrange{ \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}.
\end{equation}
Because we are evaluating this derivative only at points \( r = \pm \epsilon \ne 0 \), that derivative is
\begin{equation}\label{eqn:helmholtzGreensV2:2460}
\frac{d}{dr} e^{j k \Abs{r}}
=
j k e^{j k \Abs{r}} \mathrm{sgn}(r),
\end{equation}
leaving us with
\begin{equation}\label{eqn:helmholtzGreensV2:2480}
\begin{aligned}
\evalrange{ \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
&=
j k e^{j k \Abs{\epsilon}} \mathrm{sgn}(\epsilon) – j k e^{j k \Abs{-\epsilon}} \mathrm{sgn}(-\epsilon) \\
&=
j k e^{j k \Abs{\epsilon}} \lr{ 1 – (-1) } \\
2 j k e^{j k \Abs{\epsilon}}.
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreensV2:2500}
\lr{ \spacegrad^2 + k^2 }\int_{-\infty}^\infty e^{j k \Abs{x – x’} } V(x’) dx’
=
2 j k e^{j k \Abs{\epsilon}} V(x).
\end{equation}
Taking limits and dividing through by \( 2 j k \) proves the result.

1D Laplacian Green’s function.

Having blundered our way to what appears to be the correct Green’s function for the 1D Helmholtz operator, let’s further validate that by deriving the Green’s function for the 1D Laplacian. We should also be able to verify that it has the correct delta function semantics.

Expansion in series and taking the limit.

Expanding the Helmholtz Green’s function in series around \( k \Abs{r} \) we have
\begin{equation}\label{eqn:helmholtzGreensV2:1780}
\begin{aligned}
G(r)
&= -\frac{j}{2k} \lr{ 1 + j k \Abs{r} + O((k \Abs{r})^2) } \\
&= -\frac{j}{2} \lr{ \inv{k} + j \Abs{r} + \inv{k} O((k \Abs{r})^2) } \\
\end{aligned}
\end{equation}
This means that to first order in \( k \), we have
\begin{equation}\label{eqn:helmholtzGreensV2:1800}
G(r) + \frac{j}{2k} = \frac{\Abs{r}}{2}.
\end{equation}
As before, we are free to add constant terms to the Green’s function for the Laplacian, and we conclude that the 1D Green’s function for the Laplacian is
\begin{equation}\label{eqn:helmholtzGreensV2:1820}
\boxed{
G(r) = \frac{\Abs{r}}{2}.
}
\end{equation}

Alternatively, we may use the real sine form of the Green’s function, which has a nice expansion around \( k = 0 \), and arrive at the same result.

Observe that
\begin{equation}\label{eqn:helmholtzGreensV2:2520}
\frac{d^2}{dr^2} \frac{\Abs{r}}{2} =
\frac{d}{dr} \frac{\mathrm{sgn}(r)}{2} =
\delta(r),
\end{equation}
which verifies that this is a valid Green’s function for the 1D Laplacian.

Verifying the Green’s function with convolution.

We can also operate on the convolution with the Laplacian, to verify correctness. We are interested in evaluating
\begin{equation}\label{eqn:helmholtzGreensV2:1840}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’ = \int V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr.
\end{equation}
If all goes well, this should evaluate to \( V(x) \), indicating that \( \spacegrad^2 \Abs{x – x’}/2 = \delta(x – x’) \). As a first step, we expect \( \spacegrad^2 G = 0 \), for \( x \ne x’ \). Consider first \( r > 0 \), where
\begin{equation}\label{eqn:helmholtzGreensV2:1860}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} r
= 1,
\end{equation}
and for \( r < 0 \) where
\begin{equation}\label{eqn:helmholtzGreensV2:1880}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} (-r)
= -1.
\end{equation}
This means that, away from the origin \( d\Abs{r}/dr = \mathrm{sgn}(r) \), and \( d^2 \Abs{r}/dr^2 = 0\). We can conclude that, for some non-zero positive epsilon that we will eventually let approach zero, we have
\begin{equation}\label{eqn:helmholtzGreensV2:1900}
\begin{aligned}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’
&= \int_{-\epsilon}^\epsilon V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr \\
&= \int_{-\epsilon}^\epsilon \lr{
\frac{d}{dr} \lr{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }
– \frac{dV(x + r)}{dr} \frac{d}{dr} \frac{\Abs{r}}{2}
}
dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
– \int_{-\epsilon}^\epsilon
\lr{
\frac{d}{dr} \lr{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }
–
\frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2}
} dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
-\evalrange{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
+
\int_{-\epsilon}^\epsilon \frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2} dr \\
&=
\inv{2} \lr{ V(x + \epsilon) + V(x – \epsilon) } \\
&-\quad \frac{\epsilon}{2}\lr{
\frac{dV(x + \epsilon)}{dr}
–
\frac{dV(x – \epsilon)}{dr}
} \\
&+
\frac{\epsilon^2}{2} \lr{
\frac{d^2V(x + \epsilon)}{dr^2}
+
\frac{d^2V(x + \epsilon)}{dr^2}
}.
\end{aligned}
\end{equation}
In the limit we have
\begin{equation}\label{eqn:helmholtzGreensV2:1920}
\boxed{
\spacegrad^2 \int G(x, x’) V(x’) dx’ = \inv{2} \lr{ V(x^+) + V(x^-) }.
}
\end{equation}

If the test (or driving) function is continuous at \( x’ = x \), then this is exactly the delta-function semantics that we expect of a Green’s function. It’s interesting that this check provides us with precise semantics for the Green’s function for discontinuous functions too.

A trilogy in six+ parts: 1D Laplacian Green’s function

September 25, 2025 math and physics play , , , , , , , , ,

[Click here for a PDF version of this post, and the others in this series.]

Having blundered our way to what appears to be the correct Green’s function for the 1D Helmholtz operator, let’s further validate that by deriving the Green’s function for the 1D Laplacian. We should also be able to verify that it has the correct delta function semantics.

The 1D Laplacian Green’s function.

Expanding the Helmholtz Green’s function in series around \( k \Abs{r} \) we have
\begin{equation}\label{eqn:helmholtzGreens:1780}
\begin{aligned}
G(r)
&= -\frac{j}{2k} \lr{ 1 + j k \Abs{r} + O((k \Abs{r})^2) } \\
&= -\frac{j}{2} \lr{ \inv{k} + j \Abs{r} + \inv{k} O((k \Abs{r})^2) } \\
\end{aligned}
\end{equation}
This means that to first order in \( k \), we have
\begin{equation}\label{eqn:helmholtzGreens:1800}
G(r) + \frac{j}{2k} = \frac{\Abs{r}}{2}.
\end{equation}
As before, we are free to add constant terms to the Green’s function for the Laplacian, and we conclude that the 1D Green’s function for the Laplacian is
\begin{equation}\label{eqn:helmholtzGreens:1820}
\boxed{
G(r) = \frac{\Abs{r}}{2}.
}
\end{equation}

Observing the delta-function semantics of our Laplacian Green’s function through convolution.

We can now attempt to validate that this has the desired delta function semantics, operating on the convolution with the Laplacian. We are interested in evaluating
\begin{equation}\label{eqn:helmholtzGreens:1840}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’ = \int V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr.
\end{equation}
If all goes well, this should evaluate to \( V(x) \), indicating that \( \spacegrad^2 \Abs{x – x’}/2 = \delta(x – x’) \). As a first step, we expect \( \spacegrad^2 G = 0 \), for \( x \ne x’ \). Consider first \( r > 0 \), where
\begin{equation}\label{eqn:helmholtzGreens:1860}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} r
= 1,
\end{equation}
and for \( r < 0 \) where
\begin{equation}\label{eqn:helmholtzGreens:1880}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} (-r)
= -1.
\end{equation}
This means that, away from the origin \( d\Abs{r}/dr = \mathrm{sgn}(r) \), and \( d^2 \Abs{r}/dr^2 = 0\). We can conclude that, for some non-zero positive epsilon that we will eventually let approach zero, we have
\begin{equation}\label{eqn:helmholtzGreens:1900}
\begin{aligned}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’
&= \int_{-\epsilon}^\epsilon V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr \\
&= \int_{-\epsilon}^\epsilon \lr{
\frac{d}{dr} \lr{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }
– \frac{dV(x + r)}{dr} \frac{d}{dr} \frac{\Abs{r}}{2}
}
dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
– \int_{-\epsilon}^\epsilon
\lr{
\frac{d}{dr} \lr{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }
–
\frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2}
} dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
-\evalrange{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
+
\int_{-\epsilon}^\epsilon \frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2} dr \\
&=
\inv{2} \lr{ V(x + \epsilon) + V(x – \epsilon) } \\
&-\quad \frac{\epsilon}{2}\lr{
\frac{dV(x + \epsilon)}{dr}
–
\frac{dV(x – \epsilon)}{dr}
} \\
&+
\frac{\epsilon^2}{2} \lr{
\frac{d^2V(x + \epsilon)}{dr^2}
+
\frac{d^2V(x + \epsilon)}{dr^2}
}.
\end{aligned}
\end{equation}
In the limit we have
\begin{equation}\label{eqn:helmholtzGreens:1920}
\boxed{
\spacegrad^2 \int G(x, x’) V(x’) dx’ = \inv{2} \lr{ V(x^+) + V(x^-) }.
}
\end{equation}

If the test (or driving) function is continuous at \( x’ = x \), then this is exactly the delta-function semantics that we expect of a Green’s function. It’s interesting that this check provides us with precise semantics for the Green’s function for discontinuous functions too.

Extracting the delta-function semantics of the Laplacian Green’s function directly.

There’s a more direct, but less satisfying way to do this same computation. We can compute \( d^2 G(r)/dr^2 \). We need the trick
\begin{equation}\label{eqn:helmholtzGreens:1940}
\Abs{r} = r \Theta(r) – r \Theta(-r),
\end{equation}
and the identification \(\Theta'(r) = \delta(r) \). We find
\begin{equation}\label{eqn:helmholtzGreens:1960}
\begin{aligned}
\Abs{r}’
&= \Theta(r) – \Theta(-r) + r \delta(r) – r (-1) \delta(-r) \\
&= \Theta(r) – \Theta(-r) + 2 r \delta(r).
\end{aligned}
\end{equation}
To give \( r \delta(r) \) meaning, we can apply it to a test function
\begin{equation}\label{eqn:helmholtzGreens:1980}
\int r \delta(r) f(r) dr = \evalbar{ r f(r) }{r = 0} = 0,
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreens:2000}
\Abs{r}’ = \Theta(r) – \Theta(-r).
\end{equation}
Now we can take second derivatives
\begin{equation}\label{eqn:helmholtzGreens:2020}
\Abs{r}” = \delta(r) + \delta(-r) = 2 \delta(r).
\end{equation}
This means that
\begin{equation}\label{eqn:helmholtzGreens:2040}
\boxed{
\frac{d^2}{dr^2} \frac{\Abs{r}}{2} = \delta(r).
}
\end{equation}

A trilogy in five+ parts: Confirming an error in the derived 1D Helmholtz Green’s function.

September 24, 2025 math and physics play , , , , , , , ,

[Click here for a PDF version of this, and previous, posts in this series].

The discontinuity in the derived 1D Helmholtz Green’s function is somewhat surprising. Let’s try to verify that this works or find what does. The first thing to check is that
\begin{equation}\label{eqn:helmholtzGreens:1220}
\lr{ \spacegrad^2 + k^2} G(x,x’) = 0,
\end{equation}
at locations where \( x \ne x’ \). Since we are avoiding the origin (where the annoying sign function kicks in), means that we want to evaluate:
\begin{equation}\label{eqn:helmholtzGreens:1240}
\lr{ k^2 + \frac{d^2}{dx^2} } e^{j k \Abs{x – x’}},
\end{equation}
and expect that this will be zero. Let’s make a change of variables \( r = x’ – x \), and evaluate
\begin{equation}\label{eqn:helmholtzGreens:1260}
\lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}},
\end{equation}
assuming that we are omitting a neighbourhood around \( r = 0 \) where the absolute value causes trouble. For \( r > 0 \)
\begin{equation}\label{eqn:helmholtzGreens:1280}
\begin{aligned}
\frac{d}{dr} e^{j k \Abs{r}}
&= \frac{d}{dr} e^{j k r} \\
&= j k e^{j k r},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreens:1300}
\begin{aligned}
\frac{d^2}{dr^2} e^{j k \Abs{r}}
&= \frac{d}{dr} j k e^{j k r} \\
&= (j k)^2 e^{j k r}.
\end{aligned}
\end{equation}
Similarly, for \( r < 0 \), we have \begin{equation}\label{eqn:helmholtzGreens:1320} \frac{d^2}{dr^2} e^{j k \Abs{r}} = (-jk)^2 e^{j k \Abs{r}}. \end{equation} In both cases, provided we are in a neighbourhood that omits \( r \ne 0 \), we have \begin{equation}\label{eqn:helmholtzGreens:1340} \lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}} = 0, \end{equation} as desired. The takeaway is that we have \begin{equation}\label{eqn:helmholtzGreens:1360} \begin{aligned} \lr{ \spacegrad^2 + k^2 }\int_{-\infty}^\infty G(x, x’) V(x’) dx’ &= \int_{\Abs{x’ – x} \le \epsilon} V(x’) \lr{ \frac{d^2}{d{x’}^2} + k^2 } G(x, x’) dx’ \\ &= \int_{-\epsilon}^\epsilon V(x + r) \lr{ \frac{d^2}{dr^2} + k^2 } G(r = x’ – x) dr \\ \end{aligned} \end{equation} for some arbitrarily small value of \( \epsilon \). Observe that after bringing the operator into the integral, we also made a change of variables, first to \( x’ \) for the Laplacian, and then to \( r = x’ – x \). We’d like the 1D equivalent of Green’s theorem to reduce this, so let’s work that out first. \begin{equation}\label{eqn:helmholtzGreens:1380} \begin{aligned} \int dx\, v \frac{d^2 u}{dx^2} – \int dx\, u \frac{d^2 v}{dx^2} &= \int dx\, \lr{ \frac{d}{dx} \lr{ v \frac{du}{dx} } – \frac{dv}{dx} \frac{du}{dx} } – \int dx\, \lr{ \frac{d}{dx} \lr{ u \frac{dv}{dx} } – \frac{du}{dx} \frac{dv}{dx} } \\ &= \int dx\, \frac{d}{dx} \lr{ v \frac{du}{dx} } – \int dx\, \frac{d}{dx} \lr{ u \frac{dv}{dx} } \\ &= v \frac{du}{dx} – u \frac{dv}{dx}, \end{aligned} \end{equation} so \begin{equation}\label{eqn:helmholtzGreens:1400} \boxed{ \int_a^b dx\, v \frac{d^2 u}{dx^2} = \int_a^b dx\, u \frac{d^2 v}{dx^2} + \evalrange{v \frac{du}{dx}}{a}{b} – \evalrange{u \frac{dv}{dx}}{a}{b}. } \end{equation} Let’s try applying that to the function \( G(r) = e^{j k \Abs{r} } \), and see what happens. That is \begin{equation}\label{eqn:helmholtzGreens:1420} \begin{aligned} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’ &= \int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr \\ &\quad + \evalrange{ V(x + r) \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon} – \evalrange{ e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon}. \end{aligned} \end{equation} If we can assume that \( V \) and it’s first and second derivatives are all continuous over this small interval, then the first integral is approximately \begin{equation}\label{eqn:helmholtzGreens:1440} \begin{aligned} \int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr &\sim \lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) } \int_{-\epsilon}^\epsilon e^{j k \Abs{r}} dr \\ &= \frac{2 j}{k} \lr{ 1 – e^{ j k \epsilon} } \lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) } \\ &\rightarrow 0. \end{aligned} \end{equation} Similarly, with \( dV/dr \) continuity condition, that last term is also zero. We are left, for \( \epsilon \) sufficiently small, we are left with \begin{equation}\label{eqn:helmholtzGreens:1460} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’ = V(x) \evalrange{ \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}. \end{equation} but this is an extremely problematic derivative around the origin. The core problem is evaluating \begin{equation}\label{eqn:helmholtzGreens:1480} \frac{d}{dr} e^{j k \Abs{r}} = j k e^{j k \Abs{r} } \frac{d\Abs{r}}{dr}. \end{equation} In conventional mathematics, we’d have to say that this is undefined at the origin. In physics, on the other hand, where we play fast and loose with the mathematics, we express the absolute value in terms of Heavyside theta functions \begin{equation}\label{eqn:helmholtzGreens:1500} \Abs{r} = r \Theta(r) – r \Theta(-r). \end{equation} We may now take derivatives \begin{equation}\label{eqn:helmholtzGreens:1520} \begin{aligned} \Abs{r}’ &= \Theta(r) – \Theta(-r) + r \delta(r) + r \delta(-r) \\ &= \mathrm{sgn}(r) + 2 r \delta(r). \end{aligned} \end{equation} Evaluating \( e^{j k \Abs{r} } \Abs{r}’ \) over the \( [-\epsilon, \epsilon] \) range, we have \begin{equation}\label{eqn:helmholtzGreens:1540} \begin{aligned} \evalrange{ e^{j k \Abs{r} } \Abs{r}’ }{-\epsilon}{\epsilon} &= e^{j k \epsilon} \lr{ \evalrange{ \mathrm{sgn}(r) + 2 r \delta(r) }{-\epsilon}{\epsilon} } \\ &= e^{j k \epsilon} \lr{ 2 + 2 \epsilon \delta(\epsilon) }. \end{aligned} \end{equation} Again, playing fast and loose, we evaluate this range before taking the limit, where \( \delta(\epsilon) = 0 \) for \( \epsilon > 0 \). We are left with
\begin{equation}\label{eqn:helmholtzGreens:1560}
\lim_{\epsilon \rightarrow 0} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’
=
2 j k V(x),
\end{equation}
provided \( V \) and its first and second derivatives are continuous.

Under those constraints, the implication is that one valid Green’s function for the 1D Helmholtz operator is
\begin{equation}\label{eqn:helmholtzGreens:1580}
G(r) = -\frac{j}{2k} e^{j k \Abs{r} }.
\end{equation}
The \( \mathrm{sgn}(r) \) scale factor that was part of the Green’s function that we derived using contour integration does not appear to be required.

What happens if we retain the sign function factor? Doing so, we have
\begin{equation}\label{eqn:helmholtzGreens:1600}
\begin{aligned}
\lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} \mathrm{sgn}(x – x’) e^{j k \Abs{x – x’}} V(x’) dx’
&=
-\int_{-\epsilon}^\epsilon \mathrm{sgn}(r) e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr \\
&\quad
–
\evalrange{ V(x + r) \frac{d}{dr} \mathrm{sgn}(r) e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
+
\evalrange{ \mathrm{sgn}(r) e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon}.
\end{aligned}
\end{equation}
This time, we note that
\begin{equation}\label{eqn:helmholtzGreens:1620}
\int_{-\epsilon}^\epsilon \mathrm{sgn}(r) e^{j k \Abs{r}} dr = 0, \quad \forall \epsilon \ne 0,
\end{equation}
even without evaluating the limit. However, we have problems with the other two terms. The last term doesn’t zero out as desired, instead
\begin{equation}\label{eqn:helmholtzGreens:1640}
\evalrange{ \mathrm{sgn}(r) e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon} \rightarrow 2 V'(x).
\end{equation}
To evaluate the \( V(x) \) factor, we write
\begin{equation}\label{eqn:helmholtzGreens:1660}
\mathrm{sgn}(r) = \Theta(r) – \Theta(-r),
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreens:1680}
\begin{aligned}
\mathrm{sgn}(r)’
&= \delta(r) + \delta(-r) \\
&= 2 \delta(r).
\end{aligned}
\end{equation}
That means that
\begin{equation}\label{eqn:helmholtzGreens:1700}
\begin{aligned}
\frac{d}{dr} \lr{ \mathrm{sgn}(r) e^{j k \Abs{r}} }
&=
2 \delta(r) e^{j k \Abs{r}} + j k \mathrm{sgn}(r) e^{j k \Abs{r}} \lr{ \mathrm{sgn}(r) + 2 r \delta(r) } \\
&=
e^{j k \Abs{r}} \lr{ 2 \delta(r) + j k \lr{ 1 + 2 r \mathrm{sgn}(r) \delta(r)} } \\
&=
j k e^{j k \Abs{r}},
\end{aligned}
\end{equation}
for \( r \ne 0 \), so
\begin{equation}\label{eqn:helmholtzGreens:1760}
–
\evalrange{ V(x + r) \frac{d}{dr} \mathrm{sgn}(r) e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
\rightarrow V(x) j k \lr{ e^{j k \epsilon} – e^{j k \epsilon} } = 0.
\end{equation}

All in, we are left with
\begin{equation}\label{eqn:helmholtzGreens:1720}
\lim_{\epsilon \rightarrow 0} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} \frac{-j}{2k} \mathrm{sgn}(x – x’) e^{j k \Abs{x – x’}} V(x’) dx’
= -\frac{j}{k} V'(x),
\end{equation}
but for a Green’s function, we expected just \( V(x) \).

It seems that the sign factor in the contour integration result is definitively wrong. That result was
\begin{equation}\label{eqn:helmholtzGreens:300b}
G(u) = -\frac{j \mathrm{sgn}(u)}{2k} e^{j k \Abs{u}},
\end{equation}
but what we really want is
\begin{equation}\label{eqn:helmholtzGreens:1740}
\boxed{
G(u) = -\frac{j}{2k} e^{j k \Abs{u}}.
}
\end{equation}

Unfortunately, I don’t see any errors in the original contour integration, so I’m at a loss where things went wrong.

Circular and spherical area, volume, and boundary integrals.

February 4, 2023 math and physics play , , , , , , ,

[Click here for a PDF version of this post]

Motivation

Maverick posed the following question on the bivector discord

“I saw your blog post on curvilinear coordinates in geometric calculus. I saw your derivation of the volume of a sphere using this technique and decided for practice by doing a surface integral to calculate the area of sphere using the quantity $\partial{\theta} \wedge \partial{\phi}~ dA$ is there a way to integrate this without simply taking the magnitude of this quantity and then integrating or are we limited to only integrating quantities that are 1 dimensional like scalars and pseudoscalars”.

My initial response was that, sure, we should be able to compute bivector and trivector valued integrals. However, in retrospect, the reality is a bit more subtle.

We aren’t limited to using the magnitudes of the differential forms, but not all multivector integral are interesting.
In the original blog post, I must have computed the area of the circle using a bivector valued area element, or the volume of a sphere using a trivector valued volume element. However, if I did the volume that way, I probably cheated and computed 8 times the value of the first octant volume (which is positive), vs. the entire integral, which is zero.  [EDIT: the original post, now linked above, has been corrected.]

Let’s compute the circular area and circumference, and the spherical volume and surface area using multivector valued integrals, and see where we end up having to resort to scalar integrals.

Circular example.

The polar parameterization of points in circular region is
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:20}
\Bx = r \Be_1 e^{i\theta},
\end{equation}
where \( i = \Be_1 \Be_2 \).
Our differentials are
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:40}
\begin{aligned}
d\Bx_r &= \Be_1 e^{i\theta} \,dr \\
d\Bx_\theta &= r \Be_2 e^{i\theta} \,d\theta.
\end{aligned}
\end{equation}
Our “volume” element, is a 2D pseudoscalar
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:60}
\begin{aligned}
dA &= d\Bx_r \wedge d\Bx_\theta \\
&= r \gpgradetwo{ \Be_1 e^{i\theta} \Be_2 e^{i\theta} } \, dr d\theta \\
&= r \gpgradetwo{ \Be_1 \Be_2 e^{-i\theta} e^{i\theta} } \, dr d\theta \\
&= r i \, dr d\theta.
\end{aligned}
\end{equation}
This, as I probably pointed out in my previous blog post, can be integrated to find the area of the circle
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:80}
\begin{aligned}
A &= \int_{r = 0}^R \int_{\theta = 0}^{2\pi} r i \, dr d\theta \\
&= \frac{R^2}{2} 2 \pi i \\
&= \pi R^2 i.
\end{aligned}
\end{equation}
However, we got lucky, as the two-form area element was strictly positive (i.e.: the Jacobean for a polar change of coordinates is strictly positive.)

However, we can’t find the circumference of a circle my integrating \( d\Bx_\theta \) around that circular path, because \( d\Bx_\theta \) has an orientation, and we will get zero (given the symmetry of the problem) if we integrate all the way around
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:100}
\begin{aligned}
\int_{\theta = 0}^{2\pi} d\Bx_\theta &= \int_{\theta = 0}^{2\pi} r \Be_2 e^{i\theta} \,d\theta \\
&= r \Be_2 \evalrange{ \frac{e^{i\theta}}{i} }{0}{2\pi} \\
&= \frac{r \Be_2}{i} \times 0.
\end{aligned}
\end{equation}
If we want the circumference of a circle, we have to sum all the contributions of \( d\Bx_\theta \) that are colinear with \( \thetacap = \Be_2 e^{i\theta} \)
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:120}
\begin{aligned}
C &= \int_{\theta = 0}^{2\pi} \thetacap \cdot \, d\Bx_\theta \\
&= \int_{\theta = 0}^{2\pi} \thetacap \cdot \lr{ r \thetacap \, d\theta } \\
&= 2 \pi r.
\end{aligned}
\end{equation}
This is a plain old boring scalar integral, because the vector valued path integral isn’t terribly interesting.

Spherical example.

For a spherical parameterization, our position vector is
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:140}
\Bx = r \Be_1 e^{i \phi} \sin\theta + r \Be_3 \cos\theta,
\end{equation}
so the differentials are
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:160}
\begin{aligned}
d\Bx_r &= \lr{ \Be_1 e^{i \phi} \sin\theta + \Be_3 \cos\theta } \,dr = \rcap \, dr \\
d\Bx_\theta &= \lr{ r \Be_1 e^{i \phi} \cos\theta – r \Be_3 \sin\theta }\, d\theta = r \thetacap \,d\theta \\
d\Bx_\phi &= r \Be_2 e^{i \phi} \sin\theta \, d\phi = r \sin\theta \phicap.
\end{aligned}
\end{equation}

The oriented area element on the surface of the sphere is
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:180}
\begin{aligned}
dA &= d\Bx_\theta \wedge d\Bx_\phi \\
&= r^2 \gpgradetwo{ \lr{ \Be_1 e^{i \phi} \cos\theta – \Be_3 \sin\theta } \Be_2 e^{i \phi} \sin\theta } \,d\theta d\phi \\
&= r^2 \sin\theta \lr{ i \cos\theta – \Be_{32} e^{i \phi} \sin\theta } \,d\theta d\phi .
\end{aligned}
\end{equation}
Integrating this over the surface will give us zero, with the first integrand killed by the \( \theta \) integral, and the second by the \( \phi \) integral. As pointed out in the original question, we must integrate the absolute value of this two-form in order to find the surface area of the sphere, just as we had to integrate the absolute value of \( d\Bx_\theta \) for the circle to find the circumference.

Let’s perform that integration to verify that we get the expected result. We will first simplify our bivector valued oriented area element. Observe that \( dA \wedge \rcap = dA \rcap \propto I \), so \( dA \propto \rcap I \). We should be able to simplify our expression for \( dA \) by factoring out an \( \rcap \) term
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:200}
\begin{aligned}
dA &= r^2 \sin\theta \lr{ \Be_{1233} \cos\theta – \Be_{1132} e^{i \phi} \sin\theta } \,d\theta d\phi \\
&= r^2 \sin\theta I \lr{ \Be_3 \cos\theta + \Be_1 e^{i \phi} \sin\theta } \,d\theta d\phi \\
&= r^2 \sin\theta I \rcap \,d\theta d\phi.
\end{aligned}
\end{equation}
The spherical scalar area is
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:220}
\begin{aligned}
A
&= \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2 \pi} \Abs{ r^2 \sin\theta I \rcap } \,d\theta d\phi \\
&= r^2 \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2 \pi} \Abs{ \sin\theta } \,d\theta d\phi \\
&= 2 r^2 \int_{\theta = 0}^{\pi/2} \int_{\phi = 0}^{2\pi} \sin\theta \,d\theta d\phi \\
&= 2 r^2 (2 \pi) \\
&= 4 \pi r^2.
\end{aligned}
\end{equation}

Observe that to find the volume of the sphere, we also cannot just integrate the trivector valued volume element directly either. That oriented volume element is
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:240}
\begin{aligned}
dV
&= d\Bx_r \wedge dA \\
&= \rcap\, dr dA \\
&= r^2 \sin\theta I \,dr d\theta d\phi.
\end{aligned}
\end{equation}
This integrand is positive above the azimuthal plane, and negative below, so will give us zero if we integrate over the entire \( \theta \in [0, \pi] \) region. So, if we want to find the volume of a sphere, we also must use an absolute integrand.
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:260}
\begin{aligned}
V
&= \int_{r = 0}^{R} \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2 \pi} \Abs{ r^2 \sin\theta I } \,dr d\theta d\phi \\
&= 2 \int_{r = 0}^{R} \int_{\theta = 0}^{\pi/2} \int_{\phi = 0}^{2\pi} r^2 \sin\theta \,dr d\theta d\phi \\
&= 2 \frac{R^3}{3} (2 \pi) \\
&= \frac{4}{3} \pi R^3.
\end{aligned}
\end{equation}
Had the sign of our volume element been invariant over the entire integration region, as it was for the circular area computation (but not the circular boundary computation), we could have computed this as a pseudoscalar integral. For example, if we wanted to know what the oriented volume of the first quadrant of the sphere was, we could compute that directly, as
\begin{equation}\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:280}
\begin{aligned}
V_1 &= \int_{r = 0}^{R} \int_{\theta = 0}^{\pi/2} \int_{\phi = 0}^{\pi/2} r^2 \sin\theta I \,dr d\theta d\phi \\
&= \inv{6} \pi R^3 I,
\end{aligned}
\end{equation}
but if this volume integral is extended to the entire spherical region, the result is zero, not \( (4/3) \pi R^3 I \).

Only when our multivector integrand doesn’t change sign over the integration region, can we directly integrate without taking absolute values.
Again, this should not be too surprising.
This is why, in conventional scalar calculus, we generally must take the absolute value of our change of variable Jacobians, when we compute area or volume computations.