Verifying a Fourier series inverting $\frac{\tan(y)}y$ with the Whittaker W function

Question

$\def\k{\operatorname k}\def\W{\operatorname W}$

Using a Dirac $\delta(x)$ Fourier series and this post

$$\begin{align} y\cot(y)=\frac1x\implies\frac1{\sec^2(y)-x}=\int_0^\frac\pi2\delta(\tan(t)-xt)dt-\frac1{2|x-1|}=\frac14+\frac1\pi\sum_{n=1}^\infty\int_0^\frac\pi2\cos(n (\tan(t)-xt))dt-\frac1{2|x-1|}\end{align}$$

Now remember the Bateman k$_v(z)$ and Whittaker W$_{a,b}(x)$ functions:

$$\k_v(x)=\frac{\W_{\frac v2,\frac12}(2x)}{\left(\frac v2\right)!}=\frac2\pi\int_0^\frac\pi2\cos(x\tan(t)-vt)dt$$

Therefore, we hopefully solve for the smallest positive solution:

$$\frac{\tan(y)}y=x\implies y=\sec^{-1}\left(\sqrt{\frac1{\frac14+\frac12\sum\limits_{n=1}^\infty\frac{\W_{\frac{nx}2,\frac12}(2n)}{\left(\frac{nx}2\right)!}-\frac1{2|x-1|}}+x}\right)= \sec^{-1}\left(\sqrt{\frac1{\frac14+\frac12\sum\limits_{n=1}^\infty\k_{nx}(n)-\frac1{2|x-1|}}+x}\right) \tag1$$

There appears to be no other way to use the Bateman function to solve $\frac{\tan(y)}y=x$. A numerical tester with $y\to\sec^{-1}\left(\sqrt{\frac1y+x}\right)$ is inconclusive. Even though the above series is slow, it still could be a solution.

Is $(1)$ true?

Answer 1

The answer to your question is no. Any approach based on the integral

$$\int\limits_0^\frac\pi2\operatorname{\text{Ш}}_{2 \pi}(f_x(t))\,dt\tag{1}$$

of the Fourier series for the Dirac comb

$$\operatorname{\text{Ш}}_{2 \pi}(t)=\frac{1}{2 \pi}\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=-N}^N e^{i n t}\right)=\frac{1}{2 \pi}+\frac{1}{\pi}\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \cos(n t)\right)\tag{2}$$

where

$$f_x(t)=\tan(t)-x t\tag{3}$$

is at best an approximation and not an exact formula. You may be able to compensate for the two contributions of the tooth of the Dirac comb at $t=0$, but all teeth in the Dirac comb in the right-half plane contribute to the end result.

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Noting the Dirac comb $\operatorname{\text{Ш}}_{2 \pi}(t)$ is $2\pi$ periodic as illustrated in Figure (1) below, the following table illustrates the contribution of the tooth at $t=0$ and the first 5 teeth in the right-half plane for the case $f_2(t)=\tan(t)-2 t$ (corresponding to $x=2)$.

$$\begin{array}{cc} k & \int\limits_0^{\frac{\pi }{2}} \delta(f_2(t)-2 \pi k) \, dt \\ 0 & 0.225523 \\ 1 & 0.0119339 \\ 2 & 0.00413686 \\ 3 & 0.00208941 \\ 4 & 0.00125876 \\ 5 & 0.000840887 \\ \end{array}$$

Note the row for $k=0$ in the table above gives the correct result since Mathematica ignores the partial contribution of an analytic representation at $t=0$. Also note the contribution of each successive tooth in the right-half plane decreases in magnitude which is because the slope of $f_2(t)=\tan(t)-2 t$ increases rapidly as $t$ approaches $\frac{\pi}{2}$.

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Figure (1): Illustration of formula (2) for $\operatorname{\text{Ш}}_{2 \pi}(t)$ evaluated at $N=20$

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Figure (2) below illustrates $f_2(t)=\tan(t)-2 t$ in blue and $\delta(f_2(t))$ in orange using the limit representation

$$\delta(t)=\underset{N\to\infty}{\text{lim}}\left(2 N\, \text{sinc}(2\, \pi N\, t)\right)\tag{4}$$

evaluated at $N=20$. Note $\delta(f_2(t))$ has a $\delta$-function spike at $t=0$ as well as at $t\approx 1.16556$ corresponding to the two zeros of $f_2(t)$.

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Figure (2): Illustration of $f_2(t)=\tan(t)-2 t$ in blue and formula (4) for $\delta(f_2(t))$ in orange

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Figure (3) below illustrates $f_2(t)=\tan(t)-2 t$ in blue and $\delta(f_2(t)-2 \pi)$ in orange which corresponds to the first tooth of the Dirac comb in the right-half plane using the limit representation of $\delta(t)$ defined in formula (4) above evaluated at $N=20$. Note the $\delta$-function spike where $f_2(t)$ evaluates to $2 \pi$ (corresponding to the horizontal green reference line).

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Figure (3): Illustration of $f_2(t)=\tan(t)-2 t$ in blue and formula (4) for $\delta(f_2(t)-2 \pi)$ in orange

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Every time $f_x(t)$ crosses an integer multiple of $2 \pi$ the integral $\int\limits_0^\frac\pi2\operatorname{\text{Ш}}_{2 \pi}(f_x(t))\,dt$ will pick up another contribution of a tooth of the Dirac comb, and since $\underset{t\to \left(\frac{\pi }{2}\right)^-}{\text{lim}}(f_x(t))=\infty$ this includes a contribution from every tooth of the Dirac comb in the right-half plane.

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As $x$ grows larger, the integral $\int\limits_0^\frac\pi2\operatorname{\text{Ш}}_{2 \pi}(f_x(t))\,dt$ will also begin to pick up contributions from teeth of the Dirac comb $\operatorname{\text{Ш}}_{2 \pi}(t)$ in the left-half plane. For example, Figure (4) below illustrates $f_{10}(t)=\tan(t)-10 t$ in blue and $\delta(f_{10}(t)+2 \pi)$ in orange which corresponds to the first tooth of the Dirac comb in the left-half plane using the limit representation of $\delta(t)$ defined in formula (4) above evaluated at $N=20$. Note the two $\delta$-function spikes where $f_{10}(t)$ evaluates to $-2 \pi$ (corresponding to the horizontal green reference line).

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Figure (4): Illustration of $f_{10}(t)=\tan(t)-10 t$ in blue and formula (4) for $\delta(f_{10}(t)+2 \pi)$ in orange