Is there a pattern in the complex roots of the Sine Integral $\mathrm{Si}\left(\frac{z}{2}\right)$?

Question

I was working with the Sine Integral $\mathrm{Si}\left(\frac{z}{2}\right)$ and found it has an infinite number of complex roots mirrored in all four quadrants. In the first quadrant these start with:

$$ 11.9303627666221643518173848082 + 6.01119131213006773157261708415 \cdot i\\ 24.7070169417149873195705173033 + 7.37455299482618861728760608970 \cdot i\\ 37.3725098362278317680187364308 + 8.17692229518364269643530815855 \cdot i\\ 49.9976565288211086812290841074 + 8.74796328051411644007205709150 \cdot i\\ \cdots $$

Below is an extended plot of these roots in the first quadrant ($z=x+iy$):

and it appears the roots all lie on a log-shaped curve. After some attempts, the curve $y=2\log(x)+\frac12\,\log(2\,\pi)$ offers a pretty decent fit:

Q: Obviously, the above fitting is just some guess work of mine, but got curious whether anything more is known about the pattern of these roots. Searched the web extensively, however nothing comes up. Grateful for any reference and/or any steers on how to find the exact curve (if it exists) the roots reside on.

Answer 1

This is a heuristic consideration. From the asymptotic expansion of $\operatorname{Si}$ we may infer that $$ \operatorname{Si}\left( {\frac{z}{2}} \right) \approx \frac{\pi }{2} - \frac{2}{z}\cos \left( {\frac{z}{2}} \right) $$ for large values of $z$. Thus the roots of $\operatorname{Si}\left( {\frac{z}{2}} \right)$ are approximately the roots of $$ \frac{\pi }{4}z = \cos \left( {\frac{z}{2}} \right). $$ Taking real and imaginary parts and approximating the hyperbolic functions by exponentials, we arrive at the system $$ \left\{ \begin{array}{l} {\rm e}^{y/2} \displaystyle\cos \left( {\frac{x}{2}} \right) = \frac{\pi }{2}x, \\[1ex] {\rm e}^{y/2} \displaystyle\sin \left( {\frac{x}{2}} \right) = - \frac{\pi }{2}y, \\ \end{array} \right. $$ with $z=x+\mathrm{i}y$. Taking the square sum of these equations yields $$ {\rm e}^y = \frac{{\pi ^2 }}{4}(x^2 + y^2 ), $$ which has the approximate solution $$ y \approx 2\log x + 2\log (\pi /2). $$ Applying this in the second equation, we get $$ \sin \left( {\frac{x}{2}} \right) \approx - \frac{y}{x} \Longrightarrow x \approx 4\pi k + 2\arcsin \left( { - \frac{y}{x}} \right) \approx 4\pi k - \frac{{4\log x}}{x}, $$ with $k=0,1,2,\ldots$. The last equation is solved approximately by $$\boxed{ x_k \approx 4\pi k - \frac{{\log k}}{{\pi k}}.} $$ The corresponding imaginary part is then $$\boxed{ y_k \approx 2\log x_k + 2\log (\pi /2).} $$

Answer 2

In a comment, I wrote "This a nice observetion". I am totally wrong since it is $\color{red}{\text{remarkable}}$ !

Without any preconceived idea, I performed, using extremely high accuracy, the calculations for the first $2000$ roots and then, based on your work, used the model $$y=a\log(x)+b$$

The results of the linear regression are $$\begin{array}{|llll} \hline \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ a & 1.9989369 & 0.0000806 & \{1.9987789,1.9990949\} \\ b & 0.9130792 & 0.0007409 & \{0.9116262,0.9145323\} \\ \end{array}$$

For this regression, $R^2=0.9999999657$

These numbers are so close to your expression that you can probably conjecture it.

But, for a quick check, computing the $5000^{\text{nd}}$ root $$x=63435.038198600614034283457758965563050039115269153,\cdots$$ $$y=23.018708803640584372540630832612830290071615109658,\cdots$$

$$2\log(x)+\frac12\,\log(2\,\pi)=23.03448181$$ shows that, probably, a very tiny correction would be required.

Using for $x$ an estimate of $4 k \pi$, using your formula for the $5000^{\text{nd}}$ root, Newton iterates are

$$\left( \begin{array}{cc} 0 & 62831.8530717959+24.4016677711\, i \\ 1 & 62831.8527501440+23.3938122064\, i \\ 2 & 62831.8524664822+23.0360195052\, i \\ 3 & 62831.8524045073+22.9999299632\, i \\ 4 & 62831.8524033580+22.9996004068\, i \\ \end{array} \right)$$ which is quite spectacular.

So, at least for large values of $k$, the $k^{\text{th}}$ root of

$$\mathrm{Si}\left(\frac{z}{2}\right)=0$$ is $$z_{(k)} \sim 4 \pi k+i \log \left(16 \sqrt{2} \pi ^{5/2} k^2\right)$$

Edit

Assuming that for large values of $k$, the solution is given by $$x_k=4k\pi \quad \quad y_k=2\log(x_k)+a\quad\quad z_k=x_k+i\, y_k$$ consider the equation $$f_k(a)=\Bigg[\Re\left(\text{Si}\left(\frac{z_k}{2}\right)\right)\Bigg]^2+\Bigg[\Im\left(\text{Si}\left(\frac{z_k}{2}\right)\right)\Bigg]^2$$ which is numerically solved using Newton method starting with $$a_0=\frac{1}{2} \left(2 \log \left(\frac{\pi }{2}\right)+\frac{1}{2} \log (2 \pi)\right)$$ Using $k=10^n$, the results are eloquent $$\left( \begin{array}{cc} n & a_{10^n} \\ 1 & 0.90145748249383053660\\ 2 & 0.90313659211589139775\\ 3 & 0.90316500566623502304\\ 4 & 0.90316540531377253466\\ 5 & 0.90316541046509332494\\ 6 & 0.90316541052815491646\\ 7 & 0.90316541052890101619\\ 8 & 0.90316541052890963203\\ 9 & 0.90316541052890973083\\ 10 & 0.90316541052890973083\\ 11 & 0.90316541052890973084\\ \end{array} \right)$$

This number is also identified by the $ISC$ as $2 \log \left(\frac{\pi }{2}\right)$

Update

After @Gary's answer, let $$x_k=4 \pi k-\frac{\log (k)}{\alpha \, k}\quad \quad y_k=2\log(x_k)+2 \log \left(\frac{\pi }{2}\right)\quad\quad z_k=x_k+i\, y_k$$ and compute $\alpha$ with $k=5^n$ starting with $\alpha_0=2$ $$\left( \begin{array}{cc} n & \alpha_{5^n} \\ 1 &1.40577 \\ 2 &1.94402 \\ 3 &2.22709 \\ 4 &2.40189\\ 5 &2.52059 \\ 6 &2.60646 \\ 7 &2.67146\\ 8 &2.72269 \\ 9 &2.76164 \\ 10 &2.82973 \\ \end{array} \right)$$

Using as an empirical model $$\alpha_{5^n}=a-\frac b{c+5^n}$$ the fit is almost perfect $(R^2=0.999987)$ and the parameters are $$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & \color{red}{3.16500} & 0.01670 & \{\color{red}{3.12551,3.20448\}} \\ b & 4.05097 & 0.16182 & \{3.66833,4.43362\} \\ c & 1.30543 & 0.07768 & \{1.12174,1.48912\} \\ \end{array}$$