Super-logarithm on the imaginary line

[andydude]

I just found a way to calculate slog on the imaginary axis!
It depends very much on Jay's observation that slog is imaginary-periodic.

Let \( S(x) = \text{slog}_e(i x) \). S is periodic with period \( 2\pi \), because \( S(x + 2\pi) = \text{slog}(i (x + 2\pi)) = \text{slog}(i x + 2i\pi) = \text{slog}(i x) = S(x) \). Since S is periodic, we can use Fourier series to represent it. Let \( R(x) = S(-i\ln(x)) \), then \( R(e^{ix}) = S(x) \). The Taylor series coefficients of R will then be the Fourier series coefficients of S. In terms of the super-logarithm, \( R(x) = S(-i\ln(x)) = \text{slog}(-ii\ln(x)) = \text{slog}(\ln(x)) = \text{slog}(x) - 1 \). This means the Fourier series coefficients of \( \text{slog}(ix) \) are the Taylor series coefficients of \( \text{slog}(x) - 1 \) which we already know. In other words, \( \text{slog}(ix) = \text{slog}(e^{ix}) - 1 \), so:

\(
\text{slog}_e(ix)_4 = -2
+ \frac{12}{13}e^{ix}
+ \frac{16}{65}e^{2ix}
- \frac{12}{65}e^{3ix}
+ \frac{1}{65}e^{4ix}
\)

The nice thing about this is that it seems to bypass the radius of convergence problem near z=i since its a Fourier series and not a Taylor series. Is this right?

I've included a plot with multiple approximations, which seem to converge much faster than doing analytic continuation section-by-section. The top line is the imaginary part, and the bottom line is the real part, and the "y" axis is \( \text{slog}(ix) \):

PDF version



Andrew Robbins