What's a good approximation to the zeros of the cosine integral? I. Logarithmic integral
The logarithmic integral $\rm{li}(z)$ has a unique positive zero at $z \approx 1.451363$ called the Ramanujan-Soldner constant.

II. Cosine integral
The cosine integral $\rm{Ci}(x)$, on the other hand, has infinitely many zeros,

The smallest is nameless and is $x_0 \approx 0.61650548$. For the next zeros, I notice they can be approximated by rational multiples of $\pi$,
$$\begin{array}{|c|c|c|}
\hline
n&x_n&n\pi\\
\hline
1&3.38&3.14\\
2&6.42&6.28\\
3&9.52&9.42\\
4&12.64&12.56\\
\hline
\end{array}$$
III. Sine integral
The sine integral $\rm{Si}(y)$ equated to $\frac{\pi}2\approx 1.57$ also has infinitely many zeros. 

The smallest is again nameless and is $y_0 \approx  1.92644766$. The next zeros are,
$$\begin{array}{|c|c|c|}
\hline
n&y_n&(2n+1)\tfrac{\pi}2\\
\hline
1&4.89&4.71\\
2&7.97&7.85\\
3&11.08&10.99\\
4&14.20&14.13\\
\hline
\end{array}$$
IV. Questions


*

*Do the expressions $x_n - n\pi$ and $y_n - (2n+1)\frac{\pi}2$ converge to some non-zero constant?

*Are there better approximations to the roots $x_n$ and $y_n$?

 A: If you look here, you will see that the zeros are given by
$$a+\frac{1}{a}-\frac{16}{3\, a^3}+\frac{1673}{15\, a^5}-\frac{507746}{105
  \, a^7}+O\left(\frac{1}{a^9}\right)$$ where $a=k \pi$ for the cosine integral and $a=\left(k+\frac 12\right)\pi$ for the sine integral.
If you want a more compact form, you could use
$$a+\frac{240 a^2+3739}{a(240 a^2+5019)}$$ I built it from the above.
A: The roots of the Cardinal Sine occur at $n\pi$, and these correspond to the alternating minima and maxima of the Sine Integral. The roots of the latter will be found asymptotically in the middle and the constant is zero. (For high $n$, $\dfrac1{2n\pi+x}$ tends to a constant over a period.)
More quantitatively, using a linear approximation for $x\in[0,2\pi]$,
$$\frac{\sin(2n\pi+x)}{2n\pi+x}\approx\frac{\sin(2n\pi+x)}{2n\pi}\left(1-\frac x{2n\pi}\right)$$
and
$$\int_0^x\frac{\sin(2n\pi+t)}{2n\pi+t}dt\approx\frac{(x-2n\pi) \cos x - \sin(x)}{(2n\pi)^2}.$$
This expression cancels when
$$\cos x=\frac{x\cos(x)-\sin(x)}{2n\pi}=O\left(\frac1n\right)$$ so that the bias on the root vanishes like $\dfrac1n$. The higher order terms in the development will not change this behavior.
