Convergence of Fourier sine transform of $1$

Question

While reading Paul J. Nahin's "Hot molecules, cold electrons", while solving the heat equation for a semi-infinite mass with infinite thickness, when comparing the result with the initial conditions, he writes $1=\int_{0}^{\infty}B(\lambda)\sin(\lambda x)d\lambda$, then, recalling the Fourier sine transform $f(x)=\int_{0}^{\infty}F(\lambda)\sin(\lambda x)d\lambda$, from which $F(\lambda)=2/\pi\int_{0}^{\infty}f(x)\sin(\lambda x) dx$, he claims that in this case, since $f(x)=1$, we have $B(\lambda)=\frac{2}{\pi}\int_{0}^{\infty}\sin(\lambda x)dx$, but this integral clearly diverges. How is this result supposed to be intended?

Answer 1

That integral diverges indeed, and setting $B$ equal to it makes no sense to me. If one wants a quick fix, one can regularize the integral and take the limit, that is, $$\frac{\pi}{2}B(\lambda)=\lim_{\epsilon \downarrow 0} \int_0^\infty e^{-\epsilon x}\sin(\lambda x)\,dx=\lim_{\epsilon \downarrow 0}\frac{\lambda}{\epsilon^2+\lambda^2}=\frac{1}{\lambda}.$$ This formally shows that $B(\lambda)$ should be equal to $\frac{2}{\pi \lambda}$.

This argument can also be made rigorous in a distributional setting. Namely, if we extend 1 to an odd function on $\mathbb{R}$, it becomes the signum function, and its Fourier transform will coincide with $1/i$ times the sine transform. It is well-known that the Fourier transform of the signum function, when understood as a tempered distribution, is equal to a constant times the principal value distribution associated to $\frac{1}{i\lambda}$, which agrees with the function $\frac{1}{i\lambda}$ outside the origin. Multiplying with $i$, then yields the distribution $B$.

Answer 2

The Fourier $\sin$ transform

$$\frac{2}{\pi} \int\limits_0^{\infty} \text{sgn}(x)\, \sin(\lambda x) \, dx=\frac{2}{\pi \lambda}\tag{1}$$

and inverse Fourier $\sin$ transform

$$\int\limits_0^{\infty} \frac{2}{\pi \lambda}\, \sin(\lambda x) \, d\lambda =\text{sgn}(x)\tag{2}$$

are related to the Fourier transform

$$\mathcal{F}_x[\text{sgn}(x)](\lambda)=\frac{1}{\pi} \int\limits_{-\infty}^{\infty} \text{sgn}(x)\, e^{i \lambda x} \, dx=\frac{2 i}{\pi \lambda}\tag{3}$$

and inverse Fourier transform

$$\mathcal{F}_{\lambda}^{-1}\left[\frac{2 i}{\pi \lambda}\right](x)=\frac{1}{2} \int\limits_{-\infty}^{\infty} \frac{2 i}{\pi \lambda}\, e^{-i \lambda x} \, d\lambda =\text{sgn}(x)\tag{4}.$$

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For the Fourier transform of $\frac{1}{x}$, see row 311 at Fourier_transform: Distributions, one-dimensional which indicates the following in the remarks column.

Note that 1/x is not a distribution. It is necessary to use the Cauchy principal value when testing against Schwartz functions.

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Note the Fourier transform and inverse Fourier transform can generally be defined as

$$F(y)=\underset{\{a, b\}}{\mathcal{F}_x}[f(x)](y)=\sqrt{(2 \pi)^{a-1}} \sqrt{|b|} \int\limits_{-\infty}^{\infty} f(x)\, e^{i b y x} \, dx\tag{5}$$

and

$$f(x)=\underset{\{a, b\}}{\mathcal{F}_y}^{-1}[f(y)](x)=\sqrt{(2 \pi)^{-a-1}} \sqrt{|b|} \int\limits_{-\infty}^{\infty} F(y)\, e^{-i b x y} \, dy\tag{6}$$

where $\{a, b\}$ are the Fourier parameters, and the definitions of Fourier transform versus inverse Fourier transform are somewhat arbitrary since

$$\underset{\{a, b\}}{\mathcal{F}_x}[f(x)](y)=\underset{\{-a, -b\}}{\mathcal{F}_x}^{-1}[f(x)](y)\tag{7}.$$

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Formulas (1) to (4) are based on the Fourier parameters $\{a, b\}=\{1-\frac{2 \log(\pi)}{\log (2 \pi)}, 1\}$.

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The inverse Fourier transform integral in formula (4) above can be evaluated as a Cauchy principal value, but the Fourier transform integral in formula (3) above only converges in a distributional sense analogous to $\mathcal{F}_x[1](\omega)=\int_{-\infty}^{\infty} 1\, e^{-2 \pi i \omega x} \, dx=\delta(\omega)$ which is based on the Fourier parameters $\{a, b\}=\{0, -2 \pi\}$.

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Now consider the function

$$f(x)=\text{sgn}(x)\tag{8}$$

with Fourier transform

$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^{\infty} \text{sgn}(x)\, e^{-2 \pi i \omega x} \, dx=-\frac{i}{\pi \omega}\tag{9}$$

where formula (9) above is based on the Fourier parameters $\{a, b\}=\{0, -2 \pi\}$.

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I believe the function $f(x)=\text{sgn}(x)$ is recoverable from the inverse Fourier transform

$$f(x)=\mathcal{F}_{\omega}^{-1}[F(\omega)](x)=\int\limits_{-\infty}^{\infty} -\frac{i}{\pi \omega}\, e^{2 \pi i x \omega} \, dx=\text{sgn}(x)\tag{10}$$

via the Fourier inversion theorem which I'll attempt to illustrate below.

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More specifically, I believe the function $f(x)=\text{sgn}(x)$ is recoverable from $F(\omega)=-\frac{i}{\pi \omega}$ via the nested Fourier series representation

$$f(x)=\frac{i}{4}\ \lim_{N, f\to\infty}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1}\\ \sum\limits_{k=1}^{2 f (2 n-1)} \left(\left(4 (-1)^k \cos\left(\frac{\pi k}{2 n-1}\right)-1\right) F\left(\frac{k}{4 n-2}\right) \sin\left(\frac{\pi k x}{2 n-1}\right)\\+F\left(\frac{2 k-1}{8 n-4}\right) \sin\left(\frac{\pi (2 k-1) x}{4 n-2}\right)\right)\right),\quad f(-x)=-f(x)\tag{11}$$

where $\mu(n)$ is the Möbius function and the evaluation frequency $f$ in the inner sum over $k$ is assumed to be a positive integer.

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Formula (11) above is a refinement of formula (12) in my related MSO question which provides information on its derivation.

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Figure (1) below illustrates $f(x)=\text{sgn}(x)$ in blue overlaid by formula (11) for $fx)=\text{sgn}(x)$ in orange where formula (11) is evaluated using the upper evaluation limits $f=4$ and $N=20$. Note the evaluation frequency $f=4$ is clearly visible in Figure (1) below which is typical when evaluating formula (11) above for a function with a discontinuity (see Gibbs phenomenon). I believe the offset in the slope of the evaluation of formula (11) above depicted in Figure (1) below goes to zero as $N\to\infty$ (see Figure (2) below).

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Figure (1): Illustration of $\text{sgn}(x)$ in blue overlaid by formula (11) for $\text{sgn}(x)$ in orange

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In the case of $f(x)=\text{sgn}(x)$ and $F(\omega)=-\frac{i}{\pi \omega}$, formula (11) above simplifies to

$$\text{sgn}(x)=\frac{i}{4 \pi}\ \lim_{N\to\infty}\left(\sum\limits_{n=1}^N \mu(2 n-1)\\ \left(2 \log\left(1+e^{\frac{i \pi (x-1)}{2 n-1}}\right)-\log\left(1-e^{\frac{i \pi x}{2 n-1}}\right)+2 \log\left(1+e^{\frac{i \pi (x+1)}{2 n-1}}\right)-2 \log\left(1+e^{-\frac{i \pi (x-1)}{2 n-1}}\right)\\-2 \log\left(1+e^{-\frac{i \pi (x+1)}{2 n-1}}\right)+\log\left(1-e^{\frac{i \pi x}{1-2 n}}\right)-2 \tanh^{-1}\left(e^{\frac{i \pi x}{4 n-2}}\right)+2 \tanh^{-1}\left(e^{\frac{i \pi x}{2-4 n}}\right)\right)\right)\tag{12}$$

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Figure (2) below illustrates $\text{sgn}(x)$ in blue overlaid by formula (12) for $\text{sgn}(x)$ in orange where formula (12) is evaluated using the upper evaluation limit $N=10,000$.

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Figure (2): Illustration of $\text{sgn}(x)$ in blue overlaid by formula (12) for $\text{sgn}(x)$ in orange

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The remainder of this answer clarifies the convergence (in a distributional sense) of the Fourier transform integral in formula (3) above.

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Consider the limit representation

$$\mathcal{F}_x[\text{sgn}(x)](\lambda)=\frac{1}{\pi} \lim_{T\to\infty}\left(\int\limits_{-T}^{T} \text{sgn}(x)\, e^{i \lambda x} \, dx\right)=\lim_{T\to\infty}\left(\frac{2 i\, (1-\cos (\lambda T))}{\pi \lambda}\right)\tag{13}$$

of the Fourier transform in formula (3) above and note the related inverse Fourier transform

$$\mathcal{F}_{\lambda}^{-1}\left[\lim_{T\to\infty}\left(\frac{2 i\, (1-\cos(\lambda T))}{\pi \lambda}\right)\right](x)=\frac{1}{2} \int_{-\infty}^{\infty} \lim_{T\to\infty}\left(\frac{2 i\, (1-\cos(\lambda T))}{\pi \lambda}\right) e^{-i \lambda x} \, d\lambda \\=\frac{1}{2}\, \lim_{T\to\infty}\left((\text{sgn}(T-x)-\text{sgn}(T+x)+2\, \text{sgn}(x))\right)=\text{sgn}(x)\tag{14}$$

is consistent with the result of the inverse Fourier transform in formula (4) above.